Bài 72: Giải các hệ phương trình
\(a)\,\left\{ \matrix{
x + y = 20 \hfill \cr
{\log _4}x + {\log _4}y = 1 + {\log _4}9; \hfill \cr} \right.\)
\(b)\,\left\{ \matrix{
x + y = 1 \hfill \cr
{4^{ – 2x}} + {4^{ – 2y}} = 0,5 \hfill \cr} \right.\)
a) Điều kiện: \(x > 0; y > 0\).
\(\eqalign{
& \,\left\{ \matrix{
x + y = 20 \hfill \cr
{\log _4}x + {\log _4}y = 1 + {\log _4}9 \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{
x + y = 20 \hfill \cr
{\log _4}xy = {\log _4}36 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x + y = 20 \hfill \cr
xy = 36 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x = 2 \hfill \cr
y = 18 \hfill \cr} \right.\,\,\,\,\text{ hoặc }\,\,\,\,\,\left\{ \matrix{
x = 18 \hfill \cr
y = 2 \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {\left( {2;18} \right);\,\left( {18;2} \right)} \right\}\)
b) Từ phương trình thứ nhất suy ra \(y = 1 – x\), thay vào phương trình thứ hai ta được:
\({4^{ – 2x}} + {4^{ – 2\left( {1 – x} \right)}} = 0,5 \Leftrightarrow \,\,{4^{ – 2x}} + {4^{ – 2 + 2x}} = {1 \over 2}\)
Đặt \(t = {4^{2x\,}}\,\left( {t > 0} \right)\) ta được:
\(\eqalign{
& {1 \over t} + {t \over {16}} = {1 \over 2} \Leftrightarrow 16 + {t^2} = 8t \Leftrightarrow {\left( {t – 4} \right)^2} = 0 \cr&\Leftrightarrow t = 4 \cr
& \Leftrightarrow {4^{2x}} = 4 \Leftrightarrow 2x = 1 \Leftrightarrow x = {1 \over 2} \cr} \)
Với \(x = {1 \over 2}\) ta có \(y = 1 – x = 1 – {1 \over 2} = {1 \over 2}\)
Vậy \(S = \left\{ {\left( {{1 \over 2};{1 \over 2}} \right)} \right\}\)
Bài 73: Giải hệ phương trình:
\(a)\,\,\left\{ \matrix{
{3^{ – x}}{.2^y} = 1152 \hfill \cr
{\log _{\sqrt 5 }}\left( {x + y} \right) = 2; \hfill \cr} \right.\)
\(b)\,\left\{ \matrix{
{x^2} – {y^2} = 2 \hfill \cr
{\log _2}\left( {x + y} \right) – {\log _3}\left( {x – y} \right) = 1 \hfill \cr} \right.\)
a) Điều kiện: \(x + y > 0\).
Từ phương trình thứ hai suy ra: \(x + y = {\left( {\sqrt 5 } \right)^2} = 5 \Rightarrow y = 5 – x\) thay vào phương trình thứ nhất ta được:
\({3^{ – x}}{.2^{\left( {5 – x} \right)}} = 1152 \Leftrightarrow {6^{ – x}}.32 = 1152 \Leftrightarrow {6^{ – x}} = 36 \)
\(\Leftrightarrow x = – 2\)
Với \(x = -2\) ta có \(y = 5 – (-2) =7\).
Vậy \(S = \left\{ {\left( { – 2;7} \right)} \right\}\)
b) Điều kiện
\(\left\{ \matrix{
x + y > 0 \hfill \cr
x – y > 0 \hfill \cr} \right.\)
Ta có:
\(\left\{ \matrix{
{x^2} – {y^2} = 2 \hfill \cr
{\log _2}\left( {x + y} \right) – {\log _3}\left( {x – y} \right) = 1 \hfill \cr} \right.\)
Đặt u = \({\log _2}\left( {x + y} \right)\) và v = \({\log _2}\left( {x – y} \right)\)
Ta được hệ
Advertisements (Quảng cáo)
\(\left\{ \matrix{
u + v = 1 \hfill \cr
u – v.{\log _3}2 = 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
u = 1 \hfill \cr
v = 0 \hfill \cr} \right.\)
\( \Leftrightarrow \left\{ \matrix{
{\log _2}\left( {x + y} \right) = 1 \hfill \cr
{\log _2}\left( {x – y} \right) = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x + y = 2 \hfill \cr
x – y = 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x = {3 \over 2} \hfill \cr
y = {1 \over 2} \hfill \cr} \right.\)
Vậy \(S = \left\{ {\left( {{3 \over 2};{1 \over 2}} \right)} \right\}\)
Bài 74: \(\eqalign{
& a)\,{\log _2}\left( {3 – x} \right) + {\log _2}\left( {1 – x} \right) = 3; \cr
& c)\,{7^{\log x}} – {5^{\log x + 1}} = {3.5^{\log x – 1}} – 13.{7^{\log x – 1}} \cr} \)
\(\eqalign{
& b)\,{\log _2}\left( {9 – {2^x}} \right) = {10^{\log \left( {3 – x} \right)}} \cr
& d)\,{6^x} + {6^{x + 1}} = {2^x} + {2^{x + 1}} + {2^{x + 2}} \cr} \)
a) Điều kiện: \(x < 1\)
\(\eqalign{
& \,\,\,\,{\log _2}\left( {3 – x} \right) + {\log _2}\left( {1 – x} \right) = 3\cr& \Leftrightarrow {\log _2}\left( {3 – x} \right)\left( {1 – x} \right) = 3 \cr
& \Leftrightarrow \left( {3 – x} \right)\left( {1 – x} \right) = 8 \Leftrightarrow {x^2} – 4x – 5 = 0\cr& \Leftrightarrow \left[ \matrix{
x = – 1 \hfill \cr
x = 5\,\,\left( \text{loại} \right) \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ { – 1} \right\}\)
b) Điều kiện:
\(\left\{ \matrix{
3 – x > 0 \hfill \cr
9 – {2^x} > 0 \hfill \cr} \right. \Leftrightarrow x < 3\)
\(\eqalign{
& \,\,\,\,{\log _2}\left( {9 – {2^x}} \right) = {10^{\log \left( {3 – x} \right)}}\cr& \Leftrightarrow {\log _2}\left( {9 – {2^x}} \right) = 3 – x \Leftrightarrow 9 – {2^x} = {2^{3 – x}} \cr
& \Leftrightarrow 9 – {2^x} = {8 \over {{2^x}}} \Leftrightarrow {4^x} = {9.2^x} – 8 = 0\cr& \Leftrightarrow \left[ \matrix{
{2^x} = 1 \hfill \cr
{2^x} = 8 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 0 \hfill \cr
x = 3\,\,\left( \text{loại} \right) \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ 0 \right\}\)
c) Điều kiện: \(x > 0\)
Advertisements (Quảng cáo)
\(\eqalign{
& \Leftrightarrow {20.7^{\lg x – 1}} = {28.5^{\lg x – 1}} \cr
& \Leftrightarrow {\left( {{7 \over 8}} \right)^{\lg x – 1}} = {7 \over 8} \cr
& \Leftrightarrow \lg x – 1 = 1 \Leftrightarrow \lg x = 2 \Leftrightarrow x = 100 \cr} \)
Vậy \(S = \left\{ {100} \right\}\)
d) Ta có:
\(\eqalign{
& {6^x} + {6^{x + 1}} = {2^x} + {2^{x + 1}} + {2^{x + 2}} \cr
& \Leftrightarrow {6^x}\left( {1 + 6} \right) = {2^x}\left( {1 + 2 + {2^2}} \right) \cr
& \Leftrightarrow {3^x} = 1 \cr
& \Leftrightarrow x = 0 \cr} \)
Vậy \(S = \left\{ 0 \right\}\)
Bài 75: \(\eqalign{
& a)\,{\log _3}\left( {{3^x} – 1} \right).{\log _3}\left( {{3^{x + 1}} – 3} \right) = 12; \cr
& c)\,5\sqrt {{{\log }_2}\left( { – x} \right)} = {\log _2}\sqrt {{x^2}} ; \cr} \)
\(\eqalign{
& b)\,{\log _{x – 1}}4 = 1 + {\log _2}\left( {x – 1} \right); \cr
& d)\,{3^{{{\log }_4} + {1 \over 2}}} + \,{3^{{{\log }_4} – {1 \over 2}}} = \sqrt x . \cr} \)
a) Điều kiện: \(x > 0\)
Ta có: \(lo{g_3}\left( {{3^x} – 1} \right).lo{g_3}\left( {{3^{x + 1}} – 3} \right) = 12\)
\(\eqalign{
& \Leftrightarrow lo{g_3}\left( {{3^x} – 1} \right).lo{g_3}3\left( {{3^x} – 1} \right) = 12 \cr
& \Leftrightarrow lo{g_3}\left( {{3^x} – 1} \right)\left[ {1 + lo{g_3}\left( {{3^x} – 1} \right)} \right] = 12 \cr} \)
\( \Leftrightarrow \log _3^2\left( {{3^x} – 1} \right) + lo{g_3}\left( {{3^x} – 1} \right) – 12 = 0\)
\(\eqalign{
& \Leftrightarrow \left[ \matrix{
lo{g_3}\left( {{3^x} – 1} \right) = – 4 \hfill \cr
lo{g_3}\left( {{3^x} – 1} \right) = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{3^x} – 1 = {1 \over {81}} \hfill \cr
{3^x} – 1 = {3^3} = 27 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
{3^x} = {{82} \over {81}} \hfill \cr
{3^x} = 28 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\log _3}{{82} \over {81}} \hfill \cr
x = {\log _3}28 \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {{{\log }_3}28;{{\log }_3}82 – 4} \right\}\)
b) Điều kiện: \(x > 1\); \(x \ne 2\)
Ta có: \({\log _{x – 1}}4 = {1 \over {{{\log }_4}\left( {x – 1} \right)}} = {2 \over {{{\log }_2}\left( {x – 1} \right)}}\). Đặt \(t = {\log _2}\left( {x – 1} \right)\)
Ta có phương trình:
\(\eqalign{
& {2 \over t} = 1 + t \Leftrightarrow {t^2} + t – 2 = 0 \cr
& \Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr
t = – 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{\log _2}\left( {x – 1} \right) = 1 \hfill \cr
{\log _2}\left( {x – 1} \right) = – 2 \hfill \cr} \right.\left[ \matrix{
x = 3 \hfill \cr
x = {5 \over 4} \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {3;{5 \over 4}} \right\}\)
c) Điều kiện: \({\log _2}\left( { – x} \right) \ge 0 \Leftrightarrow – x \ge 1 \Leftrightarrow x \le – 1\)
\(5\sqrt {{{\log }_2}\left( { – x} \right)} = {\log _2}\sqrt {{x^2}}\)
\( \Leftrightarrow 5\sqrt {{{\log }_2}\left( { – x} \right)} = {\log _2}\left( { – x} \right)\)
\( \Leftrightarrow 5\sqrt t = t\) với \(t = {\log _2}\left( { – x} \right) \ge 0\)
\(\eqalign{
& \Leftrightarrow 25t = {t^2} \Leftrightarrow \left[ \matrix{
t = 0 \hfill \cr
t = 25 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{\log _2}\left( { – x} \right) = 0 \hfill \cr
lo{g_2}\left( { – x} \right) = 25 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = – 1 \hfill \cr
x = – {2^{25}} \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ { – 1; – {2^{25}}} \right\}\)
d) Điều kiện: \(x > 0\)
Ta có: \(\sqrt x = \sqrt {{4^{{{\log }_4}x}}} = {2^{{{\log }_4}x}}\)
Do đó \({3^{{1 \over 2} + {{\log }_4}x}} + {3^{{{\log }_4}x – {1 \over 2}}} = \sqrt x \)
\(\Leftrightarrow \left( {\sqrt 3 + {1 \over {\sqrt 3 }}} \right){3^{{{\log }_4}x}} = {2^{{{\log }_4}x}}\)
\(\eqalign{
& \Leftrightarrow {4 \over {\sqrt 3 }} = {\left( {{2 \over 3}} \right)^{{{\log }_4}x}} \Leftrightarrow {\log _4}x = {\log _{{2 \over 3}}}{4 \over {\sqrt 3 }} \cr
& \Leftrightarrow x = {4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}} \cr} \)
Vậy \(S = \left\{ {{4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}}} \right\}\)
