Câu 138: Tính:
\(E = {{\left( {13{1 \over 4} – 2{5 \over {27}} – 10{5 \over 6}} \right).230{1 \over {25}} + 46{3 \over 4}} \over {\left( {1{3 \over 7} + {{10} \over 3}} \right):\left( {12{1 \over 3} – 14{2 \over 7}} \right)}}\)
\(E = {{\left( {13{1 \over 4} – 2{5 \over {27}} – 10{5 \over 6}} \right).230{1 \over {25}} + 46{3 \over 4}} \over {\left( {1{3 \over 7} + {{10} \over 3}} \right):\left( {12{1 \over 3} – 14{2 \over 7}} \right)}}\)
\(\eqalign{
& = {{\left( {13 – 2 – 10 + {1 \over 4} – {5 \over {27}} – {5 \over 6}} \right).{{5771} \over {25}} + {{187} \over 4}} \over {\left( {{{30} \over {21}} + {{70} \over {21}}} \right):\left( {{{259} \over {21}} – {{300} \over {21}}} \right)}} \cr
& = {{\left( {1 + {{27 – 20 – 90} \over {108}}} \right).{{5751} \over {25}} + {{187} \over 4}} \over {{{100} \over {21}}:{{ – 41} \over {21}}}} \cr} \)
\(\eqalign{
& = {{\left( {{{108} \over {108}} – {{83} \over {108}}} \right).{{5751} \over {25}} + {{187} \over 4}} \over {{{100} \over {21}}.{{ – 21} \over {41}}}} \cr
& = {{{{25} \over {108}}.{{5751} \over {25}} + {{187} \over 4}} \over {{{ – 100} \over {41}}}} \cr
& = \left( {{{213} \over 4} + {{187} \over 4}} \right).{{ – 41} \over {100}} = 100.{{ – 41} \over {100}} \cr} \)
= -41
Câu 139: Tính
\(G = {{4,5:\left[ {47,375 – \left( {26{1 \over 3} – 18.0,75} \right).2,4:0,88} \right]} \over {17,81:1,37 – 23{2 \over 3}:1{5 \over 6}}}\)
\(G = {{4,5:\left[ {47,375 – \left( {26{1 \over 3} – 18.0,75} \right).2,4:0,88} \right]} \over {17,81:1,37 – 23{2 \over 3}:1{5 \over 6}}}\)
\( = {{4,5:\left[ {47,375 – \left( {{{79} \over 3} – 18.{3 \over 4}} \right).2{2 \over 5}:{{22} \over {25}}} \right]} \over {13 – {{71} \over 3}:{{11} \over 6}}}\)
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\(\eqalign{
& = {{4,5:\left[ {47{3 \over 8} – \left( {{{158} \over 6} – {{81} \over 6}} \right).{{12} \over 5}:{{22} \over {25}}} \right]} \over {13 – {{142} \over {11}}}} \cr
& = {{4,5:\left[ {47{3 \over 8} – {{77} \over 6}.{{12} \over 5}:{{22} \over {25}}} \right]} \over {{{143} \over {11}} – {{142} \over {11}}}} \cr} \)
\(\eqalign{
& = {{4,5:\left[ {47{3 \over 8} – {{154} \over 5}.{{25} \over {22}}} \right]} \over {{1 \over {11}}}} \cr
& = {{4,5:\left[ {47{3 \over 8} – 35} \right]} \over {{1 \over {11}}}} \cr} \)
\(\eqalign{
& = \left( {4,5:12{3 \over 8}} \right):{1 \over {11}} = 4,5.{8 \over {99}}.{{11} \over 1} \cr
& = {{4,5.8.11} \over {99}} = 4 \cr} \)
Câu 140: Cho x, y ∈ Q. Chứng tỏ rằng:
a) \(\left| {x + y} \right| \le \left| x \right| + \left| y \right|\)
b) \(\left| {x – y} \right| \ge \left| x \right| – \left| y \right|\)
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a) Với mọi x, y ∈ Q, ta có:
\(x \le \left| x \right|\) và \( – x \le \left| x \right|;y \le \left| y \right|\) và \(- y \le \left| y \right| \Rightarrow x + y \ge – \left( {\left| x \right| + \left| y \right|} \right)\)
Suy ra \( – \left( {\left| x \right| + \left| y \right|} \right) \le x + y \le \left| x \right| + \left| y \right|\)
Vậy \(\left| {x + y} \right| \le \left| x \right| + \left| y \right|\)
Dấu “=” xảy ra khi xy ≥ 0.
b) Theo kết quả câu a) ta có: \(\left| {\left( {x – y} \right) + y} \right| \le \left| {x – y} \right| + \left| y \right|\)
\( \Rightarrow \left| x \right| \le \left| {x – y} \right| + \left| y \right| \Rightarrow \left| x \right| – \left| y \right| \le \left| {x – y} \right|\)
Dấu “=” xảy ra khi xy ≥ 0 và \(\left| x \right| \ge \left| y \right|\)
Câu 141: Tìm giá trị nhỏ nhất của biểu thức:
\({\rm{A}} = \left| {x – 2001} \right| + \left| {x – 1} \right|\)
Vì \(\left| {1 – x} \right| = \left| {x – 1} \right|\) nên \(A = \left| {x – 2001} \right| + \left| {x – 1} \right|\)
\( \Rightarrow A = \left| {x – 2001} \right| + \left| {1 – x} \right| \ge \left| {x – 2001 + 1 – x} \right| \)
\(\Rightarrow\) A = 2000
Vậy biểu thức có giá trị nhỏ nhất A = 2000 khi x – 2001 và 1 – x cùng dấu
Vậy 1 ≤ x ≤ 2001
