Câu 34: Phân tích thành nhân tử
a. \({x^4} + 2{x^3} + {x^2}\)
b. \({x^3} – x + 3{x^2}y + 3x{y^2} + {y^3} – y\)
c. \(5{x^2} – 10xy + 5{y^2} – 20{z^2}\)
a. \({x^4} + 2{x^3} + {x^2}\) \( = {x^2}\left( {{x^2} + 2x + 1} \right) = {x^2}{\left( {x + 1} \right)^2}\)
b. \({x^3} – x + 3{x^2}y + 3x{y^2} + {y^3} – y\)
\(\eqalign{ & = \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^2}} \right) – \left( {x + y} \right) = {\left( {x + y} \right)^3} – \left( {x + y} \right) \cr & = \left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} – 1} \right] = \left( {x + y} \right)\left( {x + y + 1} \right)\left( {x + y – 1} \right) \cr} \)
c. \(5{x^2} – 10xy + 5{y^2} – 20{z^2} = 5\left( {{x^2} – 2xy + {y^2} – 4{z^2}} \right)\)
\(\eqalign{ & = 5\left[ {\left( {{x^2} – 2xy + {y^2}} \right) – 4{z^2}} \right] = 5\left[ {{{\left( {x – y} \right)}^2} – {{\left( {2z} \right)}^2}} \right] \cr & = 5\left( {x – y + 2z} \right)\left( {x – y – 2z} \right) \cr} \)
Câu 35: Phân tích thành nhân tử
a. \({x^2} + 5x – 6\)
b. \(5{x^2} + 5xy – x – y\)
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c. \(7x – 6{x^2} – 2\)
a. \({x^2} + 5x – 6\) \( = {x^2} – x + 6x – 6 = \left( {{x^2} – x} \right) + \left( {6x + 6} \right)\)
\( = x\left( {x – 1} \right) + 6\left( {x – 1} \right) = \left( {x – 1} \right)\left( {x + 6} \right)\)
b. \(5{x^2} + 5xy – x – y\) \( = \left( {5{x^2} + 5xy} \right) – \left( {x – y} \right) = 5x\left( {x + y} \right) – \left( {x + y} \right)\)
\( = \left( {x + y} \right)\left( {5x – 1} \right)\)
c. \(7x – 6{x^2} – 2\) \( = 4x – 6{x^2} – 2 + 3x = \left( {4x – 6{x^2}} \right) – \left( {2 – 3x} \right)\)
\( = 2x\left( {2 – 3x} \right) – \left( {2 – 3x} \right) = \left( {2 – 3x} \right)\left( {2x – 1} \right)\)
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Câu 36: Phân tích thành nhân tử
a. \({x^2} + 4x + 3\)
b. \(2{x^2} + 3x – 5\)
c. \(16x – 5{x^2} – 3\)
a. \({x^2} + 4x + 3\) \( = {x^2} + x + 3x + 3 = \left( {{x^2} + x} \right) + \left( {3x + 3} \right)\)
\(x\left( {x + 1} \right) + 3\left( {x + 1} \right) = \left( {x + 1} \right)\left( {x + 3} \right)\)
b. \(2{x^2} + 3x – 5\) \( = 2{x^2} – 2x + 5x – 5 = \left( {2{x^2} – 2x} \right) + \left( {5x – 5} \right)\)
\( = 2x\left( {x – 1} \right) + 5\left( {x – 1} \right) = \left( {x – 1} \right)\left( {2x + 5} \right)\)
c. \(16x – 5{x^2} – 3\) \( = 15x – 5{x^2} – 3 + x = \left( {15x – 5{x^2}} \right) – \left( {3 – x} \right)\)
\( = 5x\left( {3 – x} \right) – \left( {3 – x} \right) = \left( {3 – x} \right)\left( {5x – 1} \right)\)
Câu 37: Tìm \(x)\ biết:
a. \(5x\left( {x – 1} \right) = x – 1\)
b. \(2\left( {x + 5} \right) – {x^2} – 5x = 0\)
a. \(5x\left( {x – 1} \right) = x – 1\)
\(\eqalign{ & \Rightarrow 5x\left( {x – 1} \right) – \left( {x – 1} \right) = 0 \Rightarrow \left( {x – 1} \right)\left( {5x – 1} \right) = 0 \cr & \Rightarrow \left[ {\matrix{ {x – 1 = 0} \cr {5x – 1 = 0} \cr } \Rightarrow \left[ {\matrix{ {x = 1} \cr {x = {1 \over 5}} \cr } } \right.} \right. \cr} \)
b. \(2\left( {x + 5} \right) – {x^2} – 5x = 0\)
\(\eqalign{ & \Rightarrow 2\left( {x + 5} \right) – \left( {{x^2} + 5x} \right) = 0 \Rightarrow 2\left( {x + 5} \right) – x\left( {x + 5} \right) = 0 \cr & \Rightarrow \left( {x + 5} \right)\left( {2 – x} \right) = 0 \Rightarrow \left[ {\matrix{ {x + 5 = 0} \cr {2 – x = 0} \cr } \Rightarrow \left[ {\matrix{ {x = – 5} \cr {x = 2} \cr } } \right.} \right. \cr} \)
