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Giải các phương trình sau:
a. \(\left( {x + 2} \right)\left( {{x^2} – 3x + 5} \right) = \left( {x + 2} \right){x^2}\)
b. \({{ – 7{x^2} + 4} \over {{x^3} + 1}} = {5 \over {{x^2} – x + 1}} – {1 \over {x + 1}}\)
c. \(2{x^2} – x = 3 – 6x\)
d. \({{x – 2} \over {x + 2}} – {3 \over {x – 2}} = {{2\left( {x – 11} \right)} \over {{x^2} – 4}}\)
a. \(\left( {x + 2} \right)\left( {{x^2} – 3x + 5} \right) = \left( {x + 2} \right){x^2}\)
\(\eqalign{ & \Leftrightarrow \left( {x + 2} \right)\left( {{x^2} – 3x + 5} \right) – \left( {x + 2} \right){x^2} = 0 \cr & \Leftrightarrow \left( {x + 2} \right)\left[ {\left( {{x^2} – 3x + 5} \right) – {x^2}} \right] = 0 \cr & \Leftrightarrow \left( {x + 2} \right)\left( {{x^2} – 3x + 5 – {x^2}} \right) = 0 \cr & \Leftrightarrow \left( {x + 2} \right)\left( {5 – 3x} \right) = 0 \cr} \)
\( \Leftrightarrow x + 2 = 0\)hoặc \(5 – 3x = 0\)
+ \(x + 2 = 0 \Leftrightarrow x = – 2\)
+ \(5 – 3x = 0 \Leftrightarrow x = {5 \over 3}\)
Vậy phương trình có nghiệm x = -2 hoặc \(x = {5 \over 3}\)
b. \({{ – 7{x^2} + 4} \over {{x^3} + 1}} = {5 \over {{x^2} – x – 1}} – {1 \over {x + 1}}\) ĐKXĐ: \(x \ne – 1\)
\(\eqalign{ & \Leftrightarrow {{ – 7{x^2} + 4} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = {5 \over {{x^2} – x + 1}} – {1 \over {x + 1}} \cr & \Leftrightarrow {{ – 7{x^2} + 4} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = {{5\left( {x + 1} \right)} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} – {{{x^2} – x + 1} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} \cr & \Leftrightarrow {{ – 7{x^2} + 4} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = {{5x + 5} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} – {{{x^2} – x + 1} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} \cr & \Leftrightarrow – 7{x^2} + 4 = 5x + 5 – {x^2} + x – 1 \cr & \Leftrightarrow – 7{x^2} + {x^2} – 5x – x = 5 – 1 – 4 \cr & \Leftrightarrow – 6{x^2} – 6x = 0 \cr & \Leftrightarrow – {x^2} – x = 0 \cr & \Leftrightarrow x\left( {x + 1} \right) = 0 \cr} \)
\( \Leftrightarrow x = 0\)hoặc \(x + 1 = 0\)
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\( \Leftrightarrow x = 0\)hoặc \(x = – 1\) (loại)
Vậy phương trình có nghiệm x = 0
c.\(\eqalign{ & 2{x^2} – x = 3 – 6x \cr & \Leftrightarrow 2{x^2} – x + 6x – 3 = 0 \cr & \Leftrightarrow \left( {2{x^2} + 6x} \right) – \left( {x + 3} \right) = 0 \cr & \Leftrightarrow 2x\left( {x + 3} \right) – \left( {x + 3} \right) = 0 \cr & \Leftrightarrow \left( {x + 3} \right)\left( {2x – 1} \right) = 0 \cr} \)
\( \Leftrightarrow 2x – 1 = 0\)hoặc \(x + 3 = 0\)
+ \(2x – 1 = 0 \Leftrightarrow x = {1 \over 2}\)
+ \(x + 3 = 0 \Leftrightarrow x = – 3\)
Vậy phương trình có nghiệm x = -3 hoặc \(x = {1 \over 2}\)
d. \({{x – 2} \over {x + 2}} – {3 \over {x – 2}} = {{2\left( {x – 11} \right)} \over {{x^2} – 4}}\) ĐKXĐ: \(x \ne \pm 2\)
\(\eqalign{ & \Leftrightarrow {{x – 2} \over {x + 2}} – {3 \over {x – 2}} = {{2x – 22} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} \cr & \Leftrightarrow {{\left( {x – 2} \right)\left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} – {{3\left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} = {{2x – 22} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} \cr & \Leftrightarrow \left( {x – 2} \right)\left( {x + 2} \right) – 3\left( {x + 2} \right) = 2x – 22 \cr & \Leftrightarrow {x^2} – 2x – 2x + 4 – 3x – 6 = 2x – 22 \cr & \Leftrightarrow {x^2} – 2x – 2x – 3x – 2x + 4 – 6 + 22 = 0 \cr & \Leftrightarrow {x^2} – 9x + 20 = 0 \cr & \Leftrightarrow {x^2} – 5x – 4x + 20 = 0 \cr & \Leftrightarrow x\left( {x – 5} \right) – 4\left( {x – 5} \right) = 0 \cr & \Leftrightarrow \left( {x – 4} \right)\left( {x – 5} \right) = 0 \cr} \)
\( \Leftrightarrow x – 4 = 0\) hoặc \(x – 5 = 0\)
+ \(x – 4 = 0 \Leftrightarrow x = 4\)
+ \(x – 5 = 0 \Leftrightarrow x = 5\)
Vậy phương trình có nghiệm x = 4 hoặc x = 5
Mục lục môn Toán 8 (SBT)