Toán Lớp 8: Tìm x biết
a. 4 (x + 1 ) ² – ( 2x – 1 ) = -9
b. ( x + 1 ) ³ – ( x – 1 ) ³ – 6 ( x – 1 ) ² = -10
c. ( x – 2 ) ³ – ( x + 1 ) ( x ² – x + 1 ) + 6 ( x – 2 ) ( x + 2 ) = 0
d. ( x – 3 ) ( x ² + 3x + 9 ) + x ( x + 2 ) ( 2 – x ) = 1
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a)4{\left( {x + 1} \right)^2} – {\left( {2x – 1} \right)^2} = – 9\\
\Leftrightarrow 4\left( {{x^2} + 2x + 1} \right) – 4{x^2} + 4x – 1 + 9 = 0\\
\Leftrightarrow 4{x^2} + 8x + 4 – 4{x^2} + 4x + 8 = 0\\
\Leftrightarrow 12x + 12 = 0\\
\Leftrightarrow x = – 1\\
Vậy\,x = – 1\\
b){\left( {x + 1} \right)^3} – {\left( {x – 1} \right)^3} – 6{\left( {x – 1} \right)^2} = – 10\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 – {x^3} + 3{x^2} – 3x + 1\\
– 6\left( {{x^2} – 2x + 1} \right) = – 10\\
\Leftrightarrow 6{x^2} + 2 – 6{x^2} + 12x – 6 = – 10\\
\Leftrightarrow 12x = – 6\\
\Leftrightarrow x = – \dfrac{1}{2}\\
Vậy\,x = – \dfrac{1}{2}\\
c){\left( {x – 2} \right)^3} – \left( {x + 1} \right)\left( {{x^2} – x + 1} \right)\\
+ 6\left( {x – 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow {x^3} – 6{x^2} + 12x – 8 – {x^3} – 1 + 6\left( {{x^2} – 4} \right) = 0\\
\Leftrightarrow – 6{x^2} + 12x – 9 + 6{x^2} – 24 = 0\\
\Leftrightarrow 12x = 33\\
\Leftrightarrow x = \dfrac{{11}}{4}\\
Vậy\,x = \dfrac{{11}}{4}\\
d)\left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right) + x\left( {x + 2} \right)\left( {2 – x} \right) = 1\\
\Leftrightarrow {x^3} – {3^3} + x\left( {4 – {x^2}} \right) = 1\\
\Leftrightarrow {x^3} – 27 + 4x – {x^3} = 1\\
\Leftrightarrow 4x = 28\\
\Leftrightarrow x = 7\\
Vậy\,x = 7
\end{array}$