Toán Lớp 8: 1.Tìm x:
a, (x+3)^2 – 2.(x-5)(x+5) + (x-1)^2=4
b (x+3)^3 – 9x(x^2-x) + 8.(x-1)(x+1)=10
c, (2x-1)^3 – (2x-1)(4x+3) + 20x^2 – x( 8x^2 -1)=0
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Toán Lớp 8: 1.Tìm x:
a, (x+3)^2 – 2.(x-5)(x+5) + (x-1)^2=4
b (x+3)^3 – 9x(x^2-x) + 8.(x-1)(x+1)=10
c, (2x-1)^3 – (2x-1)(4x+3) + 20x^2 – x( 8x^2 -1)=0
Comments ( 1 )
a){\left( {x + 3} \right)^2} – 2\left( {x – 5} \right)\left( {x + 5} \right) + {\left( {x – 1} \right)^2} = 4\\
\Leftrightarrow {x^2} + 6x + 9 – 2\left( {{x^2} – 25} \right) + {x^2} – 2x + 1 = 4\\
\Leftrightarrow 2{x^2} + 4x + 10 – 2{x^2} + 50 = 4\\
\Leftrightarrow 4x = 4 – 60 = – 56\\
\Leftrightarrow x = – 14\\
Vậy\,x = – 14\\
b){\left( {x + 3} \right)^3} – 9x\left( {{x^2} – x} \right) + 8\left( {x – 1} \right)\left( {x + 1} \right) = 10\\
\Leftrightarrow {x^3} + 9{x^2} + 27x + 27\\
– 9{x^3} + 9{x^2} + 8\left( {{x^2} – 1} \right) = 10\\
\Leftrightarrow – 8{x^3} + 26{x^2} + 27x + 9 = 0\\
\Leftrightarrow x = 4,13\\
Vậy\,x = 4,13\\
c){\left( {2x – 1} \right)^3} – \left( {2x – 1} \right)\left( {4x + 3} \right) + 20{x^2} – x\left( {8{x^2} – 1} \right) = 0\\
\Leftrightarrow 8{x^3} – 12{x^2} + 6x – 1\\
– \left( {8{x^2} + 2x – 3} \right) – 8{x^3} + x = 0\\
\Leftrightarrow – 20{x^2} + 5x + 2 = 0\\
\Leftrightarrow x = \dfrac{{5 \pm \sqrt {165} }}{{40}}\\
Vậy\,x = \dfrac{{5 \pm \sqrt {165} }}{{40}}
\end{array}$