Toán Lớp 8: Tìm x:
a,( x+2 )( x^2-2x+4 ) – x( x^2-9 )=8
b, ( x-3 )^3 – ( 3x+1 )( 9x^2-3x+1)+26x(x^2-1)+9x^2=0
c,( x-2 )^3 – ( 2x-1 )( 4x^2+2x+1 ) – 8x( 1-x^2)= 5-6x^2
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Toán Lớp 8: Tìm x:
a,( x+2 )( x^2-2x+4 ) – x( x^2-9 )=8
b, ( x-3 )^3 – ( 3x+1 )( 9x^2-3x+1)+26x(x^2-1)+9x^2=0
c,( x-2 )^3 – ( 2x-1 )( 4x^2+2x+1 ) – 8x( 1-x^2)= 5-6x^2
Comments ( 2 )
a)\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right) – x\left( {{x^2} – 9} \right) = 8\\
\Leftrightarrow {x^3} + {2^3} – {x^3} + 9x = 8\\
\Leftrightarrow 9x = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
b){\left( {x – 3} \right)^3} – \left( {3x + 1} \right)\left( {9{x^2} – 3x + 1} \right) + 26x\left( {{x^2} – 1} \right) + 9{x^2} = 0\\
\Leftrightarrow {x^3} – 9{x^2} + 27x – 27 – \left( {27{x^3} + 1} \right) – 26{x^3} – 26x + 9{x^2} = 0\\
\Leftrightarrow x – 28 = 0\\
\Leftrightarrow x = 28\\
Vậy\,x = 28\\
c){\left( {x – 2} \right)^3} – \left( {2x – 1} \right)\left( {4{x^2} + 2x + 1} \right) – 8x\left( {1 – {x^2}} \right) = 5 – 6{x^2}\\
\Leftrightarrow {x^3} – 6{x^2} + 12x – 8 – 8{x^3} + 1 – 8x + 8{x^3} = 5 – 6{x^2}\\
\Leftrightarrow {x^3} + 4x – 12 = 0\\
\Leftrightarrow \left[ {x = 1,722} \right.\\
Vậy\,x = 1,722
\end{array}$