Toán Lớp 8: giúp mik bài này vs ạ
tìm x
a)8(x-1/2)(x^2+1/2x+1/4)-x(1+8x^2)+2=0
b)(x-1)^3+(2x)^2(4+2x+x^2)+3x(x+2)=16
c)(x+2)(x^2-2x+4)-x(x^2-2)=15
d)(x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15
nhanh giúp em vs ạ
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Toán Lớp 8: giúp mik bài này vs ạ
tìm x
a)8(x-1/2)(x^2+1/2x+1/4)-x(1+8x^2)+2=0
b)(x-1)^3+(2x)^2(4+2x+x^2)+3x(x+2)=16
c)(x+2)(x^2-2x+4)-x(x^2-2)=15
d)(x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15
nhanh giúp em vs ạ
Comments ( 1 )
a)8\left( {x – \dfrac{1}{2}} \right)\left( {{x^2} + \dfrac{1}{2}x + \dfrac{1}{4}} \right) – x\left( {1 + 8{x^2}} \right) + 2 = 0\\
\Leftrightarrow 8.\left( {{x^3} – \dfrac{1}{8}} \right) – x – 8{x^3} + 2 = 0\\
\Leftrightarrow 8{x^3} – 1 – x – 8{x^3} + 2 = 0\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
b){\left( {x – 1} \right)^3} + \left( {2 – x} \right)\left( {4 + 2x + {x^2}} \right) + 3x\left( {x + 2} \right) = 16\\
\Leftrightarrow {x^3} – 3{x^2} + 3x – 1 + {2^3} – {x^3} + {3^2} + 6x = 16\\
\Leftrightarrow – 3{x^2} + 9x = 0\\
\Leftrightarrow – 3x\left( {x – 3} \right) = 0\\
\Leftrightarrow x = 0;x = 3\\
Vậy\,x = 0;x = 3\\
c)\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right) – x\left( {{x^2} – 2} \right) = 15\\
\Leftrightarrow {x^3} + 8 – {x^3} + 2x = 15\\
\Leftrightarrow 2x = 7\\
x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
d){\left( {x – 3} \right)^3} – \left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right)\\
+ 9{\left( {x + 1} \right)^2} = 15\\
\Leftrightarrow {x^3} – 9{x^2} + 27x – 27 – {x^3} + 27\\
+ 9{x^2} + 18x + 9 = 15\\
\Leftrightarrow 45x = 6\\
\Leftrightarrow x = \dfrac{2}{{15}}\\
Vậy\,x = \dfrac{2}{{15}}
\end{array}$