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d, Ta có : \(\frac{x^3+4x^2-x-4}{x+4}\)
\(=\frac{x^2\left(x+4\right)-\left(x+4\right)}{x+4}=\frac{\left(x^2-1\right)\left(x+4\right)}{x+4}=x^2-1\)
- Thay \(x=-2\frac{1}{3}\) vào biểu thức trên ta được :
\(\left(-2\frac{1}{3}\right)^2-1=\frac{58}{9}\)
Vậy biểu thức có giá trị là \(\frac{58}{9}\) tại \(x=-2\frac{1}{3}\)
a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+\left(2y\right)^3-\left(x^3-y^3\right)\)
\(=x^3+8y^3-x^3+y^3\)
\(=9y^3\)
b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)
\(=x^3-x^2-x+1-x^3-8\)
\(=-x^2-x-7\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
a) A có nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x-1\ne0\\\left(1-x\right)\left(x+1\right)\ne0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)
b) Ta có:
A = \(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
A = \(\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
A = \(\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)
c) A = -1/2
<=> \(\frac{1}{2\left(x-1\right)}=-\frac{1}{2}\)
<=> 2(x - 1) = -2
<=> x - 1 = -1
<=> x = 0 (tmđk)
Vậy x = 0
