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bài 1:
2(x^2-9).4(x^2-1)
=(2x^2-18)(4x^2-4)
=8x^4-8x^2-72x^2+72
=8x^4-80x^2+72
\(Bai1:2\left(x-3\right)\left(x+3\right)+4\left(x-1\right)\left(x+1\right)\)
\(=2\left(x^2-9\right)+4\left(x^2-1\right)\)
\(=2x^2-18+4x^2-4\)
\(=6x^2-22\)
\(Bai2:-\left(6x-1\right)\left(3-2x\right)+\left(3x-2\right)\left(4x-3\right)=17\)
\(\Leftrightarrow-\left(18x-12x^2-3+2x\right)+12x^2-9x-8x+6=17\)
\(\Leftrightarrow-18x+12x^2+3-2x+12x^2-9x-8x+6=17\)
\(\Leftrightarrow24x^2-37x+9-17=0\)
\(\Leftrightarrow24x^2-37x-8=0\)
Đề sai??
1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)
2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)
3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)
\(A=\left[\frac{6x^2}{x^3-1}-\frac{2x-2}{x^2+x+1}-\frac{1}{x-1}\right]:\frac{x^2+9}{\left(x-1\right)\left(9-4x\right)}\)
\(=\left[\frac{6x^2}{x^3-1}-\frac{\left(2x-2\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)
\(=\frac{6x^2-\left(2x^2-4x+2\right)-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)
\(=\frac{5x^2-2x^2+4x-2-x-1}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)
\(=\frac{3x^2+3x-3}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)
Biểu thức A bạn viết đúng chưa?
\(2;A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(\frac{1-x}{x+2}\right)\)
\(ĐKXĐ:\hept{\begin{cases}x^2-4\ne0\\1-x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne\pm2\\x\ne1\end{cases}}\)
\(a,A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{1-x}\)
\(A=\left(\frac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{1-x}\)
\(A=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1-x}=\frac{-6}{\left(x-2\right)\left(1-x\right)}\)
b, Khi x = -4
\(A=\frac{-6}{\left(-4-2\right)\left(1+4\right)}=\frac{-6}{-6.5}=\frac{1}{5}\)
\(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)\\ =\left(x+2\right)\left(x+2-x+2\right)\\ =4\left(x+2\right)=4x+8\)
\(\left(x+2\right)^3-x.\left(x+2\right).\left(x-2\right)+6x^2\)
\(=x^3+3x^2.2+3x.2^2+2^3-x.\left(x^2-2^2\right)+6x^2\)
\(=x^3+6x^2+12x+8-\left(x^2-4\right)+6x^2\)
\(=x^3+6x^2+12x+8-x^3+4x+6x^2\)
\(=\left(x^3-x^3\right)+\left(6x^2+6x^2\right)+\left(12x+4x\right)+8\)
\(=12x^2+16x+8\)