ĐKXĐ:   \(x\ne\pm2\)

\(\frac{x-2}{x+2}\left(\frac{5x+10}{7x-14}+\frac{x-2}{3x-6}\right)=\frac{x-2}{x+2}\left(\frac{5x+10}{7\left(x-2\right)}+\frac{x-2}{3\left(x-2\right)}\right)\)

\(=\frac{x-2}{x+2}\left(\frac{3\left(5x+10\right)}{21\left(x-2\right)}+\frac{7\left(x-2\right)}{21\left(x-2\right)}\right)=\frac{x-2}{x+2}\left(\frac{15x+30}{21\left(x-2\right)}+\frac{7x-14}{21\left(x-2\right)}\right)\)

\(=\frac{x-2}{x+2}\times\frac{22x+16}{21\left(x-2\right)}=\frac{22x+16}{x+2}\)

a)   \(35^{2005}-35^{2004}=35^{2004}.\left(35-1\right)=35^{2004}.34=35^{2004}.2.17\)\(⋮\)\(17\)

c)    \(27^3+9^5=3^9+3^{10}=3^9\left(1+3\right)=3^9.4\) \(⋮\)\(4\)

hok tốt

ĐKXĐ:     \(a\ne\pm\frac{1}{2}\)

\(\left(\frac{2a-1}{2a+1}-\frac{2a-3}{2a-1}\right):\frac{2a-1}{2a+1}\)

\(=\left(\frac{\left(2a-1\right)^2}{\left(2a+1\right)\left(2a-1\right)}-\frac{\left(2a-3\right)\left(2a+1\right)}{\left(2a-1\right)\left(2a+1\right)}\right).\frac{2a+1}{2a-1}\)

\(=\left(\frac{4a^2-4a+1}{\left(2a+1\right)\left(2a-1\right)}-\frac{4a^2-4a+3}{\left(2a+1\right)\left(2a-1\right)}\right).\frac{2a+1}{2a-1}\)

\(=\frac{-2}{\left(2a+1\right)\left(2a-1\right)}.\frac{2a+1}{2a-1}=\frac{-2}{\left(2a-1\right)^2}\)

\(x^2+4x-y^2+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2-y\right)\left(x+2+y\right)\)

\(x^2+4x-y^2+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2-y\right)\left(x+2+y\right)\)

Chúc bạn học tốt!

1/ \(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\)

\(=-8\)

2/ \(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)

\(=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)

\(=-x-15\)

bn thay x vào tính ra là được

học tốt

a)\(\left(x-2\right)\left(x+2\right)+4\left(1-x^2\right)\)

\(=x^2-4+4-4x^2\)

\(=-3x^2\)

b) \(\left(2x-y\right)^2+\left(2x+y\right)^2-2\left(4x^2-y^2\right)\)

\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2\)

\(=\left(2x-y-2x-y\right)^2\)

\(=\left(-2y\right)^2\)

\(=4y^2\)

a) \(4x^2-4x-15=0\)

<=>  \(\left(2x-5\right)\left(2x+3\right)=0\)

<=>  \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\)<=>  \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

Vậy....

b)  \(x^3-6x^2-x+30=0\)

<=>  \(\left(x-5\right)\left(x-3\right)\left(x+2\right)=0\)

tự giải tiếp

a) \(a^3-b^3-3ab\left(a-b\right)\)

\(=a^3-3a^2b+3ab^2-b^3\)

\(=\left(a-b\right)^3\)

b) \(2x^3+x^2-4x-12\)

\(=2x^3-4x^2+5x^2-10x+6x-12\)

\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(2x^2+5x+6\right)\left(x-2\right)\)

c) \(x^3-3x^2+2\)

\(=x^3-x^2-2x^2+2x-2x+2\)

\(=x^2\left(x-1\right)-2x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x^2-2x-2\right)\left(x-1\right)\)

a)   \(a^3-b^3-3ab\left(a-b\right)=\left(a-b\right)^3\)

b)  \(2x^3+x^2-4x-12=2x^3-4x^2+5x^2-10x+6x-12\)

\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(2x^2+5x+6\right)\)

c) \(x^3-3x^2+2=x^3-x^2-2x^2+2\)

\(=x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^2+2x+2\right)\)

\(x^3+2\sqrt{2}x^2+2x=0\)

\(x\left(x^2+2\sqrt{2}x+2\right)=0\)

\(x\left[x^2+2\sqrt{2}x+\left(\sqrt{2}\right)^2\right]=0\)

\(x\left(x+\sqrt{2}\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\\left(x+\sqrt{2}\right)^2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\sqrt{2}\end{cases}}\)

Vậy ....

\(x^3+2\sqrt{2}x^2+2x=0\)

\(x\left(x^2+2\sqrt{2}x+2\right)=0\)

\(x\left(x+\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+\sqrt{2}\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\sqrt{2}\end{cases}}\)

Vậy \(s=\left\{0;-\sqrt{2}\right\}\)