Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) MTC : \(\left(x+1\right)\left(x^2-x+1\right)\)
Quy đồng :
\(\frac{x-1}{x^3+1}=\frac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2x}{x^2-x+1}=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x+1}=\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b ) MTC : \(10x\left(2y-x\right)\left(2y+x\right)\)
\(\frac{7}{5x}=\frac{7.2.\left(2y-x\right)\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=\frac{-4.10x.\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}=\frac{-40x\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
c ) MTC : \(\left(x+2\right)^3\)
\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)
\(\frac{3x}{x^2+4x+4}=\frac{3x}{\left(x+2\right)^2}=\frac{3x\left(x+2\right)}{\left(x+2\right)^3}\)
\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)
Lời giải:
a) $(x+3)^2-(x-3)^2=6x+18$
$\Leftrightarrow 12x=6x+18\Leftrightarrow 6x=18\Rightarrow x=3$
b) ĐK:$x\neq 2; x\neq 3$
PT $\Rightarrow x+3=\frac{5}{3-x}$
$\Rightarrow (x+3)(3-x)=5$
$\Rightarrow 9-x^2=5$
$\Rightarrow x^2=4\Rightarrow x=\pm 2$. Kết hợp với ĐKXĐ suy ra $x=-2$
c) ĐKXĐ: $x\neq \frac{\pm 3}{4}$
PT $\Leftrightarrow \frac{12x^2+30x-21}{(4x-3)(4x+3)}-\frac{(3x-7)(3x+4)}{(4x-3)(4x+3)}=\frac{(6x+5)(4x-3)}{(4x-3)(4x+3)}$
$\Rightarrow 12x^2+30x-21-(3x-7)(4x+3)=(6x+5)(4x-3)$
$\Leftrightarrow -24x^2+47x+15=0$
$\Rightarrow x=\frac{47\pm \sqrt{3649}}{48}$
d)
ĐK: $x\neq -1; x\neq 2$
PT $\Leftrightarrow \frac{4(x-2)}{(x+1)(x-2)}-\frac{2(x+1)}{(x-2)(x+1)}=\frac{x+3}{(x+1)(x-2)}$
$\Rightarrow 4(x-2)-2(x+1)=x+3$
$\Rightarrow x=13$ (t.m)
\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)
\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)
\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)
\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)
a) \(\left(x-2\right)^3-\left(x+4\right)^2\)
\(=x^3-6x^2+12x-8-\left(x^2+8x+16\right)\)
\(=x^3-6x^2+12x-8-x^2-8x-16\)
\(=x^3-7x^2+4x-24\)
b) \(\left(x-3\right)^3+\left(x+3\right)^3\)
\(=x^3-9x^2+27x-27+x^3+9x^2+27x+27\)
\(=2x^3+54x\)
\(=2x\left(x^2+27\right)\)
c) \(\left(x-2\right)^2-\left(x+2\right)^2=\left(x^2-4x+4\right)-\left(x^2+4x+4\right)\)
\(=x^2-4x+4-x^2-4x-4=-8x\)
d) \(\frac{x^2-25}{x+5}=\frac{\left(x-5\right)\left(x+5\right)}{x+5}=x-5\)
e) \(\frac{x^3-6x^2+12x-8}{x-2}=\frac{\left(x-2\right)^3}{x-2}=\left(x-2\right)^2\)
g) \(\frac{x^3-125}{x-5}=\frac{x^3-5^3}{x-5}=\frac{\left(x-5\right)\left(x^2+5x+25\right)}{x-5}=x^2+5x+25\)
\(\left(\frac{x}{2}+3\right)\left(5-6x\right)+\left(12x-2\right)\left(\frac{x}{4}+3\right)=0\)
\(\Rightarrow\frac{5x}{2}-3x^2+15-18x+3x^2+36x-\frac{x}{2}-6=0\)
\(\Rightarrow\frac{5x}{2}-\frac{x}{2}+15-6-\left(18x-36x\right)=0\)
\(\Rightarrow2x+9+18x=0\)
\(\Rightarrow20x=-9\)
\(\Rightarrow x=-\frac{9}{20}\)