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\(a,\frac{30x^3}{11y^2}.\frac{121y^5}{25x}\)
\(=>\frac{30x^3.121y^5}{11y^2.25x}=\frac{6x^2.11y^3}{5}=\frac{66x^2.y^3}{5}\)
\(b,\frac{24y^5}{7x^2}.\frac{-21x}{12y^3}\)
\(=>\frac{24y^5.\left(-21\right)x}{7x^2.12y^3}=\frac{2y^2.\left(-3\right)}{x}=-\frac{6y^2}{x}\)
\(c,\left(\frac{-18y^3}{25x^4}\right).\left(\frac{-15x^2}{9y^3}\right)\)
\(=>\frac{-18y^3.\left(-15\right)x^2}{25x^4.9y^3}=\frac{-2.\left(-3\right)}{5x^2}=\frac{6}{5x^2}\)
\(d,\frac{3x^2}{2y}.\frac{1}{4y}.\frac{5}{3y}\)
\(=>\frac{3x^2.1.5}{2y.4y.3y}=\frac{15x^2}{24y^3}=\frac{5x^2}{8y^3}\)
\(e,\frac{2x}{3}.\frac{x+1}{2x}\)
\(=>\frac{2x\left(x+1\right)}{3.2x}=\frac{x+1}{3}\)
\(g,\frac{5-x}{x-3}.\frac{2}{3}.\frac{x}{4}\)
\(=>\frac{2x\left(5-x\right)}{3.4\left(x-3\right)}=\frac{10x-2x^2}{12\left(x-3\right)}=\frac{10x-2x^2}{12x-9}\)
a) \(\frac{9x^2}{11y^2}:\frac{6x}{11y}=\frac{9x^2}{11y^2}\cdot\frac{11y}{6x}=\frac{3xy}{2}\)
b) \(\frac{x^2-49}{x-7}+x-2=\frac{\left(x-7\right)\left(x+7\right)}{x-7}+x-2=x+7+x-2=2x+5\)
c) \(\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
= \(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(3-x\right)\left(x+3\right)}\)
= \(\frac{3x-9}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)(đk: \(x-3\ne0\)=> \(x\ne3\))
\(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}\)
\(=\frac{\left(x+3\right)\left(2-x\right)^3}{\left(x+2\right)\left(x-2\right).9\left(x+3\right)}\)
\(=-\frac{\left(x-2\right)^2}{9\left(x+2\right)}\)
\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)
\(=\left(y-1\right)\left[\left(\frac{2}{5}x-\frac{2}{5}y\right)\right]\)
\(=\left(y-1\right)\frac{2}{5}\left(x-y\right)\)
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)
\(b,\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne2;\ne-3\)
\(\Leftrightarrow\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x^2-9=5\)
\(\Leftrightarrow x^2=14\)
\(x=\sqrt{14}\)
.....
a) \(\left(x+3\right)^2-\left(x-3\right)^2=6x\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=6x\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x\Leftrightarrow12x=6x\)\(\Leftrightarrow12x-6x=0\Leftrightarrow6x=0\Leftrightarrow x=0\)
Vậy phương trình có tập nghiệm S = { 0 }
b)\(-ĐKXĐ:\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-3\end{cases}}\)
- Ta có : \(\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\Leftrightarrow\frac{x-3}{x-2}-\frac{5}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-5}{\left(x-2\right)\left(x+3\right)}=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(thoaman\right)\\x=-3\left(kothoaman\right)\end{cases}}\)
Vậy phương trình có tập nghiệm S = { 3 }
a) \(\frac{30x^3}{11y^2}.\frac{121y^5}{25x}=\frac{6x^2.11y^3}{5}=\frac{66x^2y^3}{5}\)
b) \(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)^3}{9\left(x+3\right)}\)
\(=\frac{-\left(x-2\right)^2}{9\left(x+2\right)}\)
p/s: chúc bạn học tốt