\(\left(\dfrac{99x+1}{5x^2-5}+\dfrac{1}{5+5x}+\dfrac{20}{1-x}\right):\dfrac{4}{x^3y-xy}\)

\(=\left(\dfrac{99x+1}{5\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{5\left(x-1\right)\left(x+1\right)}-\dfrac{100\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\right):\dfrac{4}{xy\left(x^2-1\right)}\)

\(=\dfrac{99x+1+x-1-100x-100}{5\left(x-1\right)\left(x+1\right)}:\dfrac{4}{xy\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-100xy}{20}=-5xy=VP\)( đpcm )

a, \(4x\left(x+1\right)-5\left(x+1\right)=0\)

\(\left(x+1\right)\left(4x-5\right)\)=0

\(\left\{{}\begin{matrix}x+1=0\\4x-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right)\\4x=5\Rightarrow x=\frac{5}{4}\end{matrix}\right.\)

b, \(5x\left(x-20\right)+5x-100=0\)

\(5x\left(x-20\right)+\left(5x-100\right)=0\)

\(5x\left(x-20\right)+5\left(x-20\right)=0\)

\(\left(x-20\right)\left(5x+5\right)\)= 0

\(\left\{{}\begin{matrix}x-20=0\\5x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\5x=-5\Rightarrow x=-1\end{matrix}\right.\)

c, \(2\left(x-2\right)+\left(x-2\right)^2=0\)

Rút gọn thừa số chung

Giải phương trình

Giải phương trình

Biệt thức

Biệt thức

Nghiệm

Lời giải thu được

Vậy x= 0 và x = 2

d, \(\left(x-3\right)^2-5x-x^2=12\)

\(\left(x^2-2.x.3+3^2\right)-5x-x^2=12\)

\(x^2-6x+9-5x-x^2=12\)

\(-11x+9=12\)

\(-11x=3\)

=> \(x=-\frac{3}{11}\)

a: =>x/27+1=-2/3

=>x/27=-5/3

=>x=-45

b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)

=>x=221/50

c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)

=>x=1/15-2/3=1/15-10/15=-9/15=-3/5

d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)

=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)

=>x=-37/24

e: =>-3/7x=84/45

=>x=-196/45

f: =>11/10x=-2/3

=>x=-20/33

\(\left|5x+13\right|=2x-7\)

khi \(x>\frac{7}{2}\), biểu thức có dạng:

\(\orbr{\begin{cases}5x+13=2x-7\\5x+13=7-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=-20\\7x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{20}{3}\\x=-\frac{6}{7}\end{cases}}}\)

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

quy đồng mẫu số hộ mình

\(\dfrac{x}{4x-16}=\dfrac{x}{4\left(x-4\right)}=\dfrac{-5x}{20\left(4-x\right)}\)

\(\dfrac{1+x}{20-5x}=\dfrac{1+x}{5\left(4-x\right)}=\dfrac{4+4x}{20\left(4-x\right)}\)

\(\dfrac{x}{4x-16}=\dfrac{x}{4\left(x-4\right)}=\dfrac{5x}{20\left(x-4\right)}\\ \dfrac{1+x}{20-5x}=\dfrac{-\left(x+1\right)}{5\left(x-4\right)}=\dfrac{-4\left(x+1\right)}{20\left(x-4\right)}\)

1. ĐKXĐ: \(x\ge3\)

\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)

⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

⇔ \(4\sqrt{x-3}=20\)

⇔ \(\sqrt{x-3}=5\)

⇔ \(x-3=25\)

⇔ \(x=28\left(TMĐKXĐ\right)\)

Vậy....

2. ĐKXĐ: \(x\ge0\)

\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)

⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)

⇔ \(2\sqrt{5x}=7\)

⇔ \(\sqrt{5x}=\frac{7}{2}\)

⇔ \(5x=\frac{49}{4}\)

⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)

Vậy...