1: Ta có: \(4x^2-36=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]:\left(11+x\right)=3\)

\(\left[\left(35\frac{5}{7}-5\frac{5}{7}\right)+2\frac{3}{4}+\frac{1}{4}\right]:\left(11+x\right)=3\)

\(\left[30+3\right]:\left(11+x\right)=3\)

\(33:\left(11+x\right)=3\)
\(11+x=11\)

\(x=0\)

\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)

\(\left[\left(\frac{35\times7+5}{7}+\frac{2\times4+3}{4}\right)-\frac{5\times7+5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)

\(\left[\left(\frac{250}{7}+\frac{11}{4}\right)-\frac{40}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)

\(\left[\frac{1077}{28}-\frac{167}{28}\right]\div\left(11+x\right)=3\)

\(32,5\div\left(11+x\right)=3\)

\(11+x=32,5\div3\)

\(11+x=\frac{65}{6}\)

\(x=\frac{65}{6}-11=-\frac{1}{6}\)

\(\Rightarrow\left[\left(\frac{250}{7}+\frac{11}{4}\right)-\frac{40}{7}+\frac{1}{4}\right]:\left(11+x\right)=3\)

\(\Rightarrow\left(\frac{1000}{28}+\frac{77}{28}-\frac{160}{28}+\frac{7}{28}\right):\left(11+x\right)=3\)

\(\Rightarrow33:\left(11+x\right)=3\)

\(\Rightarrow11+x=11\)

\(\Rightarrow x=0\)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(E=\frac{\frac{4}{3\cdot7}-\frac{4}{11.15}}{1-\frac{3}{7}-\frac{3}{11}+\frac{1}{5}}-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}\right)\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{11}+\frac{1}{15}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2006}-\frac{1}{2007}\right)\)

\(=\frac{\frac{64}{385}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{2007}\right)=\frac{1}{2007}\)

Vậy : \(E=\frac{1}{2007}\)

a) x=-3/20

b) x=-5/7

a)  \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11-8}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{5}{20}-\dfrac{8}{20}\)

\(\Leftrightarrow x=\dfrac{-3}{20}\)

Vậy x= \(\dfrac{-3}{20}\)

b)  \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{8-15}{20}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Leftrightarrow x=\dfrac{1}{4}.\dfrac{-20}{7}\)

\(\Leftrightarrow x=\dfrac{-5}{7}\)

Vậy x= \(\dfrac{-5}{7}\)

a) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow x+\dfrac{2}{5}=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{11}{12}-\dfrac{8}{12}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{5}{20}-\dfrac{8}{20}=\dfrac{-3}{20}\)

b) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{-7}{20}\)

hay \(x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{1}{4}\cdot\dfrac{-20}{7}=\dfrac{-5}{7}\)