a, Ta có :  \(|2x-10|^{13}\ge0\)

              \(\left(25y+50\right)^{10}\ge0\) 

\(\Rightarrow\) \(|2x-10|^{13}+\left(25y+50\right)^{10}\ge0\)

\(\Rightarrow\) \(|2x-10|^{13}+\left(25y+50\right)^{10}=0\)  khi

\(2x-10=0\)      và      \(25y+50=0\)

\(\Rightarrow\)  \(2x=10\)    và      \(25y=-50\)

\(\Rightarrow\)     \(x=5\)      và        \(y=-2\)

Study well ! >_<

b,

2A=\(2^2-2^3+2^4-2^5+........+2^{2012}-2^{2013}\)

2A-A=\(-2^{2013}-2\)

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

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

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

\(\Leftrightarrow7x=1\)

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

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

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

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

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

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

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

`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):}`

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

=>x=3 hoặc x=-10/7

b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)

\(\Leftrightarrow x^2-12x-51+13x+39=0\)

\(\Leftrightarrow x^2+x-12=0\)

=>(x+4)(x-3)=0

=>x=-4

a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)

1,

Vì \(\left|2x-27\right|^{2007}\ge0;\left(3y+10\right)^{2008}\ge0\)

\(\Rightarrow\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\)

Mà \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)

\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2007}=0\\\left(3y+10\right)^{2008}=0\end{cases}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-10}{3}\end{cases}}}\)

2,

TH1: \(x\ge\frac{3}{5}\)

<=> 2(5x-3)-2x=14

<=> 10x-6-2x=14

<=>8x-6=14

<=>8x=20

<=>x=5/2 (thỏa mãn)

TH2: x < 3/5

<=> 2(3-5x)-2x=14

<=>6-10x-2x=14

<=>6-12x=14

<=>12x=-8

<=>x=-2/3 (thỏa mãn)

Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)

1 x=13,5 ;y=-10/3

2 kết quả x =-2/3