\(\dfrac{8}{\sqrt{x}-2}-\dfrac{\sqrt{x}+26}{x-4}\\ đk:x\ne4;x>0\\ =\dfrac{8\left(\sqrt{x}+2\right)-\sqrt{x}-26}{x-4}\\ =\dfrac{8\sqrt{x}+16-\sqrt{x}-26}{x-4}\\ =\dfrac{7\sqrt{x}-10}{x-4}\)

a: \(2x\left(x^2-3x+1\right)=2x^3-6x^2+2x\)

b: \(\left(x+2\right)^2-x^2=4x+4\)

c: \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=27\)

Đề là biểu thức hay phân thức ( nếu là biểu thức thi :)

a, \(x^2-10x+25=\left(x-5\right)^2\)

 \(x^2-3x-10=x^2+2x-5x-10=\left(x-5\right)\left(x+2\right)\)

\(4x+8=4\left(x+2\right)\)

Nếu là phân thức thì =) p/s : viết đề hẳn hoi đi :v 

a, \(\frac{x^2-10x+25}{x^2-3x-10}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+2\right)}=\frac{x-5}{x+2}\)

b, chả hiểu 

`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`

`<=> 4 + 3 + (-5x) + (-2)=0`

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

`<=>-5x=-5`

`<=>x=1`

`2,(25x^2-10x):5x +3(x-2)=4`

`<=> 5x - 2 + 3x-6=4`

`<=> 8x -8=4`

`<=> 8x=12`

`<=>x=12/8`

`<=>x=3/2`

`3,(3x+1)^2-(2x+1/2)^2=0`

`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`

`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`

`<=>( x+1/2) (5x+3/2)=0`

`@ TH1`

`x+1/2=0`

`<=>x=0-1/2`

`<=>x=-1/2`

` @TH2`

`5x+3/2=0`

`<=> 5x=-3/2`

`<=>x=-3/2 : 5`

`<=>x=-15/2`

`4, x^2+8x+16=0`

`<=>(x+4)^2=0`

`<=>x+4=0`

`<=>x=-4`

`5, 25-10x+x^2=0`

`<=> (5-x)^2=0`

`<=>5-x=0`

`<=>x=5`

a) \(\text{5x(x-2)+(2-x)=0}\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\text{x(2x-5)-10x+25=0}\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)

\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)

\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

a) \(5x\left(x-2\right)+\left(2-x\right)=0\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(x\left(2x-5\right)-10x+25=0\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)

d) \(x^4+2x^2-8=0\)

\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)

\(\Rightarrow\left(x^2+1\right)^2-3^2=0\)

\(\Rightarrow\left(x^2+1-3\right)\left(x^2+1+3\right)=0\)

\(\Rightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)

\(\left(3-x\right)\left(3+x\right)+\left(x-5\right)^2\\ =9-x^2+x^2-10x+25\\ =34-10x\)

`a, x^2-10x+25=0`

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

`<=>(x-5)^2=0`

`<=>x-5=0`

`<=>x=5`

__

`x^2 -8x+16=0`

`<=> x^2 - 2.x.4+4^2=0`

`<=>(x-4)^2=0`

`<=>x-4=0`

`<=>x=4`

__

`x^2-49=0`

`<=>x^2 - 7^2=0`

`<=>(x-7)(x+7)=0`

`<=>x-7=0` hoặc `x+7=0`

`<=> x=7` hoặc `x=-7`

__

`4x^2-25=0`

`<=> (2x)^2 -5^2=0`

`<=>(2x-5)(2x+5)=0`

`<=>2x-5=0` hoặc `2x+5=0`

`<=> 2x=5` hoặc `2x=-5`

`<=>x=5/2` hoặc `x=-5/2`

a: =>(x-5)^2=0

=>x-5=0

=>x=5

b: =>(x-4)^2=0

=>x-4=0

=>x=4

c: =>(x-7)(x+7)=0

=>x-7=0 hoặc x+7=0

=>x=7 hoặc x=-7

d: =>(2x-5)(2x+5)=0

=>2x-5=0 hoặc 2x+5=0

=>x=5/2 hoặc x=-5/2

Em bấm vào biểu tượng \(\sum\) trên thanh công cụ và gõ phân số để mn dễ hỗ trợ nhé!

`(x^2+x-6)/(x^2+4x+3):(x^2-10x+25)/(x^2-4x-5)(x ne -1,x ne 5,x ne -3)`

`=((x-2)(x+3))/((x+1)(x+3)):(x-5)^2/((x+1)(x-5))`

`=(x-2)/(x+1):(x-5)/(x+1)`

`=(x-2)/(x-5)`

\(\dfrac{x^2+x-6}{x^2+4x+3}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)

\(\dfrac{x^2-10x+25}{x^2-4x-5}=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+1\right)}=\dfrac{x-5}{x+1}\)

\(B=\frac{5x^2-10xy}{2\left(2y-x\right)}=\frac{5x\left(x-2y\right)}{-2\left(x-2y\right)}=\frac{5x}{-2}=\frac{-5x}{2}\)