Pt\(\Leftrightarrow\left(x^4-4x^2\right)-\left(5x^2-20\right)=0\Leftrightarrow\left(x^2-4\right)\left(x^2-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2;x=-2\\x=\sqrt{5};x=-\sqrt{5}\end{cases}}}\)

Vì x nguyên dương nên \(\Rightarrow\orbr{\begin{cases}x=2\\x=\sqrt{5}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=2\\x=\sqrt{5}\end{cases}}\)

\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)

\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)  

PS: Điều kiện xác đinh bạn tự làm nhé 

x²-5x-4x+20=0

x²-4x-5x+20=0

x(x-4)-5(x-4)=0

(x-4)(x-5)=0

\(\Rightarrow\hept{\begin{cases}x-4=0\Rightarrow x=4\\x-5=0\Rightarrow x=5\end{cases}}\)

tớ hỏi nốt nha

tính nhanh:20²-19²+18²-17²+...+2²-1²

\(x^2-9x+20=0\)

\(\Rightarrow x^2-2.x.\frac{9}{2}+\frac{81}{4}-\frac{1}{4}=0\)

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

\(\Rightarrow\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2=0\)

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

\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy \(x=5\)hoặc   \(x=4\)

Ta có \(x^2+9x+20=0\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\).

Xét 2 TH:

+) a + b = -4; ab = -5: Theo định lý Viet đảo ta có a, b là hai nghiệm của pt \(t^2+4t-5=0\Leftrightarrow\left(t-1\right)\left(t+5\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\end{matrix}\right.\)

+) a + b = -5; ab = -4: Bạn giải tương tự.

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2

\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(\Leftrightarrow\left|x\left(x^2+1\right)\right|-9\left|x^2+1\right|=0\)

\(\Leftrightarrow\left(\left|x\right|-9\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left|x\right|=9\left(x^2+1\ge1>0\right)\Leftrightarrow x=\pm9\)

Vậy ... 

\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(TH1:\left\{{}\begin{matrix}\left|x^3+x\right|=0\\\left|9x^2+9\right|=0\end{matrix}\right.\)

\(\text{Vì }9x^2\ge0\)

\(\Rightarrow9x^2+9\ge9\)

\(TH2:\left|x^3+x\right|=\left|9x^2+9\right|\)

\(\Rightarrow\left[{}\begin{matrix}x^3+x=9x^2-9\\x^3+x=9x^2+9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^3+x+9x^2+9=0\\x^3+x-9x^2-9=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\)

=>|x^3+x|=|9x^2+9|

=>x^3+x=9x^2+9 hoặc x^3+x=-9x^2-9

=>x^3-9x^2+x-9=0 hoặc x^3+9x^2+x+9=0

=>x+9=0 hoặc (x-9)(x^2+1)=0

=>x=9 hoặc x=-9