nhân chéo là đc:

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x     =14

 x      =2

Vậy x=2

Vì x là số dương nên ta Giả sử \(\hept{\begin{cases}x^2=a\\\frac{2}{x}=b\end{cases}}\) với a,b là hai số tự nhiên

Vậy \(x=\frac{2}{b}\Rightarrow x^2=\frac{4}{b^2}=a\Leftrightarrow4=ab^2\)

Do b là số tự nhiên nên \(\orbr{\begin{cases}b=1\Rightarrow a=4\\b=2\Rightarrow a=1\end{cases}}\) vậy \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)

\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}=9\)

\(\Leftrightarrow x+2=81\)

\(\Leftrightarrow x=79\)

\(\frac{x+3}{-4}=-\frac{9}{x+3}\)

\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x+3\right)^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)

Vậy ....

quy đồng

\(\left(x+3\right)^2=36\)

\(\left(x+3\right)^2-6^2=0\)

áp dụng định lí "  \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được

\(\left(x+3-6\right)\left(x+3+6\right)=0\)

\(x=3,x=-9\)

nhân chéo: 

(x+3)(x+3)=-4.(-9)

(x+3)^2   =36

Hai TH: 

TH1: x+3=6

=>   x=3

TH2: x+3=-6

=>  x=-9

Vậy x=3 hoặc x=-9

ĐK: x >0

Liên hợp:

pt <=> \(\sqrt{\frac{x^2+3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)

<=> \(\frac{\frac{x^2+3}{x}-4}{\sqrt{\frac{x^2+3}{x}}+2}=\frac{x^2+7-4\left(x+1\right)}{2\left(x+1\right)}\)

<=> \(\frac{x^2-4x+3}{x\left(\sqrt{\frac{x^2+3}{x}}+2\right)}=\frac{x^2-4x+3}{2\left(x+1\right)}\)

<=> \(\orbr{\begin{cases}x^2-4x+3=0\left(1\right)\\x\left(\sqrt{\frac{x^2+3}{x}}+2\right)=2\left(x+1\right)\left(2\right)\end{cases}}\)

(1) <=> x = 1 hoặc x = 3 (tm)

(2) <=> \(x\sqrt{\frac{x^2+3}{x}}=2\)

<=> \(x\left(x^2+3\right)=4\)

<=> \(x^3+3x-4=0\)

,<=> (x-1)(x^2 +x  +4) = 0

<=> x = 1 (tm)

Vậy x = 1 hoặc x = 3.

cách khác nhung chỉ dài thêm thôi

\(DK:x>0\)

PT\(\Leftrightarrow2\left(x+1\right)\sqrt{x^2+3}=\sqrt{x}\left(x^2+7\right)\)

Dat \(\sqrt{x^2+3}=t>0\)

PT tro thanh 

\(\sqrt{x}t^2-2\left(x+1\right)t+4\sqrt{x}=0\)

Ta co:

\(\Delta^`_t=\left(x-2\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}t_1=\frac{x+1+\left|x-2\right|}{\sqrt{x}}\\t_2=\frac{x+1-\left|x-2\right|}{\sqrt{x}}\\t_3=\frac{x+1}{\sqrt{x}}\end{cases}}\)

Sau do the vo giai nhu binh thuong :D

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)

\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)

=> PT đúng với mọi x khác \(\pm5\)

Refund QB nhìn logic :V 

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

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

\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)

\(20x=2x^2+50\)

\(20x-2x^2-50=0\)

\(2\left(10x-x^2-25\right)=0\)

\(-x^2+10x+25=0\)

\(x^2-10x+25=0\)

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

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

\(x-5=0\Leftrightarrow x=5\)

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mờ qué

bạn chụp lại đc ko?

\(ĐKXĐ:x\ne\pm3\)

\(pt\Leftrightarrow\frac{\left(x+3\right)^2-\left(x-3\right)^2}{x^2-9}=\frac{17}{x^2-9}\)

\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=17\)

Tự dừng bấm Gửi tl

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=17\)

\(\Leftrightarrow12x=17\Leftrightarrow x=\frac{17}{12}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-9}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-3^2}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{\left(x+3\right)\left(x-3\right)}=\frac{x-3}{x+3}\)

\(\left(x+3\right)^2-17=\left(x-3\right)^2\)

\(x^2+6x-8=x^2-6x+9\)

\(6x-8=-6x+9\)

\(6x=-6x+9+8\)

\(6x+6x=17\)

\(12x=17\Leftrightarrow x=\frac{17}{12}\)