\(\sqrt{x^2-\frac{7}{x^2}}+\sqrt{x-\frac{7}{x^2}}=x\)

\(\Leftrightarrow\sqrt{x^2-\frac{7}{x^2}}+\sqrt{x-\frac{7}{x^2}}-\sqrt{x-\frac{7}{x^2}}=x-\sqrt{x-\frac{7}{x^2}}\)

\(\Leftrightarrow\left(\sqrt{x^2-\frac{7}{x^2}}\right)^2=\left(x-\sqrt{x-\frac{7}{x^2}}\right)^2\)

\(\Leftrightarrow x^2-\frac{7}{x^2}=x^2-2\sqrt{x-\frac{7}{x^2}}.x+x-\frac{7}{x^2}\)

\(\Leftrightarrow2\sqrt{x-\frac{7}{x^2}}.x-x=0\)

\(\Leftrightarrow x\left(2\sqrt{x-\frac{7}{x^2}}-1=0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

=> x = 2

<=> \(\frac{2x-3}{35}\)+ \(\frac{x^2-2x}{7}\)< \(\frac{5x^2}{35}\)- \(\frac{7\left(2x-5\right)}{35}\)

<=> \(\frac{2x-3}{35}\)+ \(\frac{5\left(x^2-2x\right)}{35}\)< \(\frac{5x^2}{35}\)- \(\frac{7\left(2x-5\right)}{35}\)

<=> 2x - 3 + 5( x² - 2x ) < 5x² - 7( 2x - 5 )

<=> 2x - 3 + 5x² - 10x < 5x² - 14x + 35

<=> 2x + 5x² - 10x - 5x² + 14x < 35 + 3

<=> 6x < 38

<=> x < \(\frac{19}{3}\)

có sai thì thông cảm :)

Thực ra cũng EZ thôi :

\(\frac{6}{x^2-9}-1+\frac{4}{x^2-11}-1-\frac{7}{x^2-8}+1-\frac{3}{x^2-12}+1=0=>\)

\(\frac{15-x^2}{x^2-9}+\frac{15-x^2}{x^2-11}-\frac{15-x^2}{x^2-8}-\frac{15-x^2}{x^2-12}=0\)

=> \(\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}-\frac{1}{x^2-8}-\frac{1}{x^2-12}\right)=0\)

=>\(15-x^2=0=>x=\pm\sqrt{15}\)

Hình như còn nghiệm , any body help me ?

Đặt cái BT thứ nhất là √a thì cái BT sau là √(1/a),khi đó phương trình viết lại(a>0)

√a+√(1/a)=7/4;Bình phương 2 vế suy ra:

a+1/a+2=49/16>>>a+1/a=17/16>>>a^2+1=17/16a>>>16A^2+16-17=0(pt vô nghiệm)

Vậy phương trình vô nghiệm

pt đầu \(\Leftrightarrow x+1+\frac{1}{x+1}+x+7+\frac{7}{x+7}=x+3+\frac{3}{x+3}+x+5+\frac{5}{x+5}\)

\(\Rightarrow\frac{1}{x+1}+\frac{7}{x+7}=\frac{3}{x+3}+\frac{5}{x+5}\\ \Rightarrow\frac{8x+14}{x^2+8x+7}=\frac{8x+30}{x^2+8x+15}\)

\(\Leftrightarrow\left(4x+7\right)\left(x^2+8x+15\right)=\left(4x+15\right)\left(x^2+8x+7\right)\)

Đặt a=4x+7

      b=x² +8x+7

như vậy ta được pt mới có dạng \(a\left(b+8\right)=b\left(a+8\right)\Leftrightarrow ab+8a=ab+8b\Rightarrow a=b\)

hay\(4x+7=x^2+8x+7\Rightarrow x^2+4x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

Nhân liên hợp rồi rút gọn thì ta sẽ ra. Tôi nghĩ vậy

ĐKXĐ: \(\left\{{}\begin{matrix}x^2\ne9\\x^2\ne11\\x^2\ne8\\x^2\ne12\end{matrix}\right.\Leftrightarrow x\notin\left\{3;-3;\sqrt{11};-\sqrt{11};2\sqrt{2};-2\sqrt{2};2\sqrt{3};-2\sqrt{3}\right\}\)

Đặt \(x^2-11=a\)(Điều kiện: \(a\notin\left\{-2;0;-3;1\right\}\))

PT\(\Leftrightarrow\frac{6}{a+2}+\frac{4}{a}-\frac{7}{a+3}-\frac{3}{a-1}=0\)

\(\Leftrightarrow\frac{6}{a+2}-1+\frac{4}{a}-1+\frac{-7}{a+3}+1+\frac{-3}{a-1}+1=0\)

\(\Leftrightarrow\frac{6-a-2}{a+2}+\frac{4-a}{a}+\frac{-7+a+3}{a+3}+\frac{-3+a-1}{a-1}=0\)

\(\Leftrightarrow-\frac{a-4}{a+2}-\frac{a-4}{a}+\frac{a-4}{a+3}+\frac{a-4}{a-1}=0\)

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

\(\Leftrightarrow a-4=0\)

hay a=4

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

\(\Leftrightarrow x^2=15\)

hay \(x=\pm\sqrt{15}\)

\(x\ne\pm2\)

Đặt \(\left\{{}\begin{matrix}\frac{x+3}{x-2}=a\\\frac{x-3}{x+2}=b\end{matrix}\right.\) phương trình trở thành:

\(a^2+6b^2=7ab\)

\(\Leftrightarrow a^2-7ab+6b^2=0\)

\(\Leftrightarrow a^2-ab-6ab+6b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-6b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-6b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=6b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=\frac{6\left(x-3\right)}{x+2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(x+2\right)=\left(x-3\right)\left(x-2\right)\\\left(x+3\right)\left(x+2\right)=6\left(x-3\right)\left(x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=-5x\\x^2-7x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=6\end{matrix}\right.\)