\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(đk:x\ge-2\right)\)

Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)

\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)

PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)

+ Với a=1

\(\Rightarrow\sqrt{x+5}=1\Leftrightarrow x+5=1\Leftrightarrow x=-4\left(ktm\right)\)

+ Với b=1

\(\Rightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\left(tm\right)\)

Vậy \(S=\left\{-1\right\}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)

Thì được:

\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)

Làm tiếp

Pt <=> (x² -4)(x² -10) =72 <=> (x² -7+3)(x² - 7-3) =72 <=> (x² -7)² - 3² =72 <=> (x² -7)² = 81 <=> x² - 7 =9 hay x² - 7 = -9

<=> x² = 16 hay x² = -2 <=> x =4 ; x = -4 

     (x-2)(x+2)(x^2-10)=72

\(\Leftrightarrow\)(x²-4)(x²-10)=72

Đặt t=x²-7

Ta có phương trình ẩn t:

    (t+3)(t-3)=72

\(\Leftrightarrow\)t²-9=72

\(\Leftrightarrow\)t²=81

\(\Leftrightarrow\)t\(\in\left(9;-9\right)\)

\(\Rightarrow\)x\(\in\)(4;-4)

\(ĐK:x\ne2;x\ne-3\\ PT\Leftrightarrow\left(x-2\right)\left(x+3\right)+2\left(x+3\right)=10\left(x-2\right)+50\\ \Leftrightarrow x^2+x-6+2x+6=10x-20+50\\ \Leftrightarrow x^2-13x-30=0\\ \Leftrightarrow x^2-15x+2x-30=0\\ \Leftrightarrow\left(x-15\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\x=-2\end{matrix}\right.\left(tm\right)\)

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)

Ta có :

 \(\left(x^2-4x\right)+\left(x-2\right)^2=10\)

\(\Leftrightarrow x^2-4x+x^2-4x+4=10\)

\(\Leftrightarrow2x^2+4=10\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt{3}\)

Vậy phương trình có tập nghiệm là \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)

\(\left(x^2-4x\right)+\left(x-2\right)^2=10\)

\(\Leftrightarrow\left(x^2-4x\right)+\left(x^2-4x+4\right)=10\)

\(\Leftrightarrow x^2-4x+x^2-4x+4=10\)

\(\Leftrightarrow\left(x^2+x^2\right)-\left(4x+4x\right)+\left(4-10\right)=0\)

\(\Leftrightarrow2x^2-8x-6=0\)

Ta có \(\Delta=8^2+4.6.2=112\)

\(\Rightarrow\orbr{\begin{cases}x_1=\frac{8+\sqrt{112}}{4}=2+\sqrt{7}\\x_2=\frac{8-\sqrt{112}}{4}=2-\sqrt{7}\end{cases}}\)

ĐKXĐ: x ≠ \(\pm\) 1

Từ phương trình ban đầu suy ra:

\(x^2\left(x+1\right)^2+x^2\left(x-1\right)^2=\frac{10}{9}.\left(x^2-1\right)^2\)

⇒ \(x^4+2x^3+x^2+x^4-2x^3+x^2=\frac{10}{9}\left(x^4-2x^2+1\right)\)

⇒ \(18\left(x^4+x^2\right)=10\left(x^4-2x^2+1\right)\)

⇒ \(4x^4+19x^2-5=0\Leftrightarrow\left(x^2+5\right)\left(4x^2-1\right)=0\)

⇔ \(x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)( thỏa mãn ĐKXĐ)

Vậy ...

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{9x}{x^2-7x+10}=10\)

\(\Leftrightarrow\frac{3x^2-15x-x^2+2x+9x}{\left(x-2\right)\left(x-5\right)}=10\)

\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)

\(\Leftrightarrow8x^2-66+100=0\)

\(\Leftrightarrow4x^2-33x+50=0\)

\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)

b) [(x-7)(x-2)][(x-4)(x-5)]=72

<=> (x²-9x+14)(x²-9x+20)=72

Đặt x²-9x+17=a

=> (a+3)(a-3)=72

<=> a²-9=72

<=> a²=81

=> a=\(\left\{9;-9\right\}\)

TH1: a=9

=> x²-9x+17=9

<=> x²-9x+8=0

<=> (x-1)(x-8)=0

=> x=\(\left\{1;8\right\}\)

TH2: a=-9

=> x²-9x+17=-9

<=> x²-9x+26=0

<=> x²-9x+20,25+5,75=0

<=> (x-4,5)²+5,75=0

=> x\(\in\varnothing\)

Vậy x=\(\left\{1;8\right\}\)

\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)

\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)

\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)

=>\(-9x^4+81x^2-36=0\)

=>9x^4-81x^2+36=0

=>x^4-9x^2+4=0

=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)

=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)