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a: =>(x^2+x)^2-2(x^2+x)+(x^2+x)-2=0
=>(x^2+x-2)(x^2+x+1)=0
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
b: ĐKXĐ: x<>4; x<>1
PT =>\(\dfrac{x+3+3x-12}{x-4}=\dfrac{6}{1-x}\)
=>(4x-9)(1-x)=6(x-4)
=>4x-4x^2-9+9x=6x-24
=>-4x^2+13x-9-6x+24=0
=>-4x^2+7x+15=0
=>x=3(nhận) hoặc x=-5/4(nhận)
Đặt \(a=\frac{x+2}{x-1};b=\frac{x-2}{x+1}\)
=>\(ab=\frac{x+2}{x-1}\cdot\frac{x-2}{x+1}=\frac{x^2-4}{x^2-1}\)
Phương trình trở thành: \(a^2+b^2-6ab=0\)
=>\(a^2-6ab+9b^2=8b^2\)
=>\(\left(a-3b\right)^2=\left(2b\sqrt2\right)^2\)
=>\(\left[\begin{array}{l}a-3b=2b\sqrt2\\ a-3b=-2b\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}a=3b+2b\sqrt2=b\left(3+2\sqrt2\right)\\ a=3b-2b\sqrt2=b\cdot\left(3-2\sqrt2\right)\end{array}\right.\)
TH1: \(a=b\left(3+2\sqrt2\right)\)
=>\(\frac{x+2}{x-1}=\frac{x-2}{x+1}\cdot\left(3+2\sqrt2\right)\)
=>\(\left(x+2\right)\left(x+1\right)=\left(3+2\sqrt2\right)\cdot\left(x-2\right)\left(x-1\right)\)
=>\(\left(3+2\sqrt2\right)\left(x^2-3x+2\right)=x^2+3x+2\)
=>\(x^2\left(3+2\sqrt2\right)-x^2+\left(-9-6\sqrt2-3\right)x+6+4\sqrt2-2=0\)
=>\(x^2\left(2\sqrt2+2\right)+x\left(-12-6\sqrt2\right)+4\sqrt2+4=0\) (1)
\(\Delta=\left(-12-6\sqrt2\right)^2-4\cdot\left(2\sqrt2+2\right)\cdot\left(4\sqrt2+4\right)\)
\(=144+72+144\sqrt2-4\left(16+8\sqrt2+8\sqrt2+8\right)\)
\(=216+144\sqrt2-4\left(24+16\sqrt2\right)=216+144\sqrt2-96-64\sqrt2=120+80\sqrt2=40\left(3+2\sqrt2\right)\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{12+6\sqrt2-\sqrt{40\left(3+2\sqrt2\right)}}{2\left(2\sqrt2+2\right)}=\frac{12+6\sqrt2-2\sqrt{10}\left(\sqrt2+1\right)}{2\left(2\sqrt2+2\right)}=\frac{\left(\sqrt2+1\right)\left(6-2\sqrt{10}\right)}{4\left(\sqrt2+1\right)}=\frac{6-2\sqrt{10}}{4}=\frac{3-\sqrt{10}}{2}\\ x=\frac{12+6\sqrt2+\sqrt{40\left(3+2\sqrt2\right)}}{2\left(2\sqrt2+2\right)}=\frac{12+6\sqrt2+2\sqrt{10}\left(\sqrt2+1\right)}{2\left(2\sqrt2+2\right)}=\frac{\left(\sqrt2+1\right)\left(6+2\sqrt{10}\right)}{4\left(\sqrt2+1\right)}=\frac{6+_{}2\sqrt{10}}{4}=\frac{3+\sqrt{10}}{2}\end{array}\right.\)
TH2: \(a=b\left(3-2\sqrt2\right)\)
=>\(\frac{x+2}{x-1}=\frac{x-2}{x+1}\cdot\left(3-2\sqrt2\right)\)
=>\(\left(x+2\right)\left(x+1\right)=\left(3-2\sqrt2\right)\cdot\left(x-2\right)\left(x-1\right)\)
=>\(\left(3-2\sqrt2\right)\left(x^2-3x+2\right)=x^2+3x+2\)
=>\(x^2\left(3-2\sqrt2\right)-x^2+\left(-9+6\sqrt2-3\right)x+6-4\sqrt2-2=0\)
=>\(x^2\left(-2\sqrt2+2\right)+x\left(-12+6\sqrt2\right)-4\sqrt2+4=0\) (1)
\(\Delta=\left(-12+6\sqrt2\right)^2-4\cdot\left(-2\sqrt2+2\right)\cdot\left(-4\sqrt2+4\right)\)
\(=144+72-144\sqrt2-4\left(16-8\sqrt2-8\sqrt2+8\right)\)
\(=216-144\sqrt2-4\left(24-16\sqrt2\right)=216-144\sqrt2-96+64\sqrt2=120-80\sqrt2=40\left(3-2\sqrt2\right)\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{12-6\sqrt2-\sqrt{40\left(3-2\sqrt2\right)}}{2\left(-2\sqrt2+2\right)}=\frac{12-6\sqrt2-2\sqrt{10}\left(\sqrt2-1\right)}{2\left(-2\sqrt2+2\right)}=\frac{\left(\sqrt2-1\right)\left(6-2\sqrt{10}\right)}{4\left(-\sqrt2+1\right)}=\frac{-6+2\sqrt{10}}{4}=\frac{-3+\sqrt{10}}{2}\\ x=\frac{12-6\sqrt2+\sqrt{40\left(3-2\sqrt2\right)}}{2\left(-2\sqrt2+2\right)}=\frac{12-6\sqrt2+2\sqrt{10}\left(\sqrt2-1\right)}{2\left(-2\sqrt2+2\right)}=\frac{\left(\sqrt2-1\right)\left(6+2\sqrt{10}\right)}{4\left(-\sqrt2+1\right)}=\frac{-6-_{}2\sqrt{10}}{4}=\frac{-3-\sqrt{10}}{2}\end{array}\right.\)
1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)
Suy ra: \(x^2+4x+4+2x-4=x^2\)
\(\Leftrightarrow6x=0\)
hay \(x=0\left(nhận\right)\)
2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)
Suy ra: \(x+6-2x+12=3x+6\)
\(\Leftrightarrow-x-3x=6-18=-12\)
hay \(x=3\left(nhận\right)\)
Lời giải:
1. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)
\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)
2. ĐKXĐ: $x\neq \pm 6$
PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)
\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)
\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)
Có (x+1)/(x-2)+x/(x+2)=(6-x)/(x^2-4)+1
<=>(x+1)(x+2)/(x-2)(x+2)+x(x-2)/(x-2)(x+2)=(6-x)/(x-2)(x+2)+(x-2)(x+2)/(x-2)(x+2)
=>(x+1)(x+2)+x(x-2)=(6-x)+(x-2)(x+2)
<=>x^2+3x+2+x^2-2x=6-x+x^2-4
<=>2x^2+x+2=x^2-x+2
<=>x^2+2x=0
<=>x(x+2)=0
suy ra :x=0 hoặc x=-2
Vậy...
ĐKXĐ: $x \geq 2$
\(\Leftrightarrow2\left(x-4\right).\sqrt{x-2}-2\left(x-4\right)+\left(x-2\right)\sqrt{x+1}-2\left(x-2\right)+6x-18=0\\ \Leftrightarrow2.\left(x-4\right).\dfrac{x-3}{\sqrt{x-2}+1}+\left(x-2\right).\dfrac{x-3}{\sqrt{x+1}+2}+6.\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=0\right)\\ \Leftrightarrow x=3\)
Vì \(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=\dfrac{2\left(x-4\right)+4.\sqrt{x-2}+4}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2\\ =\dfrac{2\left(x-2\right)+4.\sqrt{x-2}}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2>0\)
Vậy....
1) Dễ thấy x= 0 không là nghiệm của phương trình nên
P T ⇔ x + 1 x − 1 x + 1 x + 4 = 6
Đặt t = x + 1 x ta được t − 1 t + 4 = 6 ⇔ t 2 + 3 t − 10 = 0 ⇔ t = 2 t = − 5
Với t = 2 ⇒ x + 1 x = 2 ⇔ x 2 − 2 x + 1 = 0 ⇔ x = 1
Với t = − 5 ⇒ x + 1 x = − 5 ⇔ x 2 + 5 x + 1 = 0 ⇔ x = − 5 − 21 2 x = − 5 + 21 2