Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK: \(x\ne0\)
\(\frac{\left(x+2\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\frac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\frac{\left(x^3+8\right)-\left(x^3-8\right)}{x^4+4x^2+16}=\frac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow16=\frac{6}{x}\)
\(\Rightarrow x=\frac{3}{8}\)
\(ĐKXĐ:x\ne0\)
\(\frac{x+2}{x^2+2x+4}+\frac{x-2}{x^2-2x+4}=\frac{32}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\frac{x+2}{x^2+2x+4}+\frac{x-2}{x^2-2x+4}-\frac{32}{x\left(x^4+4x^2+16\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+2\right)\left(x^2-2x+4\right)+x\left(x-2\right)\left(x^2+2x+4\right)-32}{x\left(x+2x+4\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow x\left(x^3+8\right)+x\left(x^3-8\right)-32=0\)
\(\Leftrightarrow x\left(x^3+8+x^3-8\right)-32=0\)
\(\Leftrightarrow2x^4-32=0\)
\(\Leftrightarrow x^4=16\)
\(\Leftrightarrow x=\pm2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2;-2\right\}\)
a) ta có :x2+2x+2=(x+1)2+1>0,với mọi x
x2+2x+3=(x+1)2+2>0,với mọi x
ĐKXĐ:x\(\in\)R.Đặt x2+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
=>6(a2-1)+6a2=7a2+7a<=>6a2-6+6a2=7a2+7a<=>12a2-7a2-7a-6=0
<=>5a2-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)
<=>a=2=>x2+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)
Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
a) \(\frac{x^2-2x+2}{x^2+x+1}-\frac{x^2}{x^2+x+1}=\frac{3}{\left(x^4+x^2+1\right)x}\)
\(\Leftrightarrow\frac{x^2-2x+2}{x^2-x+1}.x\left(x^2-x+1\right)\left(x^2+x+1\right)-\frac{x^2}{x^2+x+1}.x\left(x^2-x+1\right)\left(x^2+x+1\right)\)\(=\frac{3}{\left(x^4+x^2+1\right)x}.x\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow x\left(x^2-2x+2\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)-x^3\left(x^2-x+1\right)\left(x^4+x^2+1\right)\)\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x=\frac{3}{2}\)
b) làm tương tự nhé
\(a.\Leftrightarrow x^2+x-6+2x^2+4x+2=x^2-6x+9-2x^2+4x\)
\(\Leftrightarrow4x^2+7x-13=0\)(pt vô nghiệm)
\(b.\Leftrightarrow x^3+3x^2+3x+1-x^2+2x+8=x^3-8+2x^2\)
\(\Leftrightarrow5x=-17\Rightarrow x=\frac{-17}{5}\)
Đặt \(t=x^2+2x+2\left(t\ge1\right)\)
\(c.\Leftrightarrow\frac{t-1}{t}+\frac{t}{t+1}=\frac{7}{6}\)\(\Leftrightarrow\frac{t^2-1+t^2}{t^2+t}=\frac{7}{6}\)\(\Leftrightarrow12t^2-6=7t^2+7t\)
\(\Leftrightarrow5t^2-7t-6=0\Rightarrow\orbr{\begin{cases}t=2\left(tm\right)\\t=\frac{-3}{5}\left(l\right)\end{cases}}\)
\(\Rightarrow x^2+2x+2=2\Rightarrow x=-2\)