Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
pT <=>\(\frac{x^4}{\left(x-2\right)^2}+\frac{x^2}{x-2}-2=0\)
đk: x khác 2
Đặt \(\frac{x^2}{x-2}=t\)
Ta có phương trình:
\(t^2+t-2=0\Leftrightarrow t^2+2t-t-2=0\Leftrightarrow t\left(t+2\right)-\left(t+2\right)=0\Leftrightarrow\left(t+2\right)\left(t-2\right)=0\)
<=> \(\orbr{\begin{cases}t=2\\t=-2\end{cases}}\)
Với t=2 ta có:
\(\frac{x^2}{x-2}=2\Leftrightarrow x^2=2x-4\Leftrightarrow x^2-2x+4=0\Leftrightarrow\left(x-1\right)^2+3=0\)vô lí
Với t=-2:
\(\frac{x^2}{x-2}=-2\Leftrightarrow x^2=-2x+4\Leftrightarrow x^2+2x=4\Leftrightarrow\left(x+1\right)^2=5\Leftrightarrow\orbr{\begin{cases}x+1=\sqrt{5}\\x+1=-\sqrt{5}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{cases}}\)(tm)
Vậy...
Bài làm:
PT:
đkxđ: \(x\ne0;x\ne2\)
Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)
BPT:
Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)
\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)
\(\Leftrightarrow\frac{-x}{2}\le0\)
\(\Rightarrow-x\le0\)
\(\Rightarrow x\ge0\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)
Vậy \(S=\left\{-1\right\}\)
b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x-1\le0\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
Vậy \(x\ge0\)
ĐKXĐ: \(x\ne\pm1\)
Ta có: \(\frac{x-1}{x+1}-\frac{x^2+x-2}{x+1}=\frac{x+1}{x-1}-x-2\)
=> \(\left(x-1\right)^2-\left(x^2+x-2\right)\left(x-1\right)=\left(x+1\right)^2-x\left(x^2-1\right)-2\left(x^2-1\right)\)
<=> x2 - 2x + 1 - x^3 + 3x - 2 = x2 + 2x + 1 - x3 + x - 2x2 + 2
<=> -x3 + x2 + x - 1 = -x3 - x2 + 3x + 3
<=> -x3 + x2 + x - 1 + x3 + x2 - 3x - 3 = 0
<=> 2x2 - 2x - 4 = 0
<=> x2 - x - 2 = 0
<=> x2 - 2x + x - 2 = 0
<=> (x + 1)(x - 2) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
Vậy S = {-1; 2}
kl lại. \(\orbr{\begin{cases}x=-1\left(ktm\right)\\x=2\end{cases}}\)
Vậy S = {2}
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
<=> \(\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{1}{x+1}=0\)
<=> \(\frac{2}{\left(x-1\right)^2\left(x+1\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}+\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}=0\)
<=> \(2+3x-3+x^2-2x+1=0\)
<=> x2 + x = 0
<=> x(x + 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy S = {0; -1}
a) ĐKXĐ : \(x\ne\pm a\).
Với \(a=-3\) khi đó ta có pt :
\(A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left(-9+1\right)}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(-3-x\right)}{\left(-3-x\right)\left(-3+x\right)}+\frac{24}{\left(-3-x\right)\left(-3+x\right)}=0\)
\(\Rightarrow x^2-9-\left(-3x-x^2-9-3x\right)+24=0\)
\(\Leftrightarrow2x^2+6x+24=0\)
\(\Leftrightarrow x^2+3x+12=0\) ( vô nghiệm )
Phần b) tương tự.
\(A=\frac{x+a}{a-x}-\frac{x-a}{a+x}=\frac{a\left(3x+1\right)}{a^2-x^2}\)
\(=\frac{x+a}{a-x}+\frac{x-a}{a+x}=\frac{a\left(3+1\right)}{\left(a-x\right)\left(a+x\right)}\)
\(=\frac{\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+1\right)}=\frac{a\left(3a+1\right)}{\left(a+x\right)\left(a-x\right)}\)
\(\Leftrightarrow\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)=a\left(3a+1\right)\)
\(\Leftrightarrow x^2+2ax+a^2-ax-x^2-a^2+ax=3a^2+a\)
\(\Leftrightarrow2ax=3a^2+a\)
\(\Leftrightarrow x=\frac{3a^2+a}{2a}\left(a\ne0\right)\)
a) Khi x=-3 => \(x=\frac{3\cdot\left(-3\right)^2-3}{2\left(-3\right)}=-13\)
b) a=1
\(\Leftrightarrow x=\frac{3\cdot1^2+1}{2\cdot1}=2\)
a) \(ĐKXĐ:x\ne\pm3\)
Với a = -3
\(\Leftrightarrow A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}=\frac{24}{9-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}+\frac{24}{x^2-9}=0\)
\(\Leftrightarrow\frac{-\left(x-3\right)^2-\left(x+3\right)^2+24}{x^2-9}=0\)
\(\Leftrightarrow-x^2+6x-9-x^2-6x-9+24=0\)
\(\Leftrightarrow-2x^2+6=0\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\pm\sqrt{3}\)(tm)
Vậy với \(a=-3\Leftrightarrow x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
b) \(ĐKXĐ:x\ne\pm1\)
Với a = 1
\(\Leftrightarrow A=\frac{x+1}{1-x}-\frac{x-1}{1+x}=\frac{3+1}{1-x^2}\)
\(\Leftrightarrow\frac{x+1}{1-x}-\frac{x-1}{1+x}+\frac{4}{x^2-1}=0\)
\(\Leftrightarrow\frac{-\left(x+1\right)^2-\left(x-1\right)^2+4}{x^2-1}=0\)
\(\Leftrightarrow-x^2-2x-1-x^2+2x-1+4=0\)
\(\Leftrightarrow-2x^2+2=0\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)(ktm)
Vậy với \(a=1\Leftrightarrow x\in\varnothing\)
c) \(ĐKXĐ:a\ne\pm\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào phương trình, ta đươc :
\(A=\frac{\frac{1}{2}+a}{a-\frac{1}{2}}-\frac{\frac{1}{2}-a}{a+\frac{1}{2}}=\frac{a\left(3a+1\right)}{a^2-\frac{1}{4}}\)
\(\Leftrightarrow\frac{a+\frac{1}{2}}{a-\frac{1}{2}}+\frac{a-\frac{1}{2}}{a+\frac{1}{2}}-\frac{3a^2+a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow\frac{\left(a+\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2-3a^2-a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow a^2+a+\frac{1}{4}+a^2-a+\frac{1}{4}-3a^2-a=0\)
\(\Leftrightarrow-a^2-a+\frac{1}{2}=0\)
\(\Leftrightarrow a^2+a-\frac{1}{2}=0\)
\(\Leftrightarrow\left(a+\frac{1}{2}\right)^2-\frac{3}{4}=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}\\a=-\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{-\sqrt{3}-1}{2}\end{cases}}\)(TM)
Vậy với \(x=\frac{1}{2}\Leftrightarrow a\in\left\{\frac{\sqrt{3}-1}{2};\frac{-\sqrt{3}-1}{2}\right\}\)
\(\left(\frac{1}{x-2}-\frac{1}{x+2}\right)+\left(\frac{1}{x-1}-\frac{1}{x+1}\right)=0\)
\(\frac{x+2-x+2}{x^2-4}+\frac{x+1-x+1}{x^2-1}=0\)
\(\frac{4}{x^2-4}+\frac{2}{x^2-1}=0\)
\(4x^2-4+2x^2-8=0\)
\(6x^2-12=0\)
\(x^2=2\)
\(x=\sqrt{2}\)
ĐKXĐ: x≠-2,-1,1,2
Ta có :
\(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+1}+\frac{1}{x+2}\)
<=> \(\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x-2}\)
<=>\(\frac{2}{x^2-1}=\frac{-4}{x^2-4}\)
<=> \(2x^2-8=-4x^2+4\)
<=> \(6x^2=12\)
<=> \(x^2=2\)
<=>\(\hept{\begin{cases}x=\sqrt{2}\left(TMĐK\right)\\x=-\sqrt{2}\left(TMĐK\right)\end{cases}}\)
Vậy pt trên có tập nghiệm S={\(\sqrt{2},-\sqrt{2}\)}
k mk nha mn
\(\frac{x^2-x}{x^2-x+1}-\frac{x^2-x+2}{x^2-x-2}=1\)
\(\Leftrightarrow\frac{\left(x^2-x\right)\left(x^2-x-2\right)}{\left(x^2-x+1\right)\left(x^2-x-2\right)}-\frac{\left(x^2-x+2\right)\left(x^2-x+1\right)}{\left(x^2-x-2\right)\left(x^2-x+1\right)}=1\)
\(\Leftrightarrow\frac{x^4-x^3-2x^2-x^3+x^2+2x}{\left(x^2-x+1\right)\left(x^2-x-2\right)}-\frac{x^4-x^3+x^2-x^3+x^2-x+2x^2-2x+2}{\left(x^2-x-2\right)\left(x^2-x+1\right)}=1\)
\(\Leftrightarrow\frac{x^4-x^3-2x^2-x^3+x^2+2x-x^4+x^3-x^2+x^3-x^2+x-2x^2+2x+2}{\left(x^2-x+1\right)\left(x^2-x-2\right)}=1\)
\(\Leftrightarrow\frac{-5x^2+3x+2}{x^4-2x^3+x-2}=1\)
<=> -5x2+3x+2=x4-2x3+x-2
<=> -5x2+3x-x4+2x3-x=-2-2
<=> -5x2+2x-x4+2x3=-4
<=> 2x(1+x2)+x2(-5-x2)=-4
