Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\left(x\ne-3;x\ne1\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{x+1}{x-1}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2}{\left(x+3\right)\left(x-1\right)}-\frac{x^2+4x+3}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2-x^2-4x-3-4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3x-9}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3}{x-1}=0\)
=> PT vô nghiệm
\(ĐKXĐ:\) \(x\ne0\)
Đặt \(x+\frac{1}{x}=y\) \(\left(\text{*}\right)\), thì khi đó \(x^2+\frac{1}{x^2}=y^2-2\)
Do đó, \(y^2-2-\frac{9}{2}y+7=0\)
\(\Leftrightarrow\) \(y^2-\frac{9}{2}y+5=0\)
\(\Leftrightarrow\) \(2y^2-9y+10=0\)
\(\Leftrightarrow\) \(2y^2-4y-5y+10=0\)
\(\Leftrightarrow\) \(2y\left(y-2\right)-5\left(y-2\right)=0\)
\(\Leftrightarrow\) \(\left(y-2\right)\left(2y-5\right)=0\)
\(\Leftrightarrow\) \(^{y-2=0}_{2y-5=0}\) \(\Leftrightarrow\) \(^{y=2}_{y=\frac{5}{2}}\)
\(\text{*)}\) Với trường hợp \(y=2\) thì khi đó, \(\left(\text{*}\right)\) \(\Rightarrow\) \(x+\frac{1}{x}=2\) \(\left(1\right)\)
Vì \(x\ne0\) nên từ \(\left(1\right)\) suy ra \(x^2+1=2x\) \(\Leftrightarrow\) \(x^2-2x+1=0\) \(\Leftrightarrow\) \(\left(x-1\right)^2=0\) \(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\) ( thỏa mãn điều kiện xác định)
\(\text{*)}\) Với \(y=\frac{5}{2}\) thì \(\left(\text{*}\right)\) \(\Rightarrow\) \(x+\frac{1}{x}=\frac{5}{2}\) \(\left(2\right)\)
Từ \(\left(2\right)\) \(\Rightarrow\) \(2x^2+2=5x\) (do \(x\ne0\) )
\(\Leftrightarrow\) \(2x^2-5x+2=0\)
\(\Leftrightarrow\) \(2x^2-4x-x+2=0\)
\(\Leftrightarrow\) \(2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\) \(\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\) \(^{x-2=0}_{2x-1=0}\) \(\Leftrightarrow\) \(^{x=2}_{x=\frac{1}{2}}\) (t/mãn điều kiện xác định)
Vậy, \(S=\left\{1;2;\frac{1}{2}\right\}\)
\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)
<=> 4x-16=-3x+6
<=> 4x-16+3x-6=0
<=> 7x-22=0
<=> 7x=22
<=> \(x=\frac{22}{7}\)(TMĐK)
Điều kiện x \(\ge\frac{1}{4}\)
Đặt a = \(\sqrt{x-\frac{1}{4}}\)(a \(\ge0\))
=> x = a2 + \(\frac{1}{4}\)
=> PT <=> 2a2 + \(\frac{1}{2}\)+ \(\sqrt{a^2+\frac{1}{4}+a}\)= 2
<=> \(\sqrt{a^2+\frac{1}{4}+a}\)= \(\frac{3}{2}-2a\)
<=> a2 + 0,25 + a = 4a4 + 2,25 - 6a2
<=> 4a4 - 7a2 - a + 2 = 0
<=> (a + 1)(2a - 1)(2a2 - a - 2) = 0
<=> a = 0,5
<=> x = 0,5
a) \(\frac{2x}{x+2}+\frac{x+2}{2x}=2\)
\(\Leftrightarrow4x^2+\left(x+2\right)^2=4x\left(x+2\right)\)
\(\Leftrightarrow5x^2+4x+4=4x^2+8x\)
\(\Leftrightarrow5x^2+4x+4-4x^2-8x=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow x^2-2.x.2+2^2=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=2\)
a) chắc là nhóm lại thui để sau mk làm:v
b)\(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
Đk: tự lm nhé :v
\(pt\Leftrightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}-\left(\sqrt{2x-1}-\sqrt{3}\right)=2x^2-8\)
\(\Leftrightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2x-1-3}{\sqrt{2x-1}+\sqrt{3}}=2\left(x^2-4\right)\)
\(\Leftrightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}=2\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow\frac{\frac{-2\left(x-2\right)}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}-2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)\right)=0\)
Dễ thấy: \(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)< 0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
