Bài 1:Cho hai biểu thức:\(A=\frac{x-2}{x}vàB=\frac{4x}{x+1}+\frac{x}{1-x}+\frac{2x}{x^2-1}\)
Cho P=A.B.Tìm tất cả các giá tị của m để phương trình P=m có nghiệm duy nhất
Giúp mk với nhé!
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24 tháng 4 2020
\(M=\left(\frac{x-\sqrt{x}+2}{x-1}-\frac{1}{\sqrt{x}-1}\right)\cdot\frac{x+2\sqrt{x}+1}{2x-2\sqrt{x}}\)
\(=\frac{\left(x-\sqrt{x}+2\right)-\sqrt{x}-1}{x-1}\cdot\frac{\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{x-1}\cdot\frac{\sqrt{x}+1}{2\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{2\sqrt{x}}\)
b) PT có nghiệm <=> x>0
<=>\(\sqrt{x}>0\)
<=> \(\sqrt{x}-1>-1\)
<=> x>-1
10 tháng 5 2022
a: Khi x=2/3 thì \(A=\dfrac{\dfrac{2}{3}-2}{\dfrac{2}{3}}=\dfrac{-4}{3}\cdot\dfrac{3}{2}=-2\)
b: \(B=\dfrac{4x}{x+1}-\dfrac{x}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x^2-4x-x^2-x+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x}{x+1}\)
H
12 tháng 3 2019
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{8x}{x^2-1}\right):\left(\frac{2x-2x^2-6}{x^2-1}-\frac{2}{x-1}\right)\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{8x}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{2x-2x^2-6}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1-8x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x-2x^2-6-2x-2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{4x-8x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{-2x^2-8}\)
..........
NN
12 tháng 3 2019
\(\frac{x+32}{2008}+\frac{x+31}{2009}+\frac{x+29}{2011}+\frac{x+28}{2012}+\frac{x+2056}{4}=0\) \(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+1+\frac{x+31}{2009}+1+\frac{x+29}{2011}+1\)\(+\frac{x+28}{2012}+1+\frac{x+2056}{4}-4\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+\frac{2008}{2008}+\frac{x+31}{2009}+\frac{2009}{2009}+\)\(\frac{x+29}{2011}+\frac{2011}{2011}+\frac{x+28}{2012}+\frac{2012}{2012}+\)\(\frac{x+2056}{4}-\frac{16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32+2008}{2008}+\frac{x+31+2009}{2009}\)\(+\frac{x+29+2011}{2011}+\frac{x+28+2012}{2012}\)\(+\frac{x+2056-16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+2040}{2008}+\frac{x+2040}{2009}+\frac{x+2040}{2011}\)\(+\frac{x+2040}{2012}+\frac{x+2040}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2040\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+2040=0\\\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}=0\end{cases}}\)(vô lí)
\(\Leftrightarrow\)\(x=-2040\)
Vậy phương trình có nghiệm là : x = -2040
ĐK: \(x\ne0,x\ne\pm1\).
\(B=\frac{4x}{x+1}+\frac{x}{1-x}+\frac{2x}{x^2-1}=\frac{4x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{4x^2-4x-x^2-x+2x}{\left(x+1\right)\left(x-1\right)}=\frac{3x^2-3x}{\left(x+1\right)\left(x-1\right)}=\frac{3x}{x+1}\)
\(AB=\frac{x-2}{x}.\frac{3x}{x+1}=\frac{3x-6}{x+1}\)
\(P=m\Leftrightarrow\frac{3x-6}{x+1}=m\Rightarrow m\left(x+1\right)=3x-6\)
\(\Leftrightarrow x\left(m-3\right)=-6-m\)
Với \(m=3\)thì \(0x=-9\)phương trình vô nghiệm.
Với \(m\ne3\): \(x=\frac{-6-m}{m-3}\)
Đối chiếu điều kiện:
\(x\ne0,x\ne\pm1\)suy ra \(\hept{\begin{cases}\frac{-6-m}{m-3}\ne0\\\frac{-6-m}{m-3}\ne1\\\frac{-6-m}{m-3}\ne-1\end{cases}}\Leftrightarrow\hept{\begin{cases}m\ne-6\\m\ne-\frac{3}{2}\end{cases}}\).
Vậy \(m\ne3,m\ne-6,m\ne\frac{-3}{2}\)thì thỏa mãn ycbt.