\(\dfrac{\text{x+1}}{\text{x+2}}\)+\(\dfrac{\text{x-1}}{\text{x-2}}\)=\(\dfrac{\text{2x+1}}{\text{x+1}}\)
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22 tháng 11 2021
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
22 tháng 11 2021
\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)
12 tháng 5 2021
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)
4-x=5 hoặc 4-x=-5
x=-1 hoặc x=9
1 tháng 12 2021
\(a,\left(12x^2y^2-6xy^2\right):3xy+2y=6xy^2\left(2x-1\right):3xy+2y=2y\left(2x-1\right)+2y=4xy-2y+2y=4xy\)
\(b,\dfrac{4}{x+1} + \dfrac{8}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{4\left(x-1\right)+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{x-1}\)
\(c,\dfrac{1 }{x+1}- \dfrac{1}{x-1} +\dfrac{ 2x}{x^2-1} \)
\(=\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x-1-x-1+2x}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2x-2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2}{x+1}\)
1 tháng 12 2021
\(a,=4xy-2y+2y=4xy\\ b,\dfrac{4}{x+1}+\dfrac{8}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ c,\dfrac{1}{x+1}-\dfrac{1}{x-1}+\dfrac{2x}{x^2-1}=\dfrac{x-1-x-1+2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+1}\)
18 tháng 7 2021
a,\(x\ge\dfrac{3}{2}\)
\(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)\(=>2\sqrt{x-1}=\sqrt{2x-3}\)
\(< =>4\left(x-1\right)=2x-3< =>4x-4=2x-3< =>x=0,5\left(ktm\right)\)
\(=>x\in\phi\)
b, \(đk:\left[{}\begin{matrix}x< 1\\x\ge\dfrac{3}{2}\end{matrix}\right.\)
\(=>\sqrt{\dfrac{2x-3}{x-1}}=4< =>\dfrac{2x-3}{x-1}=>4\left(x-1\right)=2x-3\)
\(< =>4x-4=2x-3< =>2x=1=>x=\dfrac{1}{2}\left(tm\right)\)
vậy,,,..
10 tháng 11 2021
\(C=\dfrac{x-\dfrac{1}{x^2}}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}\)
Đk: \(x\ne0\)
\(\Rightarrow C=\dfrac{\dfrac{x^3-1}{x^2}}{\dfrac{x^2+x+1}{x^2}}=\dfrac{x^3-1}{x^2+x+1}\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
10 tháng 11 2021
\(=\dfrac{\dfrac{x^3-1}{x^2}}{\dfrac{x^2+x+1}{x^2}}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
8 tháng 3 2022
\(\dfrac{x-2}{x+1}-\dfrac{3}{x+2}>0.\left(x\ne-1;-2\right).\\ \Leftrightarrow\dfrac{x^2-4-3x-3}{\left(x+1\right)\left(x+2\right)}>0.\\ \Leftrightarrow\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Đặt \(f\left(x\right)=\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Ta có: \(x^2-3x-7=0.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}.\\x=\dfrac{3-\sqrt{37}}{2}.\end{matrix}\right.\)
\(x+1=0.\Leftrightarrow x=-1.\\ x+2=0.\Leftrightarrow x=-2.\)
Bảng xét dấu:
\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\left(-\infty-2\right)\cup\left(\dfrac{3-\sqrt{37}}{2};-1\right)\cup\left(\dfrac{3+\sqrt{37}}{2};+\infty\right).\)
\(\sqrt{x^2-3x+2}\ge3.\\ \Leftrightarrow x^2-3x+2\ge9.\\ \Leftrightarrow x^2-3x-7\ge0.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}.\\x=\dfrac{3+\sqrt{37}}{2}.\end{matrix}\right.\)
Đặt \(f\left(x\right)=x^2-3x-7.\)
\(f\left(x\right)=x^2-3x-7.\)
\(\Rightarrow f\left(x\right)\ge0\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
\(\Rightarrow\sqrt{x^2-3x+2}\ge3\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
\(ĐK:x\ne\pm2;x\ne-1\\ PT\Leftrightarrow\dfrac{x^2-x-2+x^2+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x+1}{x+1}\\ \Leftrightarrow\dfrac{2x^2-4}{x^2-4}=\dfrac{2x+1}{x+1}\\ \Leftrightarrow\left(2x^2-4\right)\left(x+1\right)=\left(x^2-4\right)\left(2x+1\right)\\ \Leftrightarrow2x^3+2x^2-4x-4=2x^3+x^2-8x-4\\ \Leftrightarrow x^2+4x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)