Rút gọn \(E=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
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18 tháng 5 2022
\(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)
28 tháng 1 2021
1/ ĐKXĐ : \(x\ne1\)
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)
Vậy...
b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)
Vậy....
c/ ĐKXĐ : \(x\ne0\)
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)
Vậy...
4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)
\(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2+4x-3x-2=5\)
\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)
Vậy....
5,6 Tương tự nhé !
28 tháng 1 2021
1)ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)
2) ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x-x-1=0\)
\(\Leftrightarrow-29x+11=0\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\)
Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)
3) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+6x\)
\(\Leftrightarrow2x^2-12-2x^2-6x=0\)
\(\Leftrightarrow-6x-12=0\)
\(\Leftrightarrow-6x=12\)
\(\Leftrightarrow x=-2\)
Vậy: S={-2}
20 tháng 7 2022
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
5 tháng 1 2022
\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)
\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
HP
4 tháng 2 2022
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
31 tháng 5 2017
ĐKXĐ: \(x\ne\pm3,x\ne\dfrac{9}{2}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{2x-9}.\dfrac{3\left(x-3\right)-x}{x\left(x-3\right)}\right]\) : \(\dfrac{x^2-5x-6}{-2\left(x-3\right)\left(x+3\right)}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x-3}\right]:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x\left(x+3\right)-2x^2+2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}=1\)
\(E=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\left(\dfrac{2x+3}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\dfrac{x^2+7x}{x\left(x+1\right)^2}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}\)
\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}=\dfrac{2x\left(2x+5\right)}{\left(x+1\right)\left(3x+1\right)}\)