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NT
1
11 tháng 9 2016
a) \(x^2-5=0\)
\(\Leftrightarrow x^2=5\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
b) \(\left(2x-1\right)^2-\left(3x+1\right)^2=0\)
\(\Leftrightarrow\left(2x-1+3x+1\right)\left(2x-1-3x-1\right)=0\)
\(\Leftrightarrow5x\left(-x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\-x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
c) \(\frac{4}{9}\cdot x^2=-4x-9\)
\(\Leftrightarrow\left(\frac{2}{3}x\right)^2+4x+9=0\)
\(\Leftrightarrow\left(\frac{2}{3}x+3\right)^2=0\)
\(\Leftrightarrow\)\(\frac{2}{3}x+3=0\Leftrightarrow x=-\frac{9}{2}\)
26 tháng 6 2021
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
26 tháng 6 2021
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
22 tháng 2 2020
ĐKXĐ : \(x\ne\pm\frac{1}{2}\)
\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)
\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)
\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)
\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)
\(E=\frac{8x^3+1}{1+4x^2}\)
Study well
LT
22 tháng 2 2020
E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)
E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{4x^3+1}{1+4x^2}\)
22 tháng 2 2020
E=\(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{1-4x^2}\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^2+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{1+4x+4x^2-1+4x-4x^2}\)
E=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)
12 tháng 2 2021
a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)
\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)
\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)
\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
Vậy: S={0;-8}
b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)
c) Ta có: \(x^2+6x+9=4x^2\)
\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)
\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)
LV
1
13 tháng 3 2022
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
29 tháng 7 2021
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
MD
27 tháng 5 2017
Haizzzzzzzzzzz! 
ĐKXĐ: \(x\ne0;\dfrac{-1}{2};\dfrac{1}{2}\)
\(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right)\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)
=
\(\dfrac{4x\left(x+1\right)+1}{4x^2}.\left[\dfrac{\left(1-2x\right)\left(1+2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{\left(1-2x\right)\left(1+2x\right)}.\dfrac{\left(1-2x\right)^2}{1+2x}\right]\)\(-\dfrac{1}{2x}\)
= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\left(\dfrac{1-4x^2}{\left(2x+1\right)^2}-\dfrac{1-2x}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\dfrac{2x\left(1-2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
= \(\dfrac{1-2x}{2x}-\dfrac{1}{2x}=\dfrac{-2x}{2x}=1\)
22 tháng 8 2021
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
2x-4x2
=2x(1-2x)