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29 tháng 7 2021
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
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, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
NH
1 tháng 9 2020
\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)
\(\text{Vậy x=-5}\)
\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)
\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)
\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)
\(\Rightarrow-16x-8=7\)
\(\Rightarrow-16x=15\)
\(\Rightarrow x=\frac{-15}{16}\)
\(\text{Vậy }x=\frac{-15}{16}\)
\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)
\(\Rightarrow-9+8x-1=8\)
\(\Rightarrow8x=18\)
\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)
\(\text{Vậy }x=\frac{9}{4}\)
\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)
MD
27 tháng 5 2017
Haizzzzzzzzzzz! 
ĐKXĐ: \(x\ne0;\dfrac{-1}{2};\dfrac{1}{2}\)
\(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right)\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)
=
\(\dfrac{4x\left(x+1\right)+1}{4x^2}.\left[\dfrac{\left(1-2x\right)\left(1+2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{\left(1-2x\right)\left(1+2x\right)}.\dfrac{\left(1-2x\right)^2}{1+2x}\right]\)\(-\dfrac{1}{2x}\)
= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\left(\dfrac{1-4x^2}{\left(2x+1\right)^2}-\dfrac{1-2x}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\dfrac{2x\left(1-2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
= \(\dfrac{1-2x}{2x}-\dfrac{1}{2x}=\dfrac{-2x}{2x}=1\)
16 tháng 5 2022
\(=\left(\dfrac{4x\left(x+1\right)+1}{4x^2}\right)\cdot\left(\dfrac{-2x+1}{2x+1}+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-2x+1}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
\(=\dfrac{-\left(4x^2-1\right)+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x-2x}{4x^2}=-1\)
2 tháng 2 2022
\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)
\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)
