Rút gọn:
P= \(\dfrac{2}{2x+3}+\dfrac{3}{2x+1}-\dfrac{6x+5}{\left(2x-3\right)\left(2x+3\right)}\)
K
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31 tháng 5 2017
ĐKXĐ: \(x\ne\pm3,x\ne\dfrac{9}{2}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{2x-9}.\dfrac{3\left(x-3\right)-x}{x\left(x-3\right)}\right]\) : \(\dfrac{x^2-5x-6}{-2\left(x-3\right)\left(x+3\right)}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x-3}\right]:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x\left(x+3\right)-2x^2+2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}=1\)
HN
3 tháng 5 2017
\(E=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\left(\dfrac{2x+3}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\dfrac{x^2+7x}{x\left(x+1\right)^2}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}\)
\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}=\dfrac{2x\left(2x+5\right)}{\left(x+1\right)\left(3x+1\right)}\)
18 tháng 5 2022
\(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)
9 tháng 12 2021
\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)
MD
27 tháng 5 2017
Haizzzzzzzzzzz! 
ĐKXĐ: \(x\ne0;\dfrac{-1}{2};\dfrac{1}{2}\)
\(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right)\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)
=
\(\dfrac{4x\left(x+1\right)+1}{4x^2}.\left[\dfrac{\left(1-2x\right)\left(1+2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{\left(1-2x\right)\left(1+2x\right)}.\dfrac{\left(1-2x\right)^2}{1+2x}\right]\)\(-\dfrac{1}{2x}\)
= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\left(\dfrac{1-4x^2}{\left(2x+1\right)^2}-\dfrac{1-2x}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\dfrac{2x\left(1-2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
= \(\dfrac{1-2x}{2x}-\dfrac{1}{2x}=\dfrac{-2x}{2x}=1\)
TT
20 tháng 1 2019
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
⇔ \(6x^2-5x+3=2x-9x+6x^2\)
⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)
⇔ \(2x+3=0\)
⇔ \(2x=-3\)
⇔ \(x=-\dfrac{3}{2}\)
TT
20 tháng 1 2019
b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)
⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)
⇔ \(12x-92-8\left(4x+1\right)=0\)
⇔ 12x - 92 - 32x - 8 = 0
⇔ -100 - 20x = 0
⇔ 20x = -100
⇔ x = -100 : 20
⇔ x = -5
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
HP
16 tháng 3 2021
1.
ĐK: \(x\ne7;x\ne-1;x\ne3\)
\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)
TH1: \(x< -1\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\ge0\)
\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)
TH2: \(-1< x< 3\)
\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)
\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)
\(\Leftrightarrow-x^2+5x-22\ge0\)
\(\Rightarrow\) vô nghiệm
TH3: \(3< x< 7\)
Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)
\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)
TH4: \(x>7\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\le0\)
\(\Rightarrow\) vô nghiệm
Vậy ...
Các bài kia tương tự, chứ giải ra mệt lắm.
KT
23 tháng 3 2017
a)
\(Q=\left(\dfrac{2x-x^2}{2x^2+8}-\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\left(\dfrac{2}{x^2}+\dfrac{1-x}{x}\right)\\ =\left(\dfrac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}\right)\left(\dfrac{2+x-x^2}{x^2}\right)\\ =\dfrac{x\left(x-2\right)^2\left(x+2\right)\left(x+1\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\)
\(=\dfrac{\left(x^2-4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
Sửa đề: \(P=\dfrac{2}{2x+3}+\dfrac{3}{2x+1}-\dfrac{6x+5}{\left(2x+1\right)\left(2x+3\right)}\)
\(=\dfrac{4x+2+6x+9-6x-5}{\left(2x+1\right)\left(2x+3\right)}\)
\(=\dfrac{4x+6}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2}{2x+1}\)