\(\dfrac{2}{x^2-5x+6}-\dfrac{2}{x^2-9}\)
help me
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
8 tháng 3 2018
\(\Leftrightarrow\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{x\left(x+1\right)}=-1\left(đkxđ:x\ne\pm1;0;2;3\right)\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{x+1}=-1\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x+1}=-1\)
\(\Leftrightarrow\dfrac{4}{x^2-2x-3}=-1\)
\(\Leftrightarrow x^2-2x-3=-4\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\left(loai\right)\)
Vậy không có giá trị x thỏa mãn
KT
1
15 tháng 12 2021
=\(^{\dfrac{-x^2-xy}{5\left(x^2-y^2\right)}}\).\(\dfrac{3\left(x^3-y^3\right)}{x^2-xy}\)
=\(\dfrac{-3\left(x-y\right)}{5}\)
HV
27 tháng 3 2018
Tính M = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
23 tháng 1 2022
5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)
\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)
\(\Leftrightarrow x^2-x-72=0\)
=>(x-9)(x+8)=0
=>x=9 hoặc x=-8
6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0 hoặc x=5
23 tháng 1 2022
5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x
<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8
6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9
<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5
7, <=> (x-1)^2 = (3x+3)^2
<=> (x-1-3x-3)(x-1+3x+3) = 0
<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2
8, = (x^2-10x-15)(x^2-10x+25)
VN
1
22 tháng 6 2017
\(\left(\dfrac{4x}{x^2-4x+7}-1\right)+\left(\dfrac{3x}{x^2-5x+7}-1\right)=2\)
31 tháng 5 2017
ĐKXĐ: \(x\ne\pm3,x\ne\dfrac{9}{2}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{2x-9}.\dfrac{3\left(x-3\right)-x}{x\left(x-3\right)}\right]\) : \(\dfrac{x^2-5x-6}{-2\left(x-3\right)\left(x+3\right)}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x-3}\right]:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x\left(x+3\right)-2x^2+2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}=1\)
21 tháng 1 2021
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
KV
21 tháng 1 2021
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow3}\dfrac{x^2-9}{x^2-5x+6}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{x+3}{x-2}=\dfrac{3+3}{3-2}=\dfrac{6}{1}=6\)
b: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-5x}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{x\left(x-5\right)}{x-5}=\lim\limits_{x\rightarrow5}x=5\)
c: \(\lim\limits_{x\rightarrow-3}\dfrac{x^2-3x}{2x^2+9x+9}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{x\left(x-3\right)}{2x^2+6x+3x+9}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{\left(-3\right)\left(-3-3\right)}{\left(-3+3\right)\left(2\cdot\left(-3\right)+3\right)}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{18}{0\cdot\left(-3\right)}=-\infty\)
BD
20 tháng 8 2023
Sửa đề
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)
\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)
\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)
\(=\dfrac{2}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6-2x+4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\dfrac{10}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\)