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HH
8 tháng 7 2018
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
NT
2
23 tháng 6 2017
a) \(3\left(x-1\right)+2x-2x^2=0\)
\(\Leftrightarrow3x-3+2x-2x^2=0\)
\(\Leftrightarrow-2x^2+5x-3=0\)
\(\Leftrightarrow-2x^2+2x+3x-3=0\)
\(\Leftrightarrow-2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-2x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy..
b) \(x^2+8x+15=0\)
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy..
23 tháng 6 2017
Tìm x :
a) 3(x - 1 ) + 2x - 2x2 = 0
\(\Leftrightarrow3\left(x-1\right)-2x^2+2x=0\)
\(\Leftrightarrow\) 3\(\left(x-1\right)-2x\left(x-1\right)=0\)
\(\Leftrightarrow\) (x - 1 )( 3-2x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-2x=-3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy....
b) x2 + 8x + 15 = 0
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow\) (x2 + 3x ) + ( 5x + 15 ) =0
\(\Leftrightarrow x\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy....
18 tháng 12 2023
a: \(A=\left[\left(\dfrac{4x}{x+2}+\dfrac{8x^2}{4-x^2}\right)\right]:\left[\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right]\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{8x^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-1}{x\left(x-2\right)}-\dfrac{2}{x}\right)\)
\(=\dfrac{4x\left(x-2\right)-8x^2}{\left(x+2\right)\left(x-2\right)}:\dfrac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{-8x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-2\right)}{x-1-2x+4}\)
\(=\dfrac{-8x^2}{\left(x+2\right)\cdot\left(-x+3\right)}\)
\(=\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}\)
b: \(x^2+2x=15\)
=>\(x^2+2x-15=0\)
=>(x+5)(x-3)=0
=>\(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Thay x=-5 vào A, ta được:
\(A=\dfrac{8\cdot\left(-5\right)^2}{\left(-5-3\right)\left(-5+2\right)}=\dfrac{8\cdot25}{\left(-8\right)\cdot\left(-3\right)}=\dfrac{25}{3}\)
c: |A|>A
=>A<0
=>\(\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}< 0\)
=>(x-3)(x+2)<0
TH1: \(\left\{{}\begin{matrix}x-3>0\\x+2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 3\\x>-2\end{matrix}\right.\)
=>-2<x<3
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-2< x< 3\\x\notin\left\{0;2\right\}\end{matrix}\right.\)
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
NV
31 tháng 8 2015
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
2 tháng 1 2016
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
15 tháng 3 2020
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
15 tháng 3 2020
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
28 tháng 5 2022
Bài 2:
a: \(A=x^2+8x\)
\(=x^2+8x+16-16\)
\(=\left(x+4\right)^2-16\ge-16\)
Dấu '=' xảy ra khi x=-4
b: \(B=-2x^2+8x-15\)
\(=-2\left(x^2-4x+\dfrac{15}{2}\right)\)
\(=-2\left(x^2-4x+4+\dfrac{7}{2}\right)\)
\(=-2\left(x-2\right)^2-7\le-7\)
Dấu '=' xảy ra khi x=2
c: \(C=x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\ge3\)
Dấu '=' xảy ra khi x=2
e: \(E=x^2-6x+y^2-2y+12\)
\(=x^2-6x+9+y^2-2y+1+2\)
\(=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=3 và y=1
1) \(2x^3-8x=0\)
\(\Leftrightarrow2x\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
Vậy \(x\in\left\{0;\pm2\right\}\)
2) \(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=15\end{cases}}\)
Vậy \(x\in\left\{2;15\right\}\)
1
\(2x^3-8x=0\)
\(2x\left(x^2-4\right)=0\)
\(\orbr{\begin{cases}2x=0\\x^2-4=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
2
\(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(\left(2x-4\right)\left(x-15\right)=0\)
\(\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\x=0+15\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=15\end{cases}}\)