Giải PT (nhớ tìm ĐKXĐ)
5+96/x2-16=2x-1/x+4+3x-1/x-4
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19 tháng 2 2022
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)
24 tháng 4 2020
Bài 1 :
Để phương trình có 2 nghiệm x1 , x2
\(\Rightarrow\Delta'=\left(-1\right)^2-\left(2m-1\right)\ge0\)
\(\Rightarrow m\le1\)
\(\Rightarrow\) Khi đó phương trình có 2 nghiệm x1 , x2 thỏa mãn
\(\hept{\begin{cases}x_1+x_2=2\\x_1x_2=2m-1\end{cases}}\)
Mà \(3x_1+2x_2=1\Rightarrow x_1+2\left(x_1+x_2\right)=1\Rightarrow x_1+2.2=1\Rightarrow x_1=-3\)
Vì \(x_1=-3\) là 1 nghiệm của phương trình
\(\Rightarrow\left(-3\right)^2-2\left(-3\right)+2m-1=0\Rightarrow m=-7\)
24 tháng 4 2020
Bài 2 :
\(ĐKXĐ:x\ne\pm4\)
Ta có :
\(\frac{2x-1}{x+4}-\frac{3x-1}{4-x}=5+\frac{96}{x^2-16}\)
\(\Rightarrow\frac{2x-1}{x+4}+\frac{3x-1}{x-4}=5+\frac{96}{\left(x-4\right)\left(x+4\right)}\)
\(\Rightarrow\frac{2x-1}{x+4}\left(x+4\right)\left(x-4\right)+\frac{96}{\left(x-4\right)\left(x+4\right)}\left(x+4\right)\left(x-4\right)\)
\(\Rightarrow\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)=5\left(x+4\right)\left(x-4\right)+96\)
\(\Rightarrow5x^2+2x=5x^2+16\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=8\)
N
1 tháng 4 2019
a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
1 tháng 4 2019
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
5 tháng 3 2021
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
TV
2
13 tháng 2 2017
\(\frac{96}{x^2-16}-\frac{2x+1}{x+4}=\frac{1-3x}{4-x}\)
\(\frac{96}{\left(x-4\right)\left(x+4\right)}-\frac{2x+1}{x+4}=\frac{-\left(1-3x\right)}{x-4}\)
DKCD :x \(\ne\)+4
MTC: \(\left(x-4\right)\left(x+4\right)\)
Quy dong va khu mau :
\(\frac{96-\left(2x+1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(1-3x\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
96-(2x2-4x+x-4)=-(x+4-3x2-12x)
................
Tới đây bạn tự tính bình thường ra là được
Y
26 tháng 8 2018
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
19 tháng 2 2022
đk : x khác -4 ; 4
\(\Rightarrow96+\left(1-3x\right)\left(x+4\right)=\left(2x+1\right)\left(x-4\right)-5\left(x^2-16\right)\)
\(\Leftrightarrow96x+x+4-3x^2-12x=2x^2-7x-4-5x^2+80\)
\(\Leftrightarrow92x=72\Leftrightarrow x=\dfrac{72}{92}=\dfrac{18}{23}\)(tm)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
14 tháng 3 2020
a) ĐKXĐ: \(x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> \(5+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> 5(x - 4)(x + 4) + 96(x - 4) = (2x - 1)(x - 4)(4 - x) - (3x - 1)(x + 4)(4 - x)
<=> 20x2 - 16x + 64 = 18x2 + 8x
<=> 20x2 - 16x + 64 - 18x2 - 8x = 0
<=> 2x2 - 24x + 64 = 0
<=> 2(x2 - 12x + 32) = 0
<=> 2(x - 8)(x - 4) = 0
<=> (x - 8)(x - 4) = 0
<=> x - 8 = 0 hoặc x - 4 = 0
<=> x = 8 (tm) hoặc x - 4 = 0 (ktm)
=> x = 8
b) ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-2^2}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
<=> (2 + 3x)2 - 6(3x - 2) = 9x2
<=> 16 - 6x + 9x2 = 9x2
<=> 16 - 6x + 9x2 - 9x2 = 0
<=> 16 - 6x = 0
<=> -6x = 0 - 16
<=> -6x = -16
<=> x = -16/-6 = 8/3
=> x = 8/3
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)ĐKXĐ : \(x\ne\pm4\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2-11x+9x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\)( t/m )
Vậy....