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30 tháng 8 2021
a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)
TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)
TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
21 tháng 7 2017
Ta có : |1 - 5x| - 1 = 3
=> |1 - 5x| = 4
\(\Leftrightarrow\orbr{\begin{cases}1-5x=4\\1-5x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1-4\\5x=1+4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=1\end{cases}}\)
TN
18 tháng 3 2020
\(a.\frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow\frac{3\left(4x-3\right)}{3\left(x-5\right)}=\frac{29\left(x-5\right)}{3\left(x-5\right)}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow3\left(4x-3\right)-29\left(x-5\right)=0\\ \Leftrightarrow12x-9-29x+145=0\\ \Leftrightarrow-17x+136=0\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=\frac{-136}{-17}=8\)
\(b.\frac{2x-1}{5-3x}=2\\ \Leftrightarrow\frac{2x-1}{5-3x}=\frac{4}{2}\\ \Leftrightarrow\frac{2\left(2x-1\right)}{2\left(5-3x\right)}=\frac{4\left(5-3x\right)}{2\left(5-3x\right)}\\ \Leftrightarrow2\left(2x-1\right)=4\left(5-3x\right)\\ \Leftrightarrow2\left(2x-1\right)-4\left(5-3x\right)=0\\ \Leftrightarrow4x-2-20+12x=0\\ \Leftrightarrow16x-22=0\\ \Leftrightarrow16x=22\\ \Leftrightarrow x=\frac{22}{16}=\frac{11}{8}\)
\(c.\frac{4x-5}{x-1}=\frac{2+x}{x-1}\\ \Leftrightarrow4x-5=2+x\\ \Leftrightarrow4x-5-2-x=0\\ \Leftrightarrow3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\frac{7}{3}\)
TN
18 tháng 3 2020
\(d.\frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7\left(x-5\right)-3\left(x+2\right)=0\\ \Leftrightarrow7x-35-3x-6=0\\ \Leftrightarrow4x-41=0\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)
\(e.\frac{2x+5}{2x}-\frac{x}{x+5}=0\\ \Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x.2x}{2x\left(x+5\right)}=0\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=0\\ \Leftrightarrow2x^2+10x+5x+25-2x^2=0\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow15x=-25\\ \Leftrightarrow x=\frac{-25}{15}=\frac{-5}{3}\)
\(f.\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\\\Leftrightarrow\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{\left(10x-4\right).2\left(11x-4\right)}{9.2\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\\ \Leftrightarrow18\left(12x+1\right)+\left(10x-4\right).2\left(11x-4\right)=\left(20x+17\right)\left(11x-4\right)\\ \Leftrightarrow220x^2+48x+50=220x^2+107x-68\\ \Leftrightarrow48x+50=107x-68\\ \Leftrightarrow48x-107x=-68-50\\ \Leftrightarrow59x=-118\\ \Leftrightarrow x=-2\)
PT
6 tháng 2 2019
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
14 tháng 2 2023
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
18 tháng 4 2017
câu 1:
a)x-1=5-x\(\Leftrightarrow\)x+x=5+1\(\Leftrightarrow\)2x=6\(\Leftrightarrow\)x=3
Vậy tập nghiệm của PT (a) là S={3}
b)3+x=2-x\(\Leftrightarrow\)x+x=2-3\(\Leftrightarrow\)2x=-1\(\Leftrightarrow\)x=-0,5
Vậy tập nghiệm của PT (b) là:S={-0,5}
câu 2:
a) 3x+7=2x-3\(\Leftrightarrow\)3x-2x=-3-7\(\Leftrightarrow\)x=-10
Vậy tập nghiệm của PT (a) là:S={-10}
b)4-(x-2)=(3-2x)\(\Leftrightarrow\)4-x+2=3-2x\(\Leftrightarrow\)-x+2x=-4+3-2\(\Leftrightarrow\)x=-3
Vậy tập nghiệm của PT (b) là:S={-3}
Câu 3:
a)\(\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\Leftrightarrow\dfrac{7\left(5x-4\right)}{14}=\dfrac{2\left(16x+1\right)}{14}\)
\(\Leftrightarrow\)35x-28=32x+2\(\Leftrightarrow\)35x-32x=2+28\(\Leftrightarrow\)3x=30\(\Leftrightarrow\)x=10
Vậy tập nghiệm của PT (a) là :S={10}
b)\(\dfrac{12x+5}{3}=\dfrac{2x-7}{4}\Leftrightarrow\dfrac{4\left(12x+5\right)}{12}=\dfrac{3\left(2x-7\right)}{12}\)
\(\Leftrightarrow\)48x+20=6x-21\(\Leftrightarrow\)48x-6x=-20-21\(\Leftrightarrow\)42x=-41\(\Leftrightarrow\)x=\(-\dfrac{41}{42}\)
Vậy tập nghiệm của PT (b) là:S={\(-\dfrac{41}{42}\)}
8 tháng 5 2017
Đặt t=x2-2x+3(t\(\ge\)2)
PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)
<=>2t2+2t+2t2-2=9t2-9
<=>5t2-2t-7=0
<=>(t+1)(5t-7)=0
Do t\(\ge\)2
=>t+1>0 5t-7>0
Vậy pt vô nghiệm
HN
9 tháng 5 2017
\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)
Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)
Thì ta có:
\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)
\(\Leftrightarrow5t^2-t-4=0\)
\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2=1\)
\(\Leftrightarrow x=1\)
Vậy PT có 1 nghiệm là x = 1
TT
5
N
1 tháng 7 2018
\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow\left(4x-4\right)\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy ...
1 tháng 7 2018
a) \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x+7\right)^2-\left[3\left(x+2\right)\right]^2=0\)
\(\Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(2x+7-3x-6\right)\left(2x+7+3x+6\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\5x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b)\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+6\right)\left(x+2+3x-6\right)=0\)
\(\Leftrightarrow\left(8-2x\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8-2x=0\\4x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(4-x\right)=0\\4\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left(4x+14\right)^2=\left(3x+9\right)^2\)
\(\Leftrightarrow\left(4x+14+3x+9\right)\cdot\left(4x+14-3x-9\right)=0\)
\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)
hay \(x\in\left\{-\dfrac{23}{7};-5\right\}\)
\(\Leftrightarrow\left(8x+14\right)^2=\left(9x+3\right)^2\)
\(\Leftrightarrow8x+14=9x+3\)
\(\Leftrightarrow x=11\)