1/ \((2x-1)^2-3(2x-1)^2=0\)
2/ \((x-1)^2(x+1)=x+1\)
3/ \(x^4-3x^2=x^2\)
Mọi người giúp em với ạ. Em cảm ơn.
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28 tháng 7 2019
Bài Làm:
\(1,\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
\(\Leftrightarrow-2\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy ...
\(2,\left(x-1\right)^2\left(x+1\right)=x+1\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2-2x+1-1\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)
Vậy ...
\(3,x^4-3x^2=x^2\)
\(\Leftrightarrow x^4-3x^2-x^2=0\)
\(\Leftrightarrow x^4-4x^2=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc pạn hok tốt!!!
30 tháng 7 2023
a: =>6x-3x^2-5=4-3x^2-2
=>6x-5=2
=>6x=7
=>x=7/6
b: =>20x+5-12x^2-3x=6x^2-10x+3x-5
=>-12x^2+17x+5-6x^2+7x+5=0
=>-18x^2+24x+10=0
=>x=5/3 hoặc x=-1/3
30 tháng 7 2023
a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)
=>3(3x+2)-4(3x+1)=10
=>9x+6-12x-4=10
=>-3x+2=10
=>-3x=8
=>x=-8/3
b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)
=>(x-1)(x-2)-x(x+2)=-9x+10
=>x^2-3x+2-x^2-2x=-9x+10
=>-5x+2=-9x+10
=>x=2(loại)
H
18 tháng 10 2023
\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
DT
13
VN
Võ Ngọc Phương
VIP
26 tháng 8 2023
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
26 tháng 8 2023
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
1/ \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2\left(1-3\right)=0\)
\(\left(2x-1\right)^2\cdot\left(-2\right)=0\)
\(\Rightarrow\text{ }\left(2x-1\right)^2=0\)
\(2x-1=0\)
\(2x=0+1=1\)
\(x=\frac{1}{2}\)
1) \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
=> \(\left(2x-1\right)^2\left(1-3\right)=0\)
=> \(\left(2x-1\right)^2.\left(-2\right)=0\)
=> \(\left(2x-1\right)^2=0\)
=> \(2x-1=0\)
=> \(2x=1\)
=> \(x=1:2=\frac{1}{2}\)