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a: \(\left(2x-3\right)^2=\left|3-2x\right|\)
=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)
=>\(\left(2x-3\right)\left(2x-4\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)
=>\(x^2-2x+1+4x^2-4x+1=0\)
=>\(5x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)
=>Phương trình vô nghiệm
c: ĐKXĐ: x>=0
\(x-2\sqrt{x}=0\)
=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)
=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)
mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)
nên \(x\in\varnothing\)
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
a. \(2\left(x-\dfrac{1}{3}\right)+3x=\dfrac{7}{4}\)
=> \(2x-\dfrac{2}{3}+3x=\dfrac{7}{4}\)
=> \(2x+3x=\dfrac{7}{4}+\dfrac{2}{3}\)
=> \(5x=\dfrac{29}{12}\)
=> \(x=\dfrac{29}{60}\)
Câu b và c vế 2 mk ko biết dấu / của bạn là chia hay phân số
#Yiin
ý (h) sai đầu bài .
k, \(\left(x+1\right)^3+27=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
\(\Leftrightarrow x=-4\)
m, \(\left(3x+\frac{1}{2}\right)^3+\frac{1}{27}=0\)
\(\Leftrightarrow\left(3x+\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow3x+\frac{1}{2}=-\frac{1}{3}\)
\(\Leftrightarrow3x=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{18}\)
i, \(x^3-x=0\)
\(\Leftrightarrow x.\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
n, \(x^2+\frac{1}{2}x=0\)
\(\Leftrightarrow x.\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
a) (2x-1)2 - 1 = 35
=> (2x - 1)2 = 36
=> 2x - 1 = 6 hoặc 2x - 1 = -6
=> 2x = 7 hoặc 2x = -5
=> x = 7/2 hoặc x = -5/2
vậy_
ban tra loi nhanh gium