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17 tháng 12 2022
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5
4 tháng 10 2021
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
28 tháng 10 2023
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
24 tháng 12 2023
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 \(x\in\varnothing\)
TN
1
LP
17 tháng 8 2021
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
21 tháng 7 2017
\(2x+\frac{1}{2}=\frac{-5}{3}\)
\(2x=\frac{-5}{3}-\frac{1}{2}\)
\(2x=\frac{-10}{6}-\frac{3}{6}\)
\(2x=\frac{-13}{6}\)
\(x=\frac{-13}{6}:2\)
\(x=\frac{-13}{12}\)
b) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
d) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(\Leftrightarrow x=\dfrac{1}{4}:\left(-\dfrac{7}{20}\right)=-\dfrac{5}{7}\)