\(^{3^{2x+1}=5^{x-2}}\) giải pt mũ sau:
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C
2
5 tháng 2 2018
\(\left(2x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)
\(2\left(x+2\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)
\(\left(x+2\right)\left(2x-6-x+4\right)=x\left(x+5\right)\)
\(\left(x+2\right)\left(x-2\right)-x^2-5x=0\)
\(x^2-2x+2x-4-x^2-5x=0\)
\(-5x-4=0\)
\(-5x=4\)
\(\Rightarrow\)\(x=\frac{-4}{5}\)
\(\left(x-2\right)^2=\left(2x-4\right)\left(x+5\right)\)
\(\left(x-2\right)^2-2\left(x-2\right)\left(x+5\right)=0\)
\(\left(x-2\right)\left(x-2-2x-10\right)=0\)
\(\left(x-2\right)\left(-x-12\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\-x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}}\)
Bạn tự kết luận 2 câu nhé
22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
PH
1
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, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
22 tháng 4 2022
Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....
DQ
4
4 tháng 2 2018
\(\left(2x-2\right)^2=\left(x+1\right)^2+3.\left(x-2\right)\left(x+5\right)\)
\(\left(2x-2\right)^2-\left(x+1\right)^2=3.\left(x-2\right)\left(x+5\right)\)
\(\left(2x-2-x-1\right)\left(2x-2+x+1\right)=3.\left(x-2\right)\left(x+5\right)\)
\(\left(x-3\right)\left(3x-1\right)=3.\left(x-2\right)\left(x+5\right)\)
\(3x^2-x-9x+3=\left(3x-6\right)\left(x+5\right)\)
\(3x^2-10x+3=3x^2+15x-6x-30\)
\(3x^2-3x^2-10x+6x-15x+3+30=0\)
\(-19x+33=0\)
\(-19x=-33\)
\(x=\frac{33}{19}\)
D
4 tháng 2 2018
\(\Leftrightarrow\left(2x-2\right)^2-\left(x+1\right)^2-3.\left(x-2\right).\left(x+5\right)=0\)
\(\Leftrightarrow4x^2-8x+4-\left(x^2+2x+1\right)-\left(3x-6\right).\left(x+5\right)=0\)
\(\Leftrightarrow4x^2-8x+4-x^2-2x-1-\left(3x^2+15x-6x-30\right)=0\)
\(\Leftrightarrow4x^2-8x+4-x^2-2x-1-3x^2-15x+6x+30=0\)
\(\Leftrightarrow-19x+33=0\)
\(\Leftrightarrow-19x=-33\)
\(\Leftrightarrow x=\frac{33}{19}\)
Vậy...............