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2x+3=7
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HN
9 tháng 10 2021
\((2x-7)^3=8(7-2x)^2\)
⇔ \((2x-7)^3=8(2x-7)^2\) (*)
\(TH1: (2x-7)^2=0\)
Khi đó: \(2x-7=0\) ⇔ \(x=\dfrac{7}{2} \)
\(TH2:\left(2x-7\right)^2\ne0\)
Khi đó: (*) ⇔ \(2x-7=8\) (chia 2 vế cho \((2x-7)^2\))
⇔ \(x=\dfrac{15}{2} \)
Vậy \(x=\dfrac{15}{2}\); \(x=\dfrac{7}{2}\)
7 tháng 11 2021
\(\Leftrightarrow\left(2x-3\right)^7\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
20 tháng 9 2021
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
25 tháng 6 2021
`(2x+5)(2x-7)-(2x-3)^2=36`
`<=>4x^2-14x+10x-35-(4x^2-12x+9)=36`
`<=>4x^2-4x-35-4x^2+12x-9=36`
`<=>8x-44=36`
`<=>8x=80`
`<=>x=10`
Vậy `S={10}`
25 tháng 6 2021
Ta có: \(\left(2x+5\right)\left(2x-7\right)-\left(2x-3\right)^2=36\)
\(\Leftrightarrow4x^2-14x+10x-35-\left(4x^2-12x+9\right)=36\)
\(\Leftrightarrow4x^2-4x-35-4x^2+12x-9=36\)
\(\Leftrightarrow8x-44=36\)
\(\Leftrightarrow8x=80\)
hay x=10
Vậy: S={10}
11 tháng 9 2016
a ) \(\left|2x-3\right|=\left|x+5\right|\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=x+5\\2x-3=-x+5\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-x=5+3\\2x+x=5+3\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\3x=8\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=\frac{8}{3}\left(loại\right)\end{array}\right.\)
Vậy ...........
b ) \(\left|x\right|+2x=7\)
\(\Rightarrow\left[\begin{array}{nghiempt}-x+2x=7\\x+2x=7\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=7\\3x=7\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=7\left(loại\right)\\x=\frac{7}{3}\end{array}\right.\)
Vậy .........................
c ) \(\left|x\right|+2x=7\) ( giống câu b )
NT
0
NB
5
27 tháng 7 2017
x^2 -2x = 24
=> x^2 - 2x - 24=0
=>x^2 -8x+6x - 24 = 0
=> ( x^2- 8x)+( 6x-24) = 0
=> x(x-8) + 6(x-8) = 0
=> (x+6)(x-8)=0
=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)
SH
4
30 tháng 9 2016
\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)
\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)
\(\Rightarrow7^x=25^x\Rightarrow x=0\)
\(2x+3=7\)
\(2x=7-3\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
Vậy x = 2
2X+3=7
2x=7-3
2x=4
x=4:2
x=2