5x+3 + 5 x = 126 .25
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LN
1
1 tháng 1 2022
\(a,\Rightarrow\dfrac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\\ \Rightarrow\left(-3\right)^{x-4}=\left(-3\right)^3\\ \Rightarrow x-4=3\Rightarrow x=7\\ b,Sửa:\left(x-\dfrac{1}{2}\right)^2=25\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=5\\x-\dfrac{1}{2}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{5}\\x=-\dfrac{9}{5}\end{matrix}\right.\)
11 tháng 10 2021
\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
AN
1
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
a) \(5^{x+3}+5^x=126\)
\(\Rightarrow5^x\cdot\left(5^3+1\right)=126\)
\(\Rightarrow5^x\cdot\left(125+1\right)=126\)
\(\Rightarrow5^x=\dfrac{126}{126}\)
\(\Rightarrow5^x=1\)
\(\Rightarrow5^x=5^0\)
\(\Rightarrow x=0\)
b) \(\dfrac{9-3x}{2}=\dfrac{5-2x}{3}\)
\(\Rightarrow\dfrac{3\cdot\left(9-3x\right)}{6}=\dfrac{2\cdot\left(5-2x\right)}{6}\)
\(\Rightarrow3\cdot\left(9-3x\right)=2\cdot\left(5-2x\right)\)
\(\Rightarrow27-9x=10-4x\)
\(\Rightarrow-4x+9x=27-10\)
\(\Rightarrow5x=17\)
\(\Rightarrow x=\dfrac{17}{5}\)
c) \(7-4\left(x+1\right)=5\)
\(\Rightarrow4\left(x+1\right)=7-5\)
\(\Rightarrow4\left(x+1\right)=2\)
\(\Rightarrow x+1=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
H
11 tháng 4 2023
\(\dfrac{2}{5}:\dfrac{6}{25}\\ =\dfrac{2}{5}\times\dfrac{25}{6}\\ =\dfrac{2}{5}\times\dfrac{5\times5}{2\times3}=\dfrac{5}{3}\)
11 tháng 4 2023
\(\dfrac{2}{5}:\dfrac{6}{25}\)
\(=\dfrac{2}{5}\times\dfrac{25}{6}\)
\(=\dfrac{2}{5}\times\dfrac{5\times5}{2\times3}\)
\(=\dfrac{5}{3}\)
TN
10 tháng 6 2017
a)\(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)
\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)
10 tháng 6 2017
a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)
\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)
\(33-39x=0\)
\(3\left(11-13x\right)=0\)
\(11-13x=0\)
\(13x=11\)
\(x=\frac{11}{13}\)
T
10 tháng 3 2019
b)\(\frac{2x-1}{x-1}+\frac{x}{\left(x-1\right)\left(x-2\right)}=\frac{6x-2}{x-2}\)
\(< =>\left(2x-1\right)\left(x-2\right)+x-\left(6x-2\right)\left(x-1\right)=0\)
\(< =>2x^2-x-4x+2+x-\left(6x^2-2x-6x+2\right)=0\)
\(< =>2x^2-4x+2-6x^2+8x-2=0\)
\(< =>-4x^2+4x=0\)
\(< =>-4x\left(x+1\right)=0\)
\(TH1:-4x=0< =>x=0\) \(TH2:x+1=0< =>x=-1\)
\(S=\left\{-1;0\right\}\)
TM
20 tháng 7 2023
a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)
NV
20 tháng 7 2023
1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)
2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)
3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)
5
Bài giải
5x+3 +5x = 126.25
5x+53+5x=126.25
5x(53+1)=126.25
5x.126=126.25
Vậy 5x=25=52
x=2
\(5^{x+3}+5^x=126.25\)
\(5^x.5^3+5^x=3150\)
\(5^x.\left(5^3+1\right)=3150\)
\(5^x.126=3150\)
\(5^x=3150:126\)
\(5^x=25\)
\(5^x=+-\left(5\right)^2\)
\(\Leftrightarrow x=2\)