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4 tháng 11 2017
=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
B)=>(3x-1).(5x-34)=(40-5x).(25-3x)
=>15x2-102x-5x+34=1000-120x-125x+15x2
=>15x2-107x+34=1000-245x+15x2
=>15x2-15x2-107x+245x=1000-34
=>0-107x+245x=966
=>138x=966
=>x=7
12 tháng 8 2018
A,=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
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: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
30 tháng 8 2019
a) 2x.3/5 - 5x = 3/2 - 7x
6x/5 - 5x = 3/2 - 7x
-19x/5 = 3/2 - 7x
-19x = 15/2 - 35x
-19x + 35x = 15/2
16x = 15/2
x = 15/2 : 16
x = 15/32
b) (x - 1/5)2 = 4/25
(x - 1/5)2 = (+-2/5)2
x - 1/5 = +-2/5
x - 1/5 = 2/5 hoặc x - 1/5 = -2/5
x = 2/5 + 1/5 x = -2/5 + 1/5
x = 3/5 x = -1/5
20 tháng 7 2021
a,
\(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
b,
\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)
c,
S
15 tháng 8 2018
a, \(2x\left(x-5\right)-x\left(2x+3\right)=25\)
\(\Rightarrow2x^2-10x-2x^2-3x=25\)
\(\Rightarrow-13x=25\Rightarrow x=\dfrac{-25}{13}\)
b, \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\dfrac{5}{2}\)
\(\Rightarrow3y^3-y^2+y-3y^2+y-1+4y^2-3y^3=\dfrac{5}{2}\)
\(\Rightarrow2y-1=\dfrac{5}{2}\Rightarrow2y=\dfrac{7}{2}\Rightarrow y=\dfrac{7}{4}\)
c, \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2+x-x-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow-5x=3\Rightarrow x=\dfrac{-3}{5}\)