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\(125-2x=25\\ 2x=125-25\\ 2x=100\\ x=50\\ 100-\left(x+15\right)=27\\ x+15=100-27\\ x+15=73\\ x=73-15\\ x=58\)
\(125-2x=25\Leftrightarrow-2x=25-125\Leftrightarrow-2x=-100\Leftrightarrow x=\dfrac{-100}{-2}=50\)
\(100-\left(x+15\right)=27\Leftrightarrow-\left(x+15\right)=27-100\Leftrightarrow-\left(x+15\right)=-73\Leftrightarrow x+15=73\Leftrightarrow x=73-15\Leftrightarrow x=58\)
\(\left(4x-1\right)^3=27\Leftrightarrow4x-1=3\Leftrightarrow4x=4\Leftrightarrow x=1\)
\(\left(4x-1\right)^3=27\\ \Leftrightarrow\left(4x-1\right)^3=3^3\\ \Leftrightarrow4x-1=3\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
\(\left(2x+1\right)^3=-125\)
\(< =>\left(2x+1\right)^3=\left(-5\right)^3\)
\(< =>2x+1=-5\)
\(< =>2x=-5-1=-6\)
\(< =>x=-\frac{6}{2}=-3\)
Bài làm:
a) \(\left(2x+1\right)^3=-125\)
\(\Leftrightarrow\left(2x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow2x+1=-5\)
\(\Leftrightarrow2x=-6\)
\(\Rightarrow x=-3\)
b) \(\left(7-x\right)^2-\left(-11\right)=15\)
\(\Leftrightarrow\left(7-x\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=9\end{cases}}\)
c) \(2020^x-2019=-2019\)
\(\Leftrightarrow2020^x=0\)
=> ko tồn tại x thỏa mãn PT
( 2x + 1 ) \(^3\)= 125
( 2x + 1 ) = \(\sqrt[3]{125}\)
( 2x + 1 ) = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
\(\hept{\begin{cases}3|x|+15=0\\16-x^2=0\\x^3+125=0\end{cases}}\)
\(\hept{\begin{cases}3x+15=0\\16-x^2=0\\x^3+125=0\end{cases}}\)
\(\hept{\begin{cases}3x=0-15\\x^2=16-0\\x^3=0-125\end{cases}}\)
\(\hept{\begin{cases}3x=-15\\x^2=16\\x^3=-125\end{cases}}\)
\(\hept{\begin{cases}x=\left(-15\right):3\\x^2=4^2=\left(-4\right)^2\\x^3=-5^3\end{cases}}\)
\(\hept{\begin{cases}x=-5\\x=4=-4\\x=-5\end{cases}}\)
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
a) 2x . 4 = 128
2x = 128 : 4 = 32
2x = 25
=> x = 5
x15 = x
=> x = 0 hoặc x = 1 hoặc x = -1
c) (2x + 1)3 = 125 = 53
2x + 1 = 5
2x = 4
x = 2
d) (x-5)4 = (x-5)6
Th1: x - 5 = 0
=> x = 5
th2: x - 5 = 1
x = 6
Vậy x = 5 hoặc 6
a) \(132,5-2x=29,9\)
<=>\(2x=132,5-29,9\)
<=>\(2x=102,6\)
<=>\(x=102,6:2\)
<=>\(x=51,3\)
b) \(18-5x=8\)
<=>\(5x=18-8\)
<=>\(5x=10\)
<=>\(x=10:5\)
<=>\(x=2\)
c) \(2x+14=26\)
<=>\(2x=26-14\)
<=>\(2x=12\)
<=>\(x=12:2\)
<=>\(x=6\)
d) \(x-105:21=15\)
<=>\(x-5=15\)
<=>\(x=15+5\)
<=>\(x=20\)
e) \(315+\left(125-x\right)=81\)
<=>\(125-x=81-315\)
<=>\(125-x=-234\)
<=>\(x=125--234\)
<=>\(x=125+234\)
<=>\(x=359\)
K CHO MINK NHA
2x=100=>x=50
100-x-15=27=>88 nhé