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a: \(\Leftrightarrow\left(x-1\right)\left(x-2\right)\cdot x=0\)
hay \(x\in\left\{0;1;2\right\}\)
b: =>3x-1=11 hoặc 3x-1=-11
=>3x=12 hoặc 3x=-10
=>x=4 hoặc x=-10/3
c: =>2x+1=3
=>2x=2
hay x=1
d: =>x-2=2
hay x=4
e: =>x+2=7
hay x=5
\(a,2x^5+2=4\)=> \(2x^5=2\)=> \(x^5=1\)=> x = 1
\(b,(x-2)^2=144\)
=> \((x-2)^2=12^2\)
=> \(x-2=\pm12\)
=> \(\orbr{\begin{cases}x-2=12\\x-2=-12\end{cases}}\)=> \(\orbr{\begin{cases}x=14\\x=-10\end{cases}}\)
\(c,(2x+1)^2=49\)
=>\((2x+1)^2=7^2\)
=> \(2x+1=\pm7\)
=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\)=> \(\orbr{\begin{cases}2x=6\\2x=-8\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
e, Tương tự
Bài 1 . Tìm x
a) 723 - ( 7x - 152 ) = 714
7x - 152 = 723 - 714
7x - 152 = 9
7x = 9 + 152
7x = 161
x = 161 : 7
x = 23
Vậy x = 23
b) ( 2x - 130 ) : 4 + 213 = 52 + 193
( 2x - 130 ) : 4 + 213 = 218
( 2x - 130 ) : 4 = 218 - 213
( 2x - 130 ) : 4 = 5
2x - 130 = 5 . 4
2x - 130 = 20
2x = 20 + 130
2x = 150
x = 150 : 2
x = 75
Vậy x = 75
c) ( x - 6 )2 = 9
( x - 6 )2 = 32
x - 6 = 3 <=> x = 3 + 6 <=> x = 9
x - 6 = -3 <=> x = -3 + 6 <=> x = 3
a, \(\left(2x+7\right)^4=10^{11}:10^7\)
\(\Rightarrow\left(2x+7\right)^4=10^4\)
\(\Rightarrow2x+7=10\)
\(\Rightarrow2x=10-7\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\) hay \(x=1,5\)
b, \(5^{x-1}.7^{x-1}=25.49\)
\(\Rightarrow\)\(5^{x-1}.7^{x-1}=5^2.7^2\)
\(\Rightarrow\left\{{}\begin{matrix}5^{x-1}=5^2\\7^{x-1}=7^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=2\\x-1=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)
c, \(\left(x-5\right)^{2018}=9.\left(x-5\right)^{2016}\)
\(\Rightarrow\dfrac{\left(x-5\right)^{2018}}{\left(x-5\right)^{2016}}=9.\dfrac{\left(x-5\right)^{2016}}{\left(x-5\right)^{2016}}\)
\(\Rightarrow\left(x-5\right)^2=9\)
\(\Leftrightarrow\left(x-5\right)^2=3^2\)
\(\Rightarrow x-5=3\)
\(\Rightarrow x=3+5\)
\(\Rightarrow x=8\)
a)= \(3^{x-1}=3^4\)
=>\(x-1=4\)
\(x=1+4\)
\(x=5\)
b)
\(3^{2.x}=36\)
\(x=2\)
a) 2x-15=17
2x = 17+15 = 32
x = 32 : 2 =16
b) \(2^x.4=128\)
\(2^x=128:4=32\)
Mà \(32=2^5\)
Vậy x = 5
a) Ta có: \(\left(x-1\right)^5=x-1\)
\(\Rightarrow x-1\in\left\{1;0;-1\right\}\)
Nếu x - 1 = 1 => x = 2
Nếu x - 1 = 0 => x = 1
Nếu x - 1 = -1 => x = 0
b) Ta có: 64 = 43
=> 42x+1 = 43
=> 2x + 1 = 3
=> 2x = 4
=> x = 4 : 2
=> x = 2
c) Ta có: 32 = 25
=> (x-2)5 = 25
=> x - 2 = 2
=> x = 2 +2
=> x = 4