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\(E-2\overline{yzt}=\overline{xz}\)
=>1000x+100y+10z+t-200y-20z-20t=10x+z
=>990x-100y-11z-19t=0
=>\(\left(x,y,z,t\right)\in\varnothing\)
\(\left(2x-133\right):3=33\\ 2x-133=99\\ 2x=232\\ x=116\)
12 : {390 : [500 - (125 + 35 . 7)]}
= 12 : {390 : [500 - (125 + 245)]}
= 12 : [390 : (500 - 370)]
= 12 : (390 : 130)
= 12 : 3
= 4
12 : { 390 : [ 500 - ( 125 + 35. 7 ) ] }
= 12 : { 390 : [ 500 - ( 125 + 245 ) ] }
= 12 : { 390 : [ 500 - 370 ] }
= 12 : { 390 : 130 }
= 12 : 3
= 4
\(x^2-30=34\)
\(x^2=34+30\)
\(x^2=64=8^2=\left(-8\right)^2\)
Vậy \(x=8^2\) hoặc \(x=\left(-8\right)^2\)
tạm thời cái đề bài là tìm x vậy
tiếp tục thui
=) (6x-10)-(6x-3)\(⋮\)2x-1
=)6x-10-6x+3\(⋮\)2x-1
=) (6x-6x)-(10-3)\(⋮\)2x-1
=)0-7\(⋮\)2x-1
=)-7\(⋮\)2x-1=)2x-1\(\in\)Ư(-7)={-7;-1;1;7}
=)2x\(\in\){-6;0;2;8}
=)x\(\in\){-3;0;1;4}
\((3x-2)-2^5\times9=7^2\\\Rightarrow (3x-2)-32\times9=49\\\Rightarrow (3x-2)-288=49\\\Rightarrow 3x-2=49+288\\\Rightarrow 3x-2=337\\\Rightarrow 3x=337+2\\\Rightarrow 3x=339\\\Rightarrow x=\dfrac{339}{3}\\\Rightarrow x=113\)
#\(Toru\)
(3.x - 2) - 2⁵ × 9=7²
=>(3.x-2)-32 x 9=49
=>(3.x-2)-288=49
=>3.x-2=337
=>3.x=339
=>x=113
theo đề bài ta có:(5n-2) chia hết cho (2n+1)
(2n+1) chia hết cho (2n+1)
suy ra:{[5(2n+1)]-[2(5n-2)]} chia hết cho (2n+1)
hay 9 chia hết cho (2n+1)
suy ra:2n+1 e Ư(9)
Ư(9)={1;3;9)
2n+1=1 thì n=0
2n+1=3 thì n=1
2n+1=9 thì n=4
vậy n e {0;1;4}
a: Ta có: \(\left(x-47\right)-115=0\)
\(\Leftrightarrow x-47=115\)
hay x=162
b: Ta có: \(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
\(\Leftrightarrow7x-11=10\)
\(\Leftrightarrow7x=21\)
hay x=3
c: Ta có: \(x^{10}=1^x\)
\(\Leftrightarrow x^{10}=1\)
hay \(x\in\left\{1;-1\right\}\)
d: Ta có: \(x^{10}=x\)
\(\Leftrightarrow x^{10}-x=0\)
\(\Leftrightarrow x\left(x^9-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e: Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
g: Ta có: \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot18=3^{14}\cdot2\)
Suy ra: \(3^x=3^{14}\)
hay x=14