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13 sao trừ được 18 bạn xem có sai đề ko
nếu sai đề gửi đề mới mình giải cho
a, ( 19X + 2 x 52 ) : 14 = ( 13 - 18 )2 - 42
( 19X + 2 x 52 ) : 14 = ( -5 )2 - 42
( 19X + 2 x 52 ) : 14 = 25 - 16
( 19X + 2 x 25 ) : 14 = 9
( 19X + 2 x 25 ) = 9 . 14
19X + 2 x 25 = 126
19X + 50 = 126
19X = 126 - 50
19X = 76
X = 76 : 19 = 4
b, 2 x 3X = 10 x 312 + 8 x 274
2 x 3X = 10 . 531441 + 8 . 531441
2 x 3X = 2 x 3X = 531441( 10+8)
2 x 3X = 531441 . 18 = 9565938
3x = 9565938 : 2 = 4782969
ma 3^( dungf mt ra tips keets quar )
Hok tot
\(2.x^3=10.3^{12}+27^4.8\)
\(2.x^3=2.5.3^{12}+3^{12}.2^3\)
\(2.x^3=2.3^{12}.\left(5+4\right)\)
\(x^3=3^{12}.9\)
\(x^3=3^{14}\)
2. x\(^3\)= 10. 3\(^{12}\)+ 27\(^4\). 8.
2. x\(^3\)= 10. 3\(^{12}\)+ 3\(^{3^4}\). 8.
2. x\(^3\)= 10. 3\(^{12}\)+ 3\(^{12}\). 8.
2. x\(^3\)= 3\(^{12}\)( 10+ 8).
2. x\(^3\)= 3\(^{12}\). 18.
2. x\(^3\)= 531441x 18.
2. x\(^3\)= 9565938.
x\(^3\)= 9565938: 2.
x\(^3\)= 4782969.
Vậy x\(^3\)= 4782969.
a) ( 19.x + 2.52) : 14 = ( 13 - 8 )2 - 42
(19.x + 2.52) : 14 = 52 - 42
(19.x + 2.25) : 14 = 25 - 16
(19.x + 2.25) : 14 = 9
19.x + 2.25 = 9.14
19.x + 50 = 126
19.x = 126 - 50
19.x = 76
x = 76 : 19
x = 4
Vậy x = 4.
b) mk thấy x ko có já trị nào hết
c) (x - 1)10 = 1
=> (x - 1)10 = 110
=> x - 1 = 1
x = 1 + 1
x = 2
vậy x = 2
\(\left|x-2\right|=2x-8\)
\(=>\orbr{\begin{cases}x-2=2x-8\\-x+2=2x-8\end{cases}}\)
\(=>\orbr{\begin{cases}x-2x=-8+2\\-x-2x=-8-2\end{cases}}\)
\(=>\orbr{\begin{cases}-x=-6\\-3x=-10\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=\frac{10}{3}\end{cases}}\)
1) x - 36 + 12 = - x+ 10
=> x + x = 10 + 24
=> 2x = 34
=> x = 34/2 = 17
2) (x + 15) - (11 - x) = (-2)2
=> x + 15 - 11 + x = 4
=> 2x = 4 - 4
=> 2x = 0
=> x = 0
3) 40 - 4x2 = (-6)2
=> 40 - 4x2 = 36
=> 4x2 = 40 - 36
=> 4x2 = 4
=> x2 = 1
=> x = \(\pm\)1
4) (-50) + 10x2 = (-25) x |-2|
=> -50 + 10x2 = -50
=> 10x2 = -50 + 50
=> 10x2 = 0
=> x2 = 0
=> x = 0
5) |x + 1| = 2020
=> \(\orbr{\begin{cases}x+1=2020\\x+1=-2020\end{cases}}\)
=> \(\orbr{\begin{cases}x=2019\\x=-2021\end{cases}}\)
6) (x + 1)5 + 8 = 0 (xem lại đề)
7) (-20) + x3 : 16 = -24
=> x3 : 16 = -24 + 20
=> x3 : 16 = -4
=> x3 = -4 . 16
=> x3 = -64 = (-4)3
=> x = -4
9) x14 = x17
=> x14 - x17 = 0
=> x14(1 - x3) = 0
=> \(\orbr{\begin{cases}x^{14}=0\\1-x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
10) (-36) + (1 - x)2 = 0
=> (1 - x)2 = 36
=> (1 - x)2 = 62
=> \(\orbr{\begin{cases}1-x=6\\1-x=-6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\left(x-2019\right)^{x-8}=\left(x-2019\right)^{x-10}\)
\(\Leftrightarrow\frac{\left(x-2019\right)^{x-8}}{\left(x-2019\right)^{x-10}}=1\)
\(\Leftrightarrow\frac{\left(x-2019\right)^x:\left(x-2019\right)^8}{\left(x-2019\right)^x:\left(x-2019\right)^{10}}=1\)
\(\Leftrightarrow\frac{\left(x-2019\right)^x.\frac{1}{\left(x-2019\right)^8}}{\left(x-2019\right)^x.\frac{1}{\left(x-2019\right)^{10}}}=1\)
\(\Leftrightarrow\frac{\frac{1}{\left(x-2019\right)^8}}{\frac{1}{\left(x-2019\right)^{10}}}=1\)
\(\Leftrightarrow\frac{1}{\left(x-2019\right)^8}.\left(x-2019\right)^{10}=1\)
\(\Leftrightarrow\frac{\left(x-2019\right)^{10}}{\left(x-2019\right)^8}=1\)
\(\Leftrightarrow\left(x-2019\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2019=1\\x-2019=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=2018\end{cases}}}\)
Vậy...
a) 52 . x = 62 + 82
\(5^2\cdot x=36+64\)
\(5^2\cdot x=100\)
\(x=100\div5^2\)
\(x=100\div25\)
\(x=4\)
b) ( 22 + 42 ) . x + 24 . 5 . x = 102
\(\left(4+16\right)\cdot x+16\cdot5\cdot x=100\)
\(x\cdot\left(20+80\right)=100\)
\(x\cdot100=100\)
\(x=100\div100\)
\(x=1\)
c ) 24 . x = 26
\(x=2^6\div2^4\)
\(x=2^{6-4}\)
\(x=2^2\)
\(x=4\)
d) 33 . x + 23 . x = 102
\(x\cdot\left(23+27\right)=100\)
\(x\cdot50=100\)
\(x=100\div50\)
\(x=2\)
e) 78 . x = 710
\(x=7^{10}\div7^8\)
\(x=7^{10-8}\)
\(x=7^2\)
\(x=49\)
x^10 = 4.x^8
=> x^10 - 4.x^8 = 0
x^8.(x^2-4) = 0
=> x ^8 = 0 => x = 0
x^2 - 4 = 0 => x^2 = 4 => x = 2 hoặc x = -2
KL: x = 0 hoặc x = 2 hoặc x = -2