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1) \(\left(-27\right).\left(-28+128\right)=-27.100=-2700\)
2a)\(\left(x-3\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
b) \(\left(2x-1\right)^2=81\)
\(\sqrt{\left(2x-1\right)^2}=9\)
\(\left|2x-1\right|=9\)
\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
c) \(\left(2m+5\right)^3=-27\)
\(\sqrt[3]{\left(2m+5\right)^3}=-3\)
\(2m+5=-3\)
\(m=-4\)
d) \(\left(3x-2\right)^3=64\)
tương tự câu c
\(a,\Rightarrow x=3\)
\(b,\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c,\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x-5-5x+10-4x-4=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)
\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)
\(\Leftrightarrow-3x=-9\)
hay x=3
3) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
4) Ta có: \(x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) Ta có: \(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
6) Ta có: \(\left(5x-1\right)^3=125\)
\(\Leftrightarrow5x-1=5\)
\(\Leftrightarrow5x=6\)
hay \(x=\dfrac{6}{5}\)
7) Ta có: \(3^{x+1}=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
\(a,\Rightarrow x=3\\ b,\Rightarrow2x-1=2\Rightarrow x=\dfrac{3}{2}\\ c,\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ d,\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\\ e,\Rightarrow2x+5=3^2=9\Rightarrow x=2\)
`@` `\text {Ans}`
`\downarrow`
`3)`
`(2x + 1) \div 9 - 104 \div 13 = 27^2 \div 3^5`
`\Rightarrow (2x + 1) \div 9 - 8 = (3^3)^2 \div 3^5`
`\Rightarrow (2x + 1) \div 9 - 8 = 3^6 \div 3^5`
`\Rightarrow (2x + 1) \div 9 - 8 = 3`
`\Rightarrow (2x + 1) \div 9 = 3 + 8`
`\Rightarrow (2x + 1) \div 9 = 11`
`\Rightarrow 2x + 1 = 11 . 9`
`\Rightarrow 2x + 1 = 99`
`\Rightarrow 2x = 99 - 1`
`\Rightarrow 2x = 98`
`\Rightarrow x = 98 \div 2`
`\Rightarrow x = 49`
Vậy, `x = 49.`
3) \(\left(2x+1\right):9-104:13=27^2:3^5\)
\(\left(2x+1\right):9-8=729:243\)
\(\left(2x+1\right):9-8=3\)
\(\left(2x+1\right):9=8+3\)
\(\left(2x+1\right):9=11\)
( 2x + 1) = 11.9
2x+1 = 99
2x = 99-1
2x = 98
x = 98 : 2
x = 49
\(\text{x. 3 = 27}\)
\(\Rightarrow x=9\)
\(b,\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
a) x3=27
x=27:3
x=9
b)(2x-1)3=8
(2x-1)3=23
=>2x-1=2
2x=2+1
2x=3
x=3:2
x=\(\dfrac{3}{2}\)
c)(x-2)2=16
(x-2)=16:2
x-2=8
x=8+2
x=10
(2x + 1)2 = 3 . 27
(2x + 1)2 = 81
(2x + 1)2 = 92
2x + 1 = 9
2x = 9 - 1
2x = 8
x = 8 : 2
x = 4
Đáp án + Giải thích các bước giải:
(13)2x−1−(13)2=−227(13)2x-1-(13)2=-227
(13)2.x−1−19 =−227(13)2.x-1-19 =-227
(13)2x−1 =−227+19(13)2x-1 =-227+19
(13)2x−1 =127(13)2x-1 =127
(13)2x−1 =(13)3(13)2x-1 =(13)3
2x−1=32x-1=3
2x =3+12x =3+1
2x =42x =4
x =4:2x =4:2
x =2x =2
Vậy x=2