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\(7\left(3x-3\right)=\left(2x+2\right)3\)
\(21x-21=6x+6\)
\(21x-21-6x-6=0\)
\(15x-27=0\)
\(15x=27\)
\(x=\frac{27}{15}=\frac{9}{5}\)
7 ( 3 x - 3 ) = ( 2 x + 2 ) 3
4 (3 x - 3 ) = 2 x + 2
12 x - 12 = 2 x +2
2x +9x = 2x +2
2 = 9 x
=> x = 2 : 9
=> x = rỗng
Vậy x = rỗng
( 2x - 1 ) ( y - 3 ) = 29
=> (2x - 1) ; (y - 3) là ước của 29
ta có bảng
2x - 1 | -29 | -1 | 1 | 29 |
y - 3 | -1 | -29 | 29 | 1 |
x | -14 | 0 | 1 | 15 |
y | 2 | -26 | 32 | 4 |
Vậy ...
\(\left(2x-1\right)\left(y-3\right)=29\)
\(\Rightarrow\left(2x-1\right)\left(y-3\right)\inƯ\left(29\right)\in\left\{\pm1;\pm29\right\}\)
+ Xét \(\hept{\begin{cases}2x-1=-1\\2x-1=1\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}\)
+ Xét \(\hept{\begin{cases}2x-1=29\\2x-1=-29\end{cases}}\Rightarrow\hept{\begin{cases}x=15\\x=\frac{-27}{2}\end{cases}=-14}\)
+ Xét \(\hept{\begin{cases}y-3=1\\y-3=-1\end{cases}\Rightarrow}\hept{\begin{cases}y=4\\y=2\end{cases}}\)
+ Xét \(\hept{\begin{cases}y-3=29\\y-3=-29\end{cases}}\Rightarrow\hept{\begin{cases}y=32\\y=-26\end{cases}}\)
Kết luận : .....
a.3x+27=9
3x=9-27
3x=-18
x=-18:3
x=-6
Vậy x=-6
b.2x+12=3(x-7)
2x+12=3x-21
12+21=3x-2x
33=x
Vậy x=33
a) 3x + 27 = 9
=> 3x = 9 - 27
=> 3x = -18
=> x = -18 : 3
=> x = -6
b) 2x + 12 = 3(x - 7)
=> 2x + 12 = 3x - 3 . 7
=> 2x + 12 = 3x - 21
=> 3x - 2x = 21 - 12
=> x = 9
\(\left|2x+3\right|=7\)
=> Các trường hợp
TH1 : \(\left|2x+3\right|=7\)
\(\left|2x\right|=7-3\)
\(\left|2x\right|=4\)
\(\left|x\right|=4:2\)
\(\left|x\right|=2\)
TH2 : \(\left|2x+3\right|=-7\)
\(\left|2x\right|=-7-3\)
\(\left|2x\right|=-10\)
\(\left|x\right|=\left(-10\right):2\)
\(\left|x\right|=-5\)
Vậy x = { 2 ; -5 }
a.
(-2)4.17.(-3)0.(-5)6.(-12n)
=16.17.1.15625.-1
=(16.15625).[1.(-1)].17
=250000.(-1).17
=4250000
b.3(2x2-7)=33
2x2-7 =33:3
2x2-7 =11
2x2 =11+7
2x2 =18
x2 =18:2
x2 =9
x2 =\(\left(\pm3^2\right)\)
\(\Rightarrow\) TH1: x2 =32 TH2: x2 =(-3)2
\(\Rightarrow\) x =3 \(\Rightarrow\)x =-3
Vậy x\(\in\left\{3;-3\right\}\)
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
a, | x - 3/4 | = 1/2
=>\(\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{2}{4}+\frac{3}{4}\\x=-\frac{2}{4}+\frac{3}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)
Vậy....
a) \(|x-\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)
Vay : x = 5/4 hoặc x = 1/4
b)\(saide\)
\(4\left(x-1\right)-3\left(x-2\right)=-5\)
\(\Leftrightarrow4x-4-3x+6=-5\)
\(\Leftrightarrow x=-5+4-6\)
\(\Leftrightarrow x=-7\)
Vậy x=-7
Ta có: 4(x-1) - 3(x-2) = -5
(4x-4) - (3x-6) = -5
4x - 4 - 3x + 6 = -5
(4x - 3x) + (-4+6) = -5
x + 2 = -5
x = -5 - 2
x = -7
Vậy x = -7
\(x-3=2x+4\)
\(x-2x=3+4\)
\(-1x=7\)
\(x=\frac{7}{-1}\)
\(x=-7\)
\(x-3=2x+4\)
\(\Rightarrow x-2x=4+3\)
\(x.\left(1-2\right)=7\)
\(x.\left(-1\right)=7\)
\(x=-7\)
CHÚC BẠN HỌC TỐT!!!