![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{x-1}=9\)
\(9^{x-1}=9^1\)
\(x-1=1\)
\(x=1+1\)
\(x=2\)
\(\frac{6x-18}{3}+25\times2=78\)
\(\frac{6\left(x-3\right)}{3}+50=78\)
\(2\left(x-3\right)=78-50\)
\(2x-6=28\)
\(2x=28+6\)
\(2x=34\)
\(x=\frac{34}{2}\)
\(x=17\)
a) \(9^{x-1}=9\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(\left(6x-18\right):3+25.2=78\)
\(\Rightarrow3\left(2x-6\right):3+50=78\)
\(\Rightarrow2x-6=38\)
\(\Rightarrow2x=44\)
\(\Rightarrow x=22\)
Vậy \(x=22\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) -6x = 18
x = 18 : (-6)
x = -3
c) x . 2 = 4
x = 4 : 2
x = 2
mik mới học lớp 5 nên chỉ biết ít về lớp 6
Tìm số nguyên x , biết :
a) 9-25=(7-x)-(25+7)
9 - 25 = (7 - x )- 32
-16 = (7 -x)-32
7-x= -16 - 32
7-x = -48
x = 7 -(-48)
x = 55
b)-6x=18
x = -( 18 :6)
x = -3
c)x.2=4
x = 4 : 2
x = 2
Mình không biết có đúng không nữa nhưng mk mong bạn sẽ k mk
![](https://rs.olm.vn/images/avt/0.png?1311)
`x^2=3`
`=>x=\sqrt{3}\or\x=-\sqrt{3}`
`x^2=36`
`<=>x^2=(+-6)^2`
`<=>x=+-6`
`x^2=25`
`<=>x^2=(+-5)^2`
`<=>x=+-5`
`2x^2+(-20)=55`
`<=>2x^2-20=55`
`<=>2x^2=75`
`<=>x^2=75/2`
`<=>x=+-\sqrt{75/2}`
`2(x-1)^2+5^0=9`
`<=>2(x-1)^2+1=9`
`<=>2(x-1)^2=8`
`<=>(x-1)^2=4`
`<=>x-1=2\or\x-1=-2`
`<=>x=3\or\x=-1`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 9-25=7-x-(25+7)
=>7-x-25-7=-16
=>-x-25=-16
=>x+25=16
hay x=-9
b: \(10+2\left|x\right|=2\cdot\left(3^2-1\right)\)
\(\Leftrightarrow2\left|x\right|=2\cdot8-10=6\)
=>x=3 hoặc x=-3
c: -6x=18
nên x=18:(-6)=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
X2=3 x2=25
=> X=\(\pm\sqrt{3}\) => x=5
X2=36
=> x=6
2.(x-1)2+50= 9
2.(x-1)2+1= 9
2.(x-1)2= 8
(x-1)2 = 8/2
(x-1)2 = 4
(x-1)2 = (2)2
x-1=(\(\pm\)2)
TH1: x-1= 2 TH2: x-1=-2
x=2+1 x =(-2)+1
x= 3 x = -1
Vậy x\(\in\)\(\left\{3;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b. -6x=18
=> x = 18 : (-6)
=> x = -3
c. 35-3.|x|=5.(23-4)
=> 35-3.|x|=5.(8-4)
=> 35-3.|x|=5.4
=> 35-3.|x|=20
=> 3.|x|=35-20
=> 3.|x|=15
=> |x|=5
=> x \(\in\){-5; 5}
d. => 10+2.|x|=2.(9-1)
=> 10+2.|x|=2.8
=> 10+2.|x|=16
=> 2.|x|=16-10
=> 2.|x|=6
=> |x|=3
=> x \(\in\){-3; 3}
\(x^2\) + 6\(x\) + 9 = 25
\(x^2\) + 6\(x\) + 9 - 25 = 0
\(x^2\) + 6\(x\) + (9 - 25) = 0
\(x^2\) + 6\(x\) - 16 = 0
\(x^2\) - 2\(x\) + 8\(x\) - 16 = 0
(\(x^2\) - 2\(x\)) + (8\(x\) - 16) = 0
\(x\)(\(x\) - 2) + 8(\(x-2\)) = 0
(\(x\) - 2)(\(x\) + 8) = 0
\(\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy \(x\) \(\in\) {2; - 8}
cứu mình với