tìm x:
2x=8^2
3x-x=9^5
5^2x+1=125^5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
8 tháng 10 2018
bài 1
a) 155 - 10.(x+1) = 55
=>10 .(x+1) = 100
=>x + 1 = 10
=>x = 9
còn lại tương tự
HM
3
C
6 tháng 12 2023
a)125 : x = 22 - (-1)
125 : x = 4 + 1
125 : x = 5
x = 125 : 5
x = 25
-------------------------------------------------
b) 2x - 8 = -4
2x = (-4) + 8
2x = 4
x = 4 : 2
x = 2
-----------------------------------------------------------
c) Xem lại đề.
VN
Võ Ngọc Phương
VIP
6 tháng 12 2023
\(125:x=2^2-\left(-1\right)\)
\(=>125:x=4+1\)
\(=>125:x=5\)
\(=>x=125:5\)
\(=>x=25\)
_____
\(2x-8=-4\)
\(=>2x=\left(-4\right)+8\)
\(=>2x=4\)
\(=>x=4:2\)
\(=>x=2\)
_______
\(6^{2x+5}=216\)
\(=>6^{2x+5}=6^3\)
\(=>2x+5=3\)
\(=>2x=3-5\)
\(=>2x=-2\)
\(=>x=\left(-2\right):2\)
\(=>x=-1\)
\(#NqHahh\)
16 tháng 6 2022
1: \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>23x+61=2x+61
hay x=0
2: \(\dfrac{6}{x-5}+\dfrac{x+2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)
\(\Leftrightarrow6x-48+x^2-3x-10=-18-x^2+13x-40\)
\(\Leftrightarrow x^2+3x-58+x^2-13x+58=0\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0
c: \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)
\(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)=-7x^2+3x\)
\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x=0\)
\(\Leftrightarrow x^2=0\)
hay x=0
14 tháng 7 2023
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
NN
14 tháng 7 2023
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`