Tìm x hộ mình với
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![](https://rs.olm.vn/images/avt/0.png?1311)
tìm x biết
(x+12):(x+5)
(x+12):(x-5)
(3x-5):x-4
hộ mình với ai nhanh mình tick cho giải luôn hộ mình với
![](https://rs.olm.vn/images/avt/0.png?1311)
(+) \(x+12\)\(⋮\)\(x+5\)
\(\left(x+5\right)+7⋮x+5\)
Vì\(x+5⋮x+5\)nên \(7⋮x+5\)
hay \(x+5\in U\left(7\right)=\pm1,\pm7\)sau đó tìm ra tất cả x
(+)\(x+12⋮x-5\)
\(\left(x-5\right)+15⋮x-5\)
vì \(\left(x-5\right)⋮x-5\)
nên \(15⋮x-5\)
hay \(x-5\in U\left(15\right)=\pm1,\pm3,\pm5,\pm15\)sau đó tính x ra
(+)\(3x-5⋮x-4\)
\(\left(3x-12\right)+7⋮x-4\)
Vì\(3x-12⋮x-4\)
nên \(7⋮x-4\)
hay \(x-4\in U\left(7\right)=\pm1,\pm7\)sau đó tính ra x
![](https://rs.olm.vn/images/avt/0.png?1311)
\((x - 3).(2y + 1) = 7\)
Ý của bạn là chỉ yc tìm mỗi vế của biến x ạ?
\(\left(x-3\right)\cdot\left(2y+1\right)\in\text{Ư}\left(7\right)=\left\{1;7;-1;-7\right\}\)
`\Rightarrow \text {TH1:} x - 3 = 1`
`\Rightarrow x = 1 + 3`
`\Rightarrow x = 4`
`\text {TH2:} x - 3 = 7`
`\Rightarrow x = 7 + 3`
`\Rightarrow x = 10`
`\text {TH3:} x - 3 = -1`
`\Rightarrow x = -1 + 3`
`\Rightarrow x = 2`
`\text {TH4:} x - 3 = -7`
`\Rightarrow x = -7 + 3`
`\Rightarrow x = -4`
Vậy, `x \in {-4; 4; 2; 10}`
ta có
trường hợp 1:(x-3)=7
x-3=7
x=7+3
x=10
x-3=7
x=7+3
x=10
trường hợp 2:(x-3)=1
x-3=1
x=1+3
x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
3x2-75=0
<=> 3x2=75
<=> x2=25
<=> x=5
2x2-98=0
<=> 2x2=98
<=> x2=49
<=> x=7
x2-7x=0
<=> x(x-7)=0
<=> x=0 hoặc x=7
-3x2+5x=0
x(-3x+5)=0
x=0 hoặc -3x+5=0
x=0 hoặc -3x=-5
x=0 hoặc x=5/3
x2+4x+4=0
(x+2)2=0
x+2=0
x=-2
1. 3x2 - 75 = 0
<=> 3x2 = 75
<=> x2 = 25
<=> x = \(\sqrt{25}\)
<=> x = 5
2. x2 - 7x = 0
<=> x(x - 7) = 0
<=> \(\left[{}\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
3. x2 - 14x + 13 = 0
<=> x2 - 13x - x + 13 = 0
<=> x(x - 13) - (x - 13) = 0
<=> (x - 1)(x - 13) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=13\end{matrix}\right.\)
4. 2x2 - 98 = 0
<=> 2x2 = 98
<=> x2 = 49
<=> x = \(\sqrt{49}\)
<=> x = 7
5. -3x2 + 5x = 0
<=> x(-3x + 5) = 0
<=> \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)
6. x2 - 2x - 80 = 0
<=> x2 + 8x - 10x - 80 = 0
<=> x(x + 8) - 10(x + 8) = 0
<=> (x - 10)(x + 8) = 0
<=> \(\left[{}\begin{matrix}x-10=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
7. x2 = 81
<=> x2 - 92 = 0
<=> (x - 9)(x + 9) = 0
<=> \(\left[{}\begin{matrix}x-9=0\\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
8. x2 + 4x + 4 = 0
<=> x2 + 2.x.2 + 22 = 0
<=> (x + 2)2 = 0
<=> 0 = 02 - (x + 2)2
<=> (0 + x + 2)(0 - x + 2) = 0
<=> (x + 2)(-x + 2) = 0
<=> \(\left[{}\begin{matrix}x+2=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
9. 4x2 + 12x + 5 = 0
<=> 4x2 + 2x + 10x + 5 = 0
<=> 2x(2x + 1) + 5(2x + 1) = 0
<=> (2x + 5)(2x + 1) = 0
<=> \(\left[{}\begin{matrix}2x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-8=50\%\cdot x\)
\(x-8=0,5x\)
\(x-0,5x=8\)
\(x\left(1-0,5\right)=8\)
\(0,5x=8\)
\(\Rightarrow x=\frac{8}{0,5}=16\)
Vậy x cần tìm là 16
![](https://rs.olm.vn/images/avt/0.png?1311)
=>2x-95 chia hết cho 2x+93
=>2x+93-188 chia hết cho 2x+93
=>\(2x+93\in\left\{1;-1;2;-2;4;-4;47;-47;94;-94;188;-188\right\}\)
=>\(x\in\left\{-46;-47;-\dfrac{91}{2};-\dfrac{95}{2};-\dfrac{87}{2};-\dfrac{97}{2};\dfrac{1}{2};-\dfrac{197}{2};\dfrac{95}{2};-\dfrac{281}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow x\left(2y+1\right)-3\left(2y+1\right)=7\)
\(\Leftrightarrow\left(x-3\right)\left(2y+1\right)=7=1.7=7.1=-1.-7=-7.-1\)
x-3 | -7 | -1 | 1 | 7 |
2y+1 | -1 | -7 | 7 | 1 |
x | -4 | 2 | 4 | 10 |
y | -1 | -4 | 3 | 0 |
vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
2 + 4 + 6 + ... + x = 110
=> \(\frac{\left(x+2\right)\left[\left(x-2\right):2+1\right]}{2}=110\)
\(\Rightarrow\left(x+2\right).\frac{1}{2}x=220\)
\(\Rightarrow\left(x+2\right)x=440\)
Mà 20 . 22 = 440
=> x = 20
Ta có : 2 + 4 + 6 +... + x = 110
=> [(x - 2) : 2 + 1].(x + 2) : 2 = 110
=> \(\left(\frac{x}{2}-1+1\right).\left(\frac{x+2}{2}\right)=110\)
=> \(\frac{x}{2}.\frac{x+2}{2}=110\)
=> \(\frac{x\left(x+2\right)}{4}=110\)
=> x(x + 2) = 440
=> x2 + 2x = 440
=> x2 + 2x + 1 = 441
=> x2 + x + x + 1 = 441
=> x(x + 1) + (x + 1) = 441
=> (x + 1)2 = 441
=> (x + 1)2 = 212
=> x + 1 = 21
=> x = 20
Vậy x = 20
x : 689 = 38726
x = 38726 x 689
x = 26682214
x : 689 = 157
x = 157 x 689
x = 108173