(+) \(x+12\)\(⋮\)\(x+5\)

\(\left(x+5\right)+7⋮x+5\)

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

\((x - 3).(2y + 1) = 7\)

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

5; 5,01

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

<=> \(\left[{}\begin{matrix}2x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

\(x-8=50\%\cdot x\)

\(x-8=0,5x\)

\(x-0,5x=8\)

\(x\left(1-0,5\right)=8\)

\(0,5x=8\)

\(\Rightarrow x=\frac{8}{0,5}=16\)

Vậy x cần tìm là 16

\(x-8=50\%x\)

\(x-8=\frac{1}{2}x\)

\(\Leftrightarrow\text{ }\frac{1}{2}x=x-8\)

\(x\cdot\frac{1}{2}=x-8\)

\(x\cdot\frac{1}{2}-x=-8\)

\(x\left(\frac{1}{2}-1\right)=-8\)

\(x\cdot\frac{-1}{2}=-8\)

\(x=-8\text{ : }\frac{-1}{2}\)

\(x=-8\cdot\frac{2}{-1}\)

\(x=16\)

\(x-8=50\%.x\)

\(\Leftrightarrow x-8=\frac{1}{2}x\)

\(\Leftrightarrow x-\frac{1}{2}x=8\)

\(\Leftrightarrow\frac{1}{2}x=8\)

\(\Leftrightarrow x=16\)

khó v

=>2x-95 chia hết cho 2x+93

=>2x+93-188 chia hết cho 2x+93

=>\(2x+93\in\left\{1;-1;2;-2;4;-4;47;-47;94;-94;188;-188\right\}\)

=>\(x\in\left\{-46;-47;-\dfrac{91}{2};-\dfrac{95}{2};-\dfrac{87}{2};-\dfrac{97}{2};\dfrac{1}{2};-\dfrac{197}{2};\dfrac{95}{2};-\dfrac{281}{2}\right\}\)

\(\Rightarrow x\left(2y+1\right)-3\left(2y+1\right)=7\)

\(\Leftrightarrow\left(x-3\right)\left(2y+1\right)=7=1.7=7.1=-1.-7=-7.-1\)

vậy....

cảm ơn bạn Nguyễn Xuân Anh rất nhiều

12345(6787000)=-506

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

2 + 4 + 6 + ... + x = 110

Số số hạng = \(\frac{x-2}{2}+1=\frac{x-2}{2}+\frac{2}{2}=\frac{x}{2}\)

Tổng = \(\frac{\left(x+2\right)\cdot\frac{x}{2}}{2}=110\)

\(\Leftrightarrow\left(x+2\right)\cdot\frac{x}{2}=220\)

\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{2}=220\)

\(\Leftrightarrow\left(x+2\right)x=440\)

\(\Leftrightarrow\left(x+2\right)x=22\cdot20\)

\(\Rightarrow\hept{\begin{cases}x+2=20\\x=20\end{cases}}\Rightarrow x=20\)

Không hiểu chỗ nào thì ib nhé :)

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