\(\frac{\frac{1}{5}x}{3}=\frac{\frac{2}{4}}{\frac{1}{4}}\)

\(\Leftrightarrow\frac{1}{5}x:3=2\)

\(\Rightarrow\frac{1}{5}x=6\)

\(\Rightarrow x=\frac{6}{\frac{1}{5}}\)

\(x=30\)

\(\frac{1}{5}x:3=\frac{2}{3}:0,25\)

\(\frac{1}{5}x:\frac{2}{3}=3:0,25\)

\(\frac{1}{5}x:\frac{2}{3}=3:\frac{1}{4}\)

\(\frac{1}{5}x:\frac{2}{3}=3.4\)

\(\frac{1}{5}x:\frac{2}{3}=12\)

\(\frac{1}{5}x=12.\frac{2}{3}\)

\(\frac{1}{5}x=8\)

\(x=8.5\)

\(x=40\)

a. \(3xy+x+15y+15=x\left(3y+1\right)+15\left(y+1\right)=\left(x+15\right)\left(3y+1\right)\)

b.\(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3+x+y\right)\left(3-x-y\right)\)

c.\(x^3-5x^2+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x^2+1\right)\left(x-5\right)\)

d.\(x^2-2xy+y^2-1=\left(x-y\right)^2-1=\left(x-y+1\right)\left(x-y-1\right)\)

\(\left(5x-1\right)=\left(1-5x\right)^2\)

\(\left(5x-1\right)=\left(5x-1\right)^2\)

\(\left(5x-1\right)\left(1-5x+1\right)=0\)

\(\left(5x-1\right)\left(2-5x\right)=0\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{2}{5}\end{array}\right.\)

\(9-\left(x-2\right)^2=0\)

\(\left(3+x-2\right)\left(3-x+2\right)=0\)

\(\left(1+x\right)\left(5-x\right)=0\)

\(\left[\begin{array}{nghiempt}x=-1\\x=5\end{array}\right.\)

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

1. \(3-|2x+1|=-5\)

\(\Rightarrow|2x+1|=8\)

\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)

2.\(12+|3-x|=9\)

\(\Rightarrow|3-x|=-3\)

Mà \(|3-x|\ge0\forall x\)

\(\Rightarrow\)Vô lí

Vậy không có x

3.\(|x+9|=12+\left(-9\right)+2\)

\(\Rightarrow|x+9|=5\)

\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)

Vậy \(x\in\left\{-4;-14\right\}\)

4.\(5x-16=40+x\)

\(\Rightarrow5x-x=40+16\)

\(\Rightarrow4x=56\)

\(\Rightarrow x=14\)

Vậy \(x=14\)

5.\(5x-7=-21-2x\)

\(\Rightarrow5x+2x=-21+7\)

\(\Rightarrow7x=-14\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

6.\(\left(2x-1\right)\left(y-2\right)=12\)

Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)

\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Ta có bảng : (em tự xét bảng nhé)

a)12(x-1)=0

=>x-1=0

=>x=1

b) 3/5x-3/4=(-3)²-|-11|

3/5x-3/4=9-11=-2

3/5x=-2+3/4=-5/4

=>x=-5/4:3/5=-25/12

c)4(x-5/8)-3/4=0,25

4(x-5/8)=0,25+3/4=1

=>x-5/8=1/4

x=1/4+5/8=7/8

2) Số đó là

27:3/5=45