5x - 7= x - 5

 =>5x-x=-5+7

=>4x=2

=>x=2:4

=>x=1/2

vậy x=1/2

5x-7=x-5

5x-x=-5+7

4x=2

x=\(\frac{2}{4}\) =\(\frac{1}{2}\)

a: \(=\dfrac{x^2+x-2x+2-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}\)

a) \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)

Dấu "=" xảy ra khi \(\left(x+1\right)^2+2=2\Rightarrow x=-1\)

Vậy \(MinA=2\)khi \(x=-1\)

c) \(4x^2-4x+5=\left(4x^2-4x+1\right)+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi \(\left(2x-1\right)^2+4=4\Rightarrow x=\dfrac{1}{2}\)

Vậy \(MinC=4\) khi \(x=\dfrac{1}{2}\)

Bài 1

\(\dfrac{15}{100}=\dfrac{3}{20};\dfrac{25}{200}=\dfrac{1}{8};\dfrac{88}{648}=\dfrac{11}{81};\dfrac{312}{108}=\dfrac{26}{9};\dfrac{8}{1000}=\dfrac{1}{125};\dfrac{27}{315}=\dfrac{3}{35}\)

Ai làm được rồi mình tick đúng cho

Bài 2: 

a: 7/12=21/36

19/36=19/36

b: 2/3=40/60

1/4=15/60

3/5=36/60

c: 11/20=55/100

17/50=34/100

a:  để \(\dfrac{4}{x}>\dfrac{4}{5}\\ \Rightarrow x< 5\)

mà x là các số tự nhiên 

\(\Rightarrow x\in\left\{1;2;3;4\right\}\)

b: để \(\dfrac{1}{9}< \dfrac{x}{9}< \dfrac{4}{9}th\text{ì}1< x< 4\\ \Rightarrow x\in\left\{2;3\right\}\)

A) x>5 => x=1,x=2,x=3,x=4 do hai tử bằng nhau thì phân số nào có mẫu nhỏ hơn sẽ lớn hơn

B) x=2 hoặc x= 3 

C) x=7

X=12

1)x>=0

2)v8+5v2-v32+6v1/2=2v2+5v2-4v2+3v2=9v2

3)vx+1=2

x+1=4=>x=3

\(a,\Rightarrow x-2=8\\ \Rightarrow x=10\\ b,\Rightarrow x+12-17=20\\ \Rightarrow x-5=20\\ \Rightarrow x=25\\ c,\Rightarrow11-\left(4x+5\right):3=4\\ \Rightarrow\left(4x+5\right):3=7\\ \Rightarrow4x+5=21\\ \Rightarrow x=4\\ d,\Rightarrow\left(35:x+3\right)\cdot17=136\\ \Rightarrow35:x+3=8\\ \Rightarrow35:x=5\\ \Rightarrow x=7\\ e,\Rightarrow41-\left(2x-5\right)=720:8\cdot5=180\\ \Rightarrow2x-5=-139\\ \Rightarrow2x=-134\\ \Rightarrow x=-67\)

\(2,\\ a,\Rightarrow x^2=4^3:16=64:16=4=2^2=\left(-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ b,\Rightarrow\left(x-1\right)^2=9=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(3x-7\right)^5=2^5\\ \Rightarrow3x-7=2\\ \Rightarrow3x=9\Rightarrow x=3\)