a: Ta có: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{x^2-4y^2}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(\)\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)

b: \(\frac{4x+7}{2x+2}-\frac{3x+6}{2x+2}\)

\(=\frac{4x+7-3x-6}{2x+2}\)

\(=\frac{x+1}{2\left(x+1\right)}=\frac12\)

c: \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)

\(=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\cdot\left(x+3\right)}\)

\(=\frac{x\left(x+9\right)-3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)

d: \(\frac{1}{x^2+3x+2}-\frac{1}{x^2-4}\)

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

\(=\frac{x-2-\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{-3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

2/3x = -7/12 - 1/4 

2/3x = -5/6

x      = -5/6 : 2/3

x      = -5/4

mk mới hok lớp 5 thôi

khó quá đấy

bố trí một cái

a: Ta có: \(3x+5\le4x-9\)

\(\Leftrightarrow-x\le-14\)

\(\Leftrightarrow x\ge14\)

b: Ta có: \(6-2x< 6-x\)

\(\Leftrightarrow-x< 0\)

hay x>0

c: Ta có: \(7\left(x-1\right)+5>-3x\)

\(\Leftrightarrow7x-7+5+3x>0\)

\(\Leftrightarrow10x>2\)

hay \(x>\dfrac{1}{5}\)

ai giúp minh đi cần quá gấp

Ta có:

a, /x-5/+5=x (1) 

Với /x-5/<0

=> /x-5/=5-x

Thay vào (1) ta đc 

/x-5/+5=x 

=> 5-x+5=x

=> 10-x=x

=> x=5

Với /x-5/>=0 

=> x>=5

và /x-5/=x-5

thay vào (1) ta đc

0=0 thỏa mãn 

Vậy x>=5 

Các câu sau cmtt nha bạn

a) Ta có : |2x - 5| = x + 1

\(\Leftrightarrow\orbr{\begin{cases}2x-5=-x-1\\2x-5=x+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+5\\2x-x=1+5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=4\\x=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=6\end{cases}}\)

mik ko pc