a, \(\dfrac{x}{3}=\dfrac{8}{12};12:3=4\)\(;8:4=2\)

⇒\(x=\dfrac{2}{3}\)

a: \(\dfrac{x}{3}=\dfrac{8}{12}\)

nên x/3=2/3

hay x=2

b: \(x+\dfrac{1}{15}=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)

c: \(x\cdot\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{5}{7}+\dfrac{2}{7}=1\)

hay x=7/4

d: \(\Leftrightarrow x\cdot\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{1}{2}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}=-\dfrac{2}{4}+\dfrac{5}{4}=\dfrac{3}{4}\)

hay x=1

\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\left(x\ne5;-2\right).\\ \Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\\ \Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\\ \Rightarrow7x=14.\\ \Leftrightarrow x=2\left(TM\right).\)

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

giúp mình đi mọi người

5/6-x=2/3

x=5/6-2/3

x=1/6

4/3:x=2

x=4/3:2

x=2/3

a) x = 1/6

b) x = 2/3

\(\dfrac{2x}{14}-\dfrac{x}{14}=-\dfrac{21}{14}\)

\(2x-x=-21\)

\(x=-21\)

x=-21

1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)

2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)

\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)

\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)

3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)

\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)

A. (Theo mình là -4xy thì mới rút gọn được)

B = (6x + y)^2

C = (2x)^2 = 4x^2

a: \(M=\left(\dfrac{-3}{7}x^3y\right)\cdot\dfrac{7xy^3}{12}-x^2y^2\cdot\left(-\dfrac{3}{4}x^2y^2\right)\)

\(=\dfrac{-1}{4}x^4y^4+\dfrac{3}{4}x^4y^4\)

\(=\dfrac{1}{2}x^4y^4\)

b: Hệ số là 1/2

Biến là \(x^4;y^4\)

bậc là 4+4=8

c: Thay x=-1 và y=-2 vào M, ta được:

\(M=\dfrac{1}{2}\left(-1\right)^4\cdot\left(-2\right)^4=\dfrac{1}{2}\cdot16=8\)

\(\left|x+1\right|-\left|-2x-2\right|=2\)

\(\Leftrightarrow\left|x+1\right|-\left|-2\left(x+1\right)\right|=2\)

\(\Leftrightarrow\left|x+1\right|-2\left|x+1\right|=2\)

\(\Leftrightarrow-\left|x+1\right|=2\)

\(\Leftrightarrow\left|x+1\right|=-2\)

\(\Leftrightarrow\left|x+1\right|+2=0\)

Mà: \(\left|x+1\right|\ge0\forall x\Rightarrow\left|x+1\right|+2\ge2>0\)

\(\Leftrightarrow\left|x+1\right|+2=0\) (vô lí)

Vậy phương trình vô nghiệm:

\(x\in\varnothing\)

=>|x+1|-2|x+1|=2

=>-|x+1|=2

=>|x+1|=-2(vô lý)

Vậy: \(x\in\varnothing\)

\(TH_1:\left\{{}\begin{matrix}x+1\ge0\\-2x-2\ge0\end{matrix}\right.\) 

\(\left(1\right)\left(x+1\right)-\left(-2x-2\right)=2\) 

\(\Leftrightarrow x+1+2x+2=2\)

\(\Leftrightarrow3x+3=2\)  

\(\Leftrightarrow x=-\dfrac{1}{3}\left(l\right)\)

\(TH_2:\left\{{}\begin{matrix}x+1< 0\\-2x-2< 0\end{matrix}\right.\) 

\(\left(1\right)\left(x+1\right)-\left(-\left(-2x-2\right)\right)=2\) 

\(\Leftrightarrow-x-1+2x+2=2\) 

\(\Leftrightarrow x+1=2\)

\(\Leftrightarrow x=1\left(l\right)\) 

Vậy tập nghiệm rỗng.

\(\dfrac{6^x}{3^3\cdot4}=2\)

=>6^x=2*4*3^3=8*3^3=6^3

=>x=3

help mì