Trả lời câu trước r nha bn

ờ thế yêu cầu đề là j mà kêu giúp ???

minh ko hieu

a)\(x+\frac{1}{3}=\frac{3}{4}\)

\(\Rightarrow x=\frac{3}{4}-\frac{1}{3}\)

\(\Rightarrow x=\frac{5}{12}\)

b)\(x-\frac{2}{5}=\frac{5}{7}\)

\(\Rightarrow x=\frac{5}{7}+\frac{2}{5}\)

\(\Rightarrow x=1\frac{4}{35}\)

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

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

\(\Rightarrow-x=-\frac{4}{21}\)

\(\Rightarrow x=\frac{4}{21}\)

d)\(\frac{4}{7}-x=\frac{1}{3}\)

\(x=\frac{4}{7}-\frac{1}{3}\)

\(\Rightarrow x=\frac{5}{21}\)

A) x+1/3=3/4

    x        =3/4-1/3

    x        =3/4+(-1/3)

    x        =5/12

B) x-2/5=5/7

    x       =5/7+2/5

    x       =39/35

C) -x-2/3=-6/7

     -x      =-6/7+2/3

     -x      =-4/21

     x       =4/21

D) 4/7-x=1/3

           x=4/7-1/3

           x=4/7+(-1/3)

           x=5/21

1*2+2*4+3*3*2

1*2+2*4+9*2

2*(1+4+9)

2*14

28

câu b tương tự nhé

mình nhầm

đúng rồi 

Lời giải:

Đặt $A=x^{2011}+x^{2010}+....+x+1$

$Ax=x^{2012}+x^{2011}+...+x^2+x$

$\Rightarrow Ax-A=x^{2012}-1$

$\Rightarrow A=\frac{x^{2012}-1}{x-1}$

$B=x^{502}+x^{501}+...+x+1$

$Bx=x^{503}+x^{502}+....+x^2+x$

$\Rightarrow Bx-B=x^{503}-1$

$\Rightarrow B=\frac{x^{503}-1}{x-1}$

Khi đó: $A:B = \frac{x^{2012}-1}{x-1}: \frac{x^{503}-1}{x-1}=\frac{x^{2012}-1}{x^{503}-1}=\frac{(x^{503})^4-1}{x^{503}-1}$

Đặt $x^{503}=a$ thì:

$A:B=\frac{a^4-1}{a-1}=a^3+a^2+a+1$

$\Rightarrow A\vdots B$

\(=\dfrac{x^3+x^2+x^2+x}{x^2+x}\\ =\dfrac{x^2\left(x+1\right)+x\left(x+1\right)}{x\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x^2+x\right)}{x\left(x+1\right)}\\ =\dfrac{x\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}\\ =x+1\)

\(\dfrac{x^3+2x^2+x}{x^2+x}\\ =\dfrac{x^3+x^2+x^2+x}{x\left(x+1\right)}\\ =\dfrac{x^2\left(x+1\right)+x\left(x+1\right)}{x\left(x+1\right)}\\ =\dfrac{\left(x^2+x\right)\left(x+1\right)}{x\left(x+1\right)}\\ =\dfrac{x\left(x+1\right)}{x}\\ =x+1\)

=x(x^2+2x+1)/x(x+1)=(x+1)^2/(x+1)=x+1

\(\dfrac{\left(2x+1\right)}{2}-\dfrac{1}{3}=\dfrac{\left(x+1\right)}{6}\left(MC=12\right)\\ \Leftrightarrow\dfrac{\left(2x+1\right)2}{2.6}-\dfrac{1.4}{3.4}=\dfrac{\left(x+1\right)2}{6.2}\\ \Leftrightarrow\dfrac{4x+2}{12}-\dfrac{4}{12}=\dfrac{2x+2}{12}\\ \Leftrightarrow4x+2-4=2x+2\\ \Leftrightarrow4x-2x=2+2-4\\ \Leftrightarrow2x=0\\ \Leftrightarrow x=0\)

\(x^2+4y^2+9-4xy-6x+12y\)

\(=\left(x^2-4xy+4y^2\right)+\left(-6x+12y\right)+9\)

\(=\left(x-2y\right)^2-6\left(x-2y\right)+9\)

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