6:

a: a+b+c=0

=>a=-b-c; b=-a-c; c=-a-b

=>a^2=(b+c)^2; b^2=(a+c)^2; c^2=(a+b)^2

=>4bc-a^2=4bc-(b+c)^2=-(b-c)^2; 4ac-b^2=-(a-c)^2; 4ab-c^2=-(a-b)^2

\(M=\dfrac{-\left(b-c\right)^2}{bc+2a^2}\cdot\dfrac{-\left(a-c\right)^2}{ca+2b^2}\cdot\dfrac{-\left(a-b\right)^2}{ab+2c^2}\)

\(=-\dfrac{\left(b-c\right)^2}{2b^2+4bc+2c^2+bc}\cdot\dfrac{\left(a-c\right)^2}{2a^2+4ac+2c^2+ac}\cdot\dfrac{\left(a-b\right)^2}{2a^2+4ab+2b^2+ab}\)

\(=\dfrac{-\left(b-c\right)^2}{2b\left(b+2c\right)+c\left(b+2c\right)}\cdot\dfrac{\left(a-c\right)^2}{2a\left(2c+a\right)+c\left(2c+a\right)}\cdot\dfrac{\left(a-b\right)^2}{2a\left(2b+a\right)+b\left(2a+b\right)}\)

\(=\dfrac{-\left(b-c\right)^2}{\left(b+2c\right)\left(2b+c\right)}\cdot\dfrac{\left(a-c\right)^2}{\left(2a+c\right)\left(2c+a\right)}\cdot\dfrac{\left(a-b\right)^2}{\left(2b+a\right)\left(2a+b\right)}\)

\(=\dfrac{-\left(b-c\right)^2}{\left(-a-c+2c\right)\left(2b-a-b\right)}\cdot\dfrac{\left(a-c\right)^2}{\left(2a-a-b\right)\left(2c-b-c\right)}\cdot\dfrac{\left(a-b\right)^2}{\left(2b-b-c\right)\left(2a-a-c\right)}\)

\(=\dfrac{-\left(b-c\right)^2}{\left(c-a\right)\left(b-a\right)}\cdot\dfrac{\left(a-c\right)^2}{\left(a-b\right)\left(c-b\right)}\cdot\dfrac{\left(a-b\right)^2}{\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{-\left(b-c\right)^2\left(a-c\right)^2\left(a-b\right)^2}{-\left(a-b\right)^2\cdot\left[-\left(a-c\right)^2\right]\cdot\left[-\left(b-c\right)^2\right]}\)

=1

b: \(N=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}=\dfrac{\left(-c\right)\cdot\left(-a\right)\cdot\left(-b\right)}{abc}=-1\)

`a(b+1)+b(a+1)=(a+1)(b+1)`

`VT=ab+a+ab+b`

`=2ab+a+b`

`=2.1+a+b`

`=2+a+b(1)`

`VP=(a+1)(b+1)`

`=a(b+1)+1(b+1)`

`=ab+a+b+1`

`=1+a+b+1`

`=2+a+b(2)`

`(1)(2)=>VT=VP`.

Có: a(b + 1) + b(a + 1)

= ab + a + ab + b

= ab + a + b + 1 (do ab = 1)

= a(b + 1) + (b + 1)

= (a + 1)(b + 1)

Lời giải:
$7^{3x-2}-3.7^3=7^3.4$

$7^{3x-2}=3.7^3+7^3.4=7^3(3+4)=7^3.7=7^4$
$\Rightarrow 3x-2=4$

$\Rightarrow 3x=6$

$\Rightarrow x=2$

\(\left(3x-\dfrac{1}{3}\right)^2=16=4^2=\left(-4\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{3}=4\\3x-\dfrac{1}{3}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{13}{3}\\3x=-\dfrac{11}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\\x=-\dfrac{11}{9}\end{matrix}\right.\)

góc mOn=góc mOc+góc nOc

=1/2(góc aOc+góc bOc)

=1/2*180=90 độ

a) \(=\dfrac{\left(12x+5\right)\left(4x+3\right)}{30\left(x+9\right)\left(12x+5\right)}+\dfrac{\left(12x+5\right)\left(6-3x\right)}{30\left(x+9\right)\left(12x+5\right)}\)
\(=\dfrac{4x+3-3x+6}{30\left(x+9\right)}\)
\(=\dfrac{1}{30}\)

a: Kẻ OE vuông góc AB, OF vuông góc CD

Xét (O) có

AB=CD

AB,CD là hai dây

d(O;AB)=OE; d(O;CD)=OF

=>OE=OF

Xét tứ giác OEMF có

góc OEM=góc OFM=góc FME=90 độ

OE=OF

=>OEMF là hình vuông

=>MO là phân giác của góc FMO

b: ΔOCD cân tại O

mà OF là đường cao

nên F là trung điểm của CD

=>DF=CD/2=(4+2)/2=3cm

=>MF=DF-DM=3-2=1cm

=>d(O;AB)=d(O;CD)=1cm

`(2x-1/3)^2=4`

\(< =>\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}:2\\x=-\dfrac{5}{3}:2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}\cdot\dfrac{1}{2}\\x=-\dfrac{5}{3}\cdot\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

\(\left(2x-\dfrac{1}{3}\right)^2=4\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)