#)Giải :

\(\left(\frac{3}{79}\right)^{20}\cdot\left(\frac{3}{-79}\right)^{19}=\left(\frac{3}{79}\right)^{20}\cdot\left(\frac{3}{79}\right)^{19}=\left(\frac{3}{79}\right)^{39}\)

\(\left(\frac{2}{3}\right)^5:\left(\frac{2}{-3}\right)^3=\left(\frac{2}{3}\right)^5:\left(\frac{2}{3}\right)^3=\left(\frac{2}{3}\right)^2\)

Ai biết chỉ mình nha 

1. 11(x-9)=77

x-9=7

x=16

2. 4(x-3)=48

x-3=12

x=15

3.3(x-8)=81

(x-8)=27

x=35

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)........\left(1-\frac{2}{5}\right)+.......1-\frac{1}{99}\)

\(=>\frac{3}{2}.\frac{4}{5}.\frac{5}{4}\)

\(=>\frac{100}{99}=\frac{100}{2}=50\)

~Hok tốt~

\(N=x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)

\(N=x^3+y^3+6x^2y^2+3xy\left[\left(x+y\right)^2-2xy\right]\)

\(N=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2+3xy-6x^2y^2\)

\(N=x^2-xy+y^2+3xy\)

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

\(N=1\)

\(x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]\)

\(=x^2-xy+y^2+6x^2y^2+3xy-6x^2y^2\)(  Do  \(x+y=1\))

\(=\left(x+y\right)^2-2xy-xy+3xy+6x^2y^2-6x^2y^3\)

\(=\left(x+y\right)^2=1^2=1\)

4x²-24x+42=70

=>4x²-24x-28=0

=>4x²-28x+4x-28=0

=>4x(x-7)+4(x-7)=0

=>(x-7)(4x+4)=0

=>\([\frac{x=7}{x=-1}\)

4x(7x-5)-7x(4x-2)=-12

⇒28x^2-20x-28x^2+14x=-12

⇒-20x+14x=-12

⇒-6x=-12

⇒x=2

Chị tham khảo ở đây ạ:

Câu hỏi của Vũ Thảo Vy - Toán lớp 9 - Học toán với OnlineMath

Có \(\frac{a}{b}=\frac{3}{7}\)

Theo bài ra,ta có:

\(\frac{a+15}{b}=\frac{36}{49}\)

\(\frac{a}{b}+\frac{15}{b}=\frac{36}{49}\)

\(\frac{3}{7}+\frac{15}{b}=\frac{36}{49}\)

\(\frac{15}{b}=\frac{36}{49}-\frac{3}{7}\)

\(\frac{15}{b}=\frac{15}{49}\)

b=49

Mà \(\frac{a}{b}=\frac{3}{4}\Rightarrow\frac{a}{49}=\frac{3}{4}\Rightarrow a=\frac{3}{4}\times49=\frac{147}{4}\)

vậy \(\frac{a}{b}=\frac{147}{4}\)

tk mik nha

A) x² -- 2 x y + y²

B) 4x²y² +2 * xy *3 + 9

=> ( 2xy + 3 )²

a) x²-2xy+...

Áp dụng HĐT ta có:

\(x^2-2xy+y^2\)

b) 4x²y²+...+9

Áp dụng HĐT ta có :

\(\left(2xy\right)^2+2.2xy.3+3^2\)

\(\Leftrightarrow4x^2y^2+12xy+9\)