Lời giải:
Đặt $\frac{x}{3}=\frac{y}{5}=t(t\neq 0)$

$\Rightarrow x=3t; y=5t$

Khi đó:

$\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5(3t)^2+3(5t)^2}{10(3t)^2-3(5t)^2}=\frac{120t^2}{15t^2}=8$

a: \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{2\left(x-3\right)}{2-x}\)

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

\(=\dfrac{5x^2+4x+4-4+4x-x^2}{\left(2+x\right)}\cdot\dfrac{1}{2\left(x-3\right)}\)

\(=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{1}{2\left(x-3\right)}=\dfrac{4x\left(x+2\right)}{2\left(x+2\right)}\cdot\dfrac{1}{x-3}=\dfrac{2x}{x-3}\)

b: |x-2|=2

=>x-2=2 hoặc x-2=-2

=>x=0(nhận) hoặc x=4(nhận)

Khi x=0 thì \(A=\dfrac{2\cdot0}{0-3}=\dfrac{-2}{3}\)

Khi x=4 thì \(A=\dfrac{2\cdot4}{4-3}=8\)

c: A>0

=>x/x-3>0

=>x>3 hoặc x<0

=>x>3

a) Ta có: \(N=\left(-\dfrac{3}{4}xy^4\right)\cdot\left(\dfrac{6}{9}x^2y^2\right)\)

\(=\left(-\dfrac{3}{4}\cdot\dfrac{6}{9}\right)\cdot\left(x\cdot x^2\right)\cdot\left(y^4\cdot y^2\right)\)

\(=-\dfrac{1}{2}x^3y^6\)

Hệ số: \(-\dfrac{1}{2}\)

Phần biến: \(x^3;y^6\)

Bậc của đơn thức là 9

\(A=19\frac{1}{4}+\frac{1}{2}\times2\frac{1}{3}+5,75-\frac{1}{6}+74\)

MK GHI ĐẦY ĐỦ RA RÙI, BẠN TỰ BẤM MÁY TÍNH LÀM NHA ( MÌNH LƯỜI )

\(A=19\frac{1}{4}+\frac{1}{2}\times2\frac{1}{3}+5,75-\frac{1}{6}+74\)

\(A=\frac{77}{4}+\frac{1}{2}\times\frac{7}{3}+\frac{23}{4}-\frac{1}{6}+74\)

\(A=\frac{77}{4}+\frac{7}{6}+\frac{23}{4}-\frac{1}{6}+74\)

\(A=(\frac{77}{4}+\frac{23}{4})+(\frac{7}{6}-\frac{1}{6})+74\)

\(A=25+1+74\)

\(A=26+74\)

\(A=100\)

Llkogvhhvcudycvkuwyvukvdyckuyvculvycduydvclduydcyludvdulyerguc

a)a+b+c=9

=>(a+b+c)²=81

=>a²+b²+c²+2ab+2bc+2ca=81

Từ a²+b²+c²=141=>2ab+2bc+2ca=81-141=-60

=>2(ab+bc+ca)=-60=>ab+bc+ca=-30

b)x+y=1

=>(x+y)³=1

=>x³+3x²y+3xy²+y³=1

=>x³+y³+3xy(x+y)=1

=>x³+y³+3xy=1(Do x+y=1)

c)a³-3ab+2c=(x+y)³-3(x+y)(x²+y²)+2(x³+y³)

=x³+3x²y+3xy²+y³-3x³-3y³-3x²y-3xy²+2x³+2y³=0

d)đang tìm hướng giải

\(\dfrac{11\cdot9^{11}\cdot3^7-27^{10}}{4\cdot81^1}\)

\(=\dfrac{11\cdot3^{22}\cdot3^7-\left(3^3\right)^{10}}{4\cdot3^4}\)

\(=\dfrac{11\cdot3^{29}-3^{30}}{4\cdot3^4}\)

\(=\dfrac{3^{29}\cdot\left(11-3\right)}{4\cdot3^4}\)

\(=\dfrac{3^{29}\cdot8}{4\cdot3^4}\)

\(=3^{25}\cdot2\)

@ Huỳnh Thanh Phong

Sao lại lấy 3²⁹ là thừa số chung ạ

vì 3³⁰=3²⁹.3¹ á bạn