\(\frac{a}{5}=\frac{b}{6}=>\frac{a^2}{25}=\frac{b^2}{36}=>\frac{2a^2}{50}=\frac{b^2}{36}\)

áp dụng .... ta có;

\(\frac{2a^2}{50}=\frac{b^2}{36}=\frac{2a^2-b^2}{50-36}=\frac{56}{14}=4\)

từ 2a^2/50=4=>2a^2=200=>a^2=100=>a=+10

b^2/36=4=>b^2=144=>b=+12

vì a;b là 2 số dương >a=10;b=12

khi đó a+b=10+12=22

tick đúng cho tớ nhé

a+b=22

a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{3b+5d}{3a+5c}=\dfrac{3b+5d}{3bk+3dk}=\dfrac{1}{k}\)

\(\dfrac{b-2d}{a-2c}=\dfrac{b-2d}{bk-2dk}=\dfrac{1}{k}\)

=>\(\dfrac{3b+5d}{3a+5c}=\dfrac{b-2d}{a-2c}\)

b: \(\dfrac{ab}{a^2-b^2}=\dfrac{bk\cdot b}{b^2k^2-b^2}=\dfrac{k}{k^2-1}\)

\(\dfrac{cd}{c^2-d^2}=\dfrac{dk\cdot d}{d^2k^2-d^2}=\dfrac{k}{k^2-1}\)

=>ab/a^2-b^2=cd/c^2-d^2

c: \(\dfrac{a^2+b^2}{\left(a+b\right)^2}=\dfrac{b^2k^2+b^2}{\left(bk+b\right)^2}=\dfrac{k^2+1}{\left(k+1\right)^2}\)

\(\dfrac{c^2+d^2}{\left(c+d\right)^2}=\dfrac{d^2k^2+d^2}{\left(dk+d\right)^2}=\dfrac{k^2+1}{\left(k+1\right)^2}\)

=>\(\dfrac{a^2+b^2}{\left(a+b\right)^2}=\dfrac{c^2+d^2}{\left(c+d\right)^2}\)

\(\frac{a}{5}=\frac{b}{6}\Rightarrow\frac{a^2}{25}=\frac{b^2}{36}=\frac{2a^2-b^2}{50-36}=\frac{56}{14}=4\)

\(\Rightarrow\) a² = 100; b² = 144

\(\Rightarrow\) a = 10; b = 12 (vì a,b > 0)

\(\Rightarrow\) a + b = 10 + 12 = 22

bai 1

=ax5-x5-9xy-4xy-7x

=ax5-(5x+7x)-(9xy+4xy)

=5ax-12x-13xy

2

M=4a+ab-2b+2a-2b+ab

=6a+2ab-4b

n=6a+2b-ab+2a

=8a+2b-ab

m-n=6a+2ab-4b-8a-2b+ab

=3ab-2a-6b

22 tick mk nha bạn 

22 tick nhatrần cao sang