Mọi người nhớ Cmt coi đúngko nhé.

Do 2 đường thẳng y=ax+b và y'=a'x+b' cắt nhau tại 1 điểm trên trục hoành nên y = 0

Ta có y = ax+b <=> 0 = ax+b <=> -ax = b <=> x = -b/a (1)

Tương tự ta có : x = -b'/a' (2)

Từ (1) và (2) ta có : -b/a = -b'/a' hay b=a = b'/a'

=> ba'=b'a.Đúng ko zậy mina ?

Chỗ hay b= a =b'/a' là b/a = b'/a' nhé mn.

Viết  nhầm.

be quiet bitch

= 6 nha

6-nhanh+

a) \(\sqrt{81}-\sqrt{80}.\sqrt{0,2}=9-4\sqrt{5}.\frac{\sqrt{5}}{5}=9-\frac{4.5}{5}=9-4=5\)

b) \(\sqrt{\left(2-\sqrt{5}\right)^2}-\frac{1}{2}\sqrt{20}=\left|2-\sqrt{5}\right|-\frac{1}{2}.2\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)

\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

\(=3-\sqrt{3}+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}\)

\(=3-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=3-\sqrt{3}+\sqrt{3}-1\)

\(=2\)

\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)

\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)

\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}=1-\frac{1}{4}=\frac{3}{4}\)

\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)

\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)

\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}\)

\(=1-\frac{1}{4}\)

\(=\frac{3}{4}\)