Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Đây nè @Võ Hồng Phúc(Phúc bím)
a: \(=3\sqrt{5}-\left(\sqrt{5}-2\right)=2\sqrt{5}+2\)
b: \(=\left|a-b\right|-\left|b-c\right|-\left|c-d\right|\)
\(=b-a-\left(c-b\right)-\left(d-c\right)\)
=b-a-c+b-d+c
=2b-d-a
a: Sửa đề: \(\sqrt{9\left(a-1\right)^2}\)
\(=3|a-1|=3(a-1)=3a-3\)
b: \(=6\cdot\left|a-3\right|=6\left(3-a\right)=18-6a\)
c: \(=a\left|a+2\right|\)
d: \(=\left|a\right|\cdot\left|a-1\right|=a\left(a-1\right)\)
Ta có \(\left(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\right)^2\)\(\ge\)\(\left(a+c\right)^2+\left(b+d\right)^2\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\)\(\ge\)\(a^2+b^2+c^2+d^2\)\(+2\left(ac+bd\right)\)
\(\Leftrightarrow\)\(\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\)\(\ge\)\(ac+bd\)
\(\Leftrightarrow\)\(\left(a^2+b^2\right)\left(c^2+d^2\right)\)\(\ge\)\(\left(ac+bd\right)^2\)(*)
Vì (*) luôn đúng theo bđt bunhia copxki \(\Rightarrow\)đpcm
dấu ''='' xảy ra khi a/c=b/d
a) \(\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}=\sqrt{2}+\sqrt{3}\)
b) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
c) \(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)\(=2\sqrt{5}\)
d) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}=\sqrt{2}-\sqrt{3}-1-\sqrt{3}\)
\(=\sqrt{12}-\sqrt{2}-1\)
e) \(\sqrt{\left(\sqrt{3-1}^2\right)-\sqrt{3}}=\sqrt{\sqrt{2}^2-\sqrt{3}}=\sqrt{2-\sqrt{3}}\)
P/S: Ko chắc
Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3