\(a+b+c=3\\ \Leftrightarrow a\left(b+c+2\right)=ab+ac+a+b+c+1=\left(a+1\right)\left(b+c+1\right)\)

Tương tự:

\(b\left(c+a+2\right)=\left(b+1\right)\left(a+c+1\right)\\ c\left(a+b+2\right)=\left(c+1\right)\left(a+b+1\right)\)

Áp dụng BĐT cosi:

\(\left\{{}\begin{matrix}\left(a+1\right)\left(b+c+1\right)\le\dfrac{\left(a+1+b+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(b+1\right)\left(a+c+1\right)\le\dfrac{\left(b+1+a+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(c+1\right)\left(a+b+1\right)\le\dfrac{\left(c+1+a+b+1\right)^2}{2}=\dfrac{2^2}{2}=2\end{matrix}\right.\)

Cộng vế theo vế 2 BĐT trên:

\(\Leftrightarrow\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(c+a+2\right)}+\sqrt{c\left(a+b+2\right)}\le2+2+2=6\)

Dấu \("="\Leftrightarrow a=b=c=1\)

Áp dụng BĐT Bunhiacopski:

\(VT^2=\left(\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(a+c+2\right)}+\sqrt{c\left(a+b+2\right)}\right)^2\\ \le\left(a+b+c\right)\left(b+c+2+a+c+2+a+b+2\right)\\ =3\cdot\left(2\cdot3+6\right)=36\\ \Leftrightarrow VT\le\sqrt{36}=6\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{b+c+2}}{\sqrt{a}}=\dfrac{\sqrt{a+c+2}}{\sqrt{b}}=\dfrac{\sqrt{a+b+2}}{\sqrt{c}}\\a+b+c=3\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=1\)

Câu b tương tự

a/ \(\sqrt[5]{2a+b}+\sqrt[5]{2b+c}+\sqrt[5]{2c+a}\)

\(=\frac{1}{\sqrt[5]{3^4}}\left(\sqrt[5]{3^4}.\sqrt[5]{2a+b}+\sqrt[5]{3^4}.\sqrt[5]{2b+c}+\sqrt[5]{3^4}.\sqrt[5]{2c+a}\right)\)

\(\le\frac{1}{\sqrt[5]{3^4}}\left(\frac{3+3+3+3+2a+b}{5}+\frac{3+3+3+3+2b+c}{5}+\frac{3+3+3+3+2c+a}{5}\right)\)

\(=\frac{1}{\sqrt[5]{3^4}}\left(\frac{36}{5}+\frac{3\left(a+b+c\right)}{5}\right)\)

\(=\frac{1}{\sqrt[5]{3^4}}.9=3\sqrt[5]{3}\)

b/ \(\sqrt[5]{a\left(a+c\right)\left(2a+b\right)}+\sqrt[5]{b\left(b+a\right)\left(2b+c\right)}+\sqrt[5]{c\left(c+b\right)\left(2c+a\right)}\)

\(\frac{1}{\sqrt[5]{6^4}}.\left(\sqrt[5]{6^2}.\sqrt[5]{6.a.3.\left(a+c\right).2.\left(2a+b\right)}+\sqrt[5]{6^2}.\sqrt[5]{6.b.3.\left(b+a\right).2.\left(2b+c\right)}+\sqrt[5]{6^2}.\sqrt[5]{6.c.3.\left(c+b\right).2.\left(2c+a\right)}\right)\)

\(\le\frac{1}{\sqrt[5]{6^4}}.\left(\frac{6+6+6a+3\left(a+c\right)+2\left(2a+b\right)}{5}+\frac{6+6+6b+3\left(b+a\right)+2\left(2b+c\right)}{5}+\frac{6+6+6c+3\left(c+b\right)+2\left(2c+a\right)}{5}\right)\)

\(=\frac{1}{\sqrt[5]{6^4}}.\left(\frac{36}{5}+\frac{18\left(a+b+c\right)}{5}\right)\)

\(=\frac{1}{\sqrt[5]{6^4}}.18=3\sqrt[5]{6}\)

\(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

b. 

\(=\sqrt{3-\sqrt{5}}.\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+\sqrt{3+\sqrt{5}}.\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=\sqrt{3-\sqrt{5}}.\sqrt{9-5}+\sqrt{3+\sqrt{5}}.\sqrt{9-5}\)

\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{10}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}\)

\(=\sqrt{10}-\sqrt{2}+\sqrt{10}+\sqrt{2}=2\sqrt{10}\)

c.

\(\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}}=\dfrac{\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)}{\sqrt{b}.\sqrt{b}}=\dfrac{a^2-b}{b}\)

cảm ơn mọi người nhé mk ra rồi

bn gửi lên cho các bn cùng tham khảo đi! ^-^

vào câu hỏi ttự cũng có đấy =.=

Lời giải:

Do $ab+bc+ac=5$ nên:

\(a^2+5=a^2+ab+bc+ac=(a+b)(a+c)\)

\(b^2+5=b^2+ab+bc+ac=(b+c)(b+a)\)

\(c^2+5=c^2+ab+bc+ac=(c+a)(c+b)\)

Do đó:

\(A=a\sqrt{\frac{(b+c)(b+a)(c+a)(c+b)}{(a+b)(a+c)}}+b\sqrt{\frac{(a+b)(a+c)(c+a)(c+b)}{(b+c)(b+a)}}+c\sqrt{\frac{(a+b)(a+c)(b+c)(b+a)}{(c+a)(c+b)}}\)

\(=a\sqrt{(b+c)^2}+b\sqrt{(c+a)^2}+c\sqrt{(a+b)^2}=a(b+c)+b(c+a)+c(a+b)\)

\(=2(ab+bc+ac)=2.5=10\)

Hai bài giống hệt nhau về cách làm:

Cho a, b, c > 0 thoả mãn: \(a b c=\sqrt{a} \sqrt{b} \sqrt{c}=2\). Chứng minh rằng: \(\dfrac{\sqrt{a}}{a 1} \dfrac{\sqrt{... - Hoc24

`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`

`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`

`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`

`=sqrt2(sqrt5+1+sqrt5-1)`

`=sqrt{2}.2sqrt5`

`=2sqrt{10}`

`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`

`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`

`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`

`=2(7-3)`

`=8`

`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`

`=2/sqrt2=sqrt2`

`d)D=\sqrt{2+sqrt3}+sqrt{14-5sqrt3}+sqrt2`

`=>sqrt2D=sqrt{4+2sqrt3}+sqrt{28-10sqrt3}+2`

`=>sqrt2D=sqrt{(sqrt3+1)^2}+sqrt{(5-sqrt3)^2}+2`

`=>sqrt2D=8`

`=>D=4sqrt2`