Xin chao ban theo doi loi giai cua minh nhe.

Ap dung BDT AM-GM ta co:

\(3VT=\sqrt[3]{27a}+\sqrt[3]{27b}+\sqrt[3]{27c}\)

\(=\sqrt[3]{\left(a+b+c\right)a\cdot3\cdot3}+\sqrt[3]{\left(a+b+c\right)b\cdot3\cdot3}+\sqrt[3]{\left(a+b+c\right)c\cdot3\cdot3}\)

\(\le\frac{a+b+c+3a+3}{3}+\frac{a+b+c+3b+3}{3}+\frac{a+b+c+3c+3}{3}\)

\(\le\frac{a+b+c+3a+3}{3}+\frac{a+b+c+3b+3}{3}+\frac{a+b+c+3c+3}{3}\)

\(=2\left(a+b+c\right)+3=2\cdot3+3=9\)

Hay \(3VT\le9\Leftrightarrow VT\le3\)

Dau "=" khi \(a=b=c=1\)(ban tu lam ro phan nay nhe!)

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

\(=\sqrt{x^1+x}.x+3\sqrt{x^2+1}=x^2+3x+1\)

\(=\sqrt{x^1+1}.\left(x+3\right)=x^3+3x+1\)

\(=\left[\sqrt{x^1+1}.\left(x+3\right)\right]^2=\left(x^2+3x+1\right)\)

\(=x^4+6x^3+10x^2+6x+9=x^4+6x^3+11x^2+6x+1\)

\(=x=2\sqrt{2};-2\sqrt{2}\)

\(7x-11\sqrt{x}+4=0\)

\(=7x-11\sqrt{x}+4-\left(7x+4\right)=0-\left(7+4x\right)\)

\(=\left(-11\sqrt{x}\right)^2=\left(-7x-4\right)^2\)

\(=121x=49x^2+56x+16\)

\(\Rightarrow x=1;\frac{16}{49}\)

dkxd: x >= 0

7x + 4 = 11Vx

49x^2 + 16 + 56x = 121x

49x^2 - 65x + 16 = 0

49x^2 - 49x - 16x + 16 = 0

49x(x-1) - 16(x-1) = 0

(49x-16)(x-1) = 0

49x-16=0 hoac x-1=0

x=16/49 hoac x=1

\(\frac{-20+32\sqrt{7}}{9}\)

các bn trình bày bài giải cho mk nha :D

Ta có: \(\hept{\begin{cases}\sqrt{0,2}>0\\1=\sqrt{1}< \sqrt{3}\Rightarrow1-\sqrt{3}< 0\end{cases}\Rightarrow1-\sqrt{3}< \sqrt{0,2}}\)

Ta có: \(\hept{\begin{cases}\sqrt{0,5}>0\\\sqrt{3}< \sqrt{4}=2\Rightarrow\sqrt{3}-2< 0\end{cases}\Rightarrow\sqrt{0,5}>\sqrt{3}-2}\)

7 √x+2 chứ ko phải 1√x+2

\(5\sqrt[3]{x+1}+7\sqrt{x+2}+5\sqrt[3]{x+3}=0\)

ĐKXĐ: \(x\ge-2\)

\(\Rightarrow VT\ge5\sqrt[3]{-2+1}+0+5\sqrt[3]{-2+3}=0\)

Dấu = xảy ra khi \(x=-2\)