ĐKXĐ : \(\hept{\begin{cases}a\ne0\\a\le1\\a\ge1\end{cases}\Rightarrow a=1}\)

\(A=\sqrt{1-1}+\sqrt{1\left(1-1\right)}+1.\sqrt{\frac{1-1}{1}}=0\)

Cảm ơn bạn nhé nhưng sai mất rồi

minh chiu ban ah thng cam

ĐK: \(x\ge0;y\ge1;z\ge2\)

Theo AM-GM thì \(\sqrt{x}\le\frac{x+1}{2};\sqrt{y-1}\le\frac{y}{2};\sqrt{z-2}\le\frac{z-1}{2}\)

\(\Rightarrow\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{x+1}{2}+\frac{y}{2}+\frac{z-1}{2}\)

\(\Leftrightarrow\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{1}{2}\left(x+y+z\right)\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}2\sqrt{x}=x+1\\2\sqrt{y-1}=y\\2\sqrt{z-2}=z-1\end{cases}}\Leftrightarrow x=1;y=2;z=3\)

Vậy \(x=1;y=2;z=3\)

Mình cần gấp ạ!! 

\(a+b+c=3\Rightarrow b=3-a-c\)

\(\Leftrightarrow a+a\left(3-a-c\right)+2ac\left(3-a-c\right)\le\frac{9}{2}\)

\(\Leftrightarrow f\left(a\right)=\left(2c+1\right)a^2+\left(2c^2-5c-4\right)a-\frac{9}{2}\ge0\)

thấy f(a) là một tam thức bậc 2 của a có hệ số a²>=0 và lại có

\(\Delta=\left(2c^2-5c-4\right)^2-48\left(2c+1\right)=\left(2c-1\right)^2\left(c^2-4c-2\right)\le0\)

đúng do 0=<c=<3

=> f(a) >=0

dấu "=" xảy ra khi \(a=\frac{3}{2};b=1;c=\frac{1}{2}\)

Đk: \(x\ge\frac{-3}{2}\)

Bất pt <=> \(2x+3+x+2+2\sqrt{2x^2+7x+6}\le1\)

<=> \(2\sqrt{2x^2+7x+6}\le-4-3x\)

<=> \(\hept{\begin{cases}-3-4x\ge0\\4\left(2x^2+7x+6\right)\le16+24x+9x^2\end{cases}}\)

<=> \(\hept{\begin{cases}x\le-\frac{3}{4}\\x^2-4x-8\ge0\end{cases}}\)

<=> \(\hept{\begin{cases}x\le-\frac{3}{4}\\\left(x-2\right)^2\ge12\end{cases}}\)

<=> \(x\le2-\sqrt{12}\)

Đối chiếu đk: \(-\frac{3}{2}\le x\le2-\sqrt{12}\)

ĐỀ BÀI SAI RỒI !!!!!!

1 SỐ CÓ DẠNG \(\sqrt{x}\ge0\)   VÀ     \(x\ge0\left(đkxđ\right)\)

NHƯNG \(\sqrt{3}< 2\)

=> \(\sqrt{3}-2< 0\)

=> \(\sqrt{\sqrt{3}-2}\)KO TỒN TẠI !!!!!.

Đặt  \(A=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right).\sqrt{\sqrt{3}-2}\)

\(\Rightarrow A^2=\left(\sqrt{6}+\sqrt{2}\right)^2.\left(\sqrt{3}-2\right)^2.\left(\sqrt{3}-2\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right).\left(-1-4\sqrt{3}\right)\left(\sqrt{3}-2\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right).\left(-\sqrt{3}+2-12+8\sqrt{3}\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right)\left(7\sqrt{3}-10\right)\)

\(\Leftrightarrow A^2=56\sqrt{3}-80+84-40\sqrt{3}\)

\(\Leftrightarrow A^2=16\sqrt{3}+4\)

\(\Rightarrow A=\pm\sqrt{16\sqrt{3}+4}\)

\(P=x+\frac{9}{x-2}+2018\)

\(=\left(x-2\right)+\frac{9}{x-2}+2020\)

\(\ge2\sqrt{\frac{\left(x-2\right)9}{x-2}}+2020\)

\(=2\sqrt{9}+2020=2026\)

Dấu = xảy ra khi và chỉ khi \(x=5\)

Vậy \(Min_P=2026\)khi \(x=5\)

\(P=\left(x-2\right)+\frac{9}{x-2}+2020\)

\(P\ge2.\sqrt{\frac{\left(x-2\right).9}{x-2}}+2020\)

=> \(P\ge6+2020=2026\)

"=" xảy ra <=> \(x-2=\frac{9}{x-2}\)

<=> \(\left(x-2\right)^2=9\)

<=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Do \(x>2\)    => \(x=5\)

VẬY P MIN = 2026 <=> x = 5.