\(A=\sqrt{\left(x+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{7}}{2}\right)^2}+\sqrt{\left(\dfrac{1}{2}-x\right)^2+\left(\dfrac{\sqrt{7}}{2}\right)^2}\)

\(A\ge\sqrt{\left(x+\dfrac{1}{2}+\dfrac{1}{2}-x\right)^2+\left(\sqrt{7}\right)^2}=2\sqrt{2}\)

\(A_{min}=2\sqrt{2}\) khi \(x+\dfrac{1}{2}=\dfrac{1}{2}-x\Leftrightarrow x=0\)

Bạn cũng có thể bình phương A lên

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(M=A\cdot B=\dfrac{x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

=>\(M=\dfrac{x}{\sqrt{x}+2}\)

=>\(M=\dfrac{x-4+4}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{4}{\sqrt{x}+2}\)

=>\(M=\sqrt{x}+2+\dfrac{4}{\sqrt{x}+2}-4\)

=>\(M>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{4}{\sqrt{x}+2}}-4=0\)

Dấu '=' xảy ra khi \(\sqrt{x}+2=\sqrt{4}=2\)

=>\(\sqrt{x}=0\)

=>x=0(nhận)

b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

A = \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)  ; Đk \(x\) \(\ge\) 0

A = 2  - \(\dfrac{3}{\sqrt{x}+1}\) 

\(\sqrt{x}\) \(\ge\) 0 => \(\sqrt{x}\) + 1 \(\ge\) 1

\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) \(\dfrac{3}{1}\)

\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) 3

- \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -3

2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) 2 - 3

2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -1

A(min ) = -1  dấu bằng xảy ra khi  \(\sqrt{x}\) = 0 hay \(x\) = 0

Sửa đề: Tìm GTLN

\(A=\dfrac{-3\sqrt{x}-6+11}{\sqrt{x}+2}=-3+\dfrac{11}{\sqrt{x}+2}< =\dfrac{11}{2}-3=\dfrac{5}{2}\)

Dấu = xảy ra khi x=0

a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)