Áp dụng bđt cosi ta được \(4x+\frac{1}{4x}\ge2\sqrt{4x.\frac{1}{4x}}=2\)
\(x+\frac{1}{4}\ge2\sqrt{\frac{1}{4}x}=\sqrt{x}\Leftrightarrow4x+1\ge4\sqrt{x}\Leftrightarrow4\left(x+1\right)\ge4\sqrt{x}+3\Leftrightarrow-\left(4\sqrt{x}+3\right)\ge-4\left(x+1\right)\Leftrightarrow-\frac{\left(4\sqrt{x}+3\right)}{x+1}\ge-4\)Khi đó \(A\ge2-4+2016=2014\)
Dấu = xảy ra khi x=1/4

ta có: \(4x^2+9x+18\sqrt{x}+9=4x^2+9\left(\sqrt{x}+1\right)^2\),\(4x\sqrt{x}+4x=4x\left(\sqrt{x}+1\right)\)
Đặt \(a=x,b=\sqrt{x}+1\)ta có:
\(A=\frac{4a^2+9b^2}{4ab}+\frac{4ab}{4a^2+9b^2}=t+\frac{1}{t},t=\frac{4a^2+9b^2}{4ab}\)
có \(\frac{4a^2+9b^2}{4ab}=t\Rightarrow4a^2-t.4ab+9b^2=0\Leftrightarrow4.\left(\frac{a}{b}\right)^2-4t.\frac{a}{b}+9=0,\)do a khác 0.
Đặt \(\frac{a}{b}=y\Rightarrow4y^2-t.4y+9=0\), \(\Delta=16t^2-36\ge0\Leftrightarrow t\ge\frac{3}{2}\left(t>0\right)\)
xét \(f\left(t\right)=t+\frac{1}{t}\left(t\ge\frac{3}{2}\right)\)
lấy \(\frac{3}{2}< t_1< t_2\)
\(\Rightarrow f\left(t_1\right)-f\left(t_2\right)=\left(t_1-t_2\right)\left(\frac{t_1.t_2-1}{t_1.t_2}\right)< 0\)
suy ra với t càng tăng thì f(t) càng lớn vậy min \(f\left(t\right)=\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
các em tự tìm x nhé.

bài này bạn áp dụng BĐT cô si cko 2 số dương là đc.

đáp án: Min A=  2

dùng côsi ra = 1 chắc v

ê tuấn nếu cô-si thì mk nghĩ phải =2 chứ sao =1 được 

\(\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4\sqrt{x}}+\frac{4x\sqrt{x}+4\sqrt{x}}{4x^2+9x+18\sqrt{x}+9}-2=\frac{\left(-4x\sqrt{x}+4x^2+9x+22\sqrt{x}+9\right)^2}{\left(4x^2+9x+18\sqrt{x}+9\right)\left(4x\sqrt{x}+4\sqrt{x}\right)}\ge0\)

Đặt \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}\left(x>0\right)\Rightarrow M>0\)

Đặt \(y=\sqrt{x}>0\)ta có \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}=\frac{4y^4+9y^2+18y+9}{4y^3+4y^2}\)\(=\frac{3\left(4y^3+4y^2\right)+\left(4y^2-12y^3-3y^2+18y+9\right)}{4y^3+4y^2}=3+\frac{\left(2y^2-3y-3\right)^2}{4y^3+4y^2}\ge3\)

\(y>0\Rightarrow\hept{\begin{cases}4y^3+4y^2>0\\\left(2y^2-3y-3\right)^2\ge0\end{cases}\Rightarrow\frac{\left(2y-3y-3\right)^2}{4y^3+4y^2}\ge0}\)

Đẳng thức xảy ra \(\Leftrightarrow2y^2-3y-3=0\Leftrightarrow y=\frac{3+\sqrt{33}}{4}\left(y>0\right)\)

\(\Rightarrow x=\left(\frac{3+\sqrt{33}}{4}\right)^2=\frac{21+3\sqrt{33}}{8}\)

Khi đó \(A=M+\frac{1}{M}=\frac{8M}{9}+\left(\frac{M}{9}+\frac{1}{M}\right)\ge\frac{8\cdot3}{9}+2\sqrt{\frac{M}{9}\cdot\frac{1}{M}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}M=3\\\frac{M}{9}=\frac{1}{M}\end{cases}\Leftrightarrow M=3\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}}\)

Vậy \(A_{min}=\frac{10}{3}\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}\)

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

\(A=1-\left(\frac{2}{1+2\sqrt{x}}-\frac{5\sqrt{x}}{4x-1}-\frac{1}{1-2\sqrt{x}}\right):\frac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)

\(=1-\left(\frac{2\left(1-2\sqrt{x}\right)+5\sqrt{x}-1-2\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}\right):\frac{\sqrt{x}-1}{\left(1+2\sqrt{x}\right)^2}\)

\(=1-\frac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}.\frac{\left(1+2\sqrt{x}\right)^2}{\sqrt{x}-1}=1-\frac{1+2\sqrt{x}}{1-2\sqrt{x}}=2-\frac{2}{1-2\sqrt{x}}\)

để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

cái x trục căn thức trong căn đi rồi thay vô A

giải chi tiết ra jùm cái