thay điểm A vào đồ thị hàm số thì ta có:

3=-2+b

=>b=5

\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)

\(pt\Leftrightarrow4x^2+12+\sqrt{x-1}=4x\sqrt{5x-1}+4\sqrt{9-5x}\)

\(\Leftrightarrow4x^2-4+\sqrt{x-1}=4x\sqrt{5x-1}-8+4\sqrt{9-5x}-8\)

\(\Leftrightarrow4\left(x^2-1\right)+\sqrt{x-1}=\frac{16x^2\left(5x-1\right)-64}{4x\sqrt{5x-1}+8}+\frac{16\left(9-5x\right)-64}{4\sqrt{9-5x}+8}\)

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)+\frac{x-1}{\sqrt{x-1}}=\frac{80x^3-16x^2-64}{4x\sqrt{5x-1}+8}+\frac{80-80x}{4\sqrt{9-5x}+8}\)

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)+\frac{x-1}{\sqrt{x-1}}-\frac{16\left(x-1\right)\left(5x^2+4x+4\right)}{4x\sqrt{5x-1}+8}+\frac{80\left(x-1\right)}{4\sqrt{9-5x}+8}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4\left(x+1\right)+\frac{1}{\sqrt{x-1}}-\frac{16\left(5x^2+4x+4\right)}{4x\sqrt{5x-1}+8}+\frac{80}{4\sqrt{9-5x}+8}\right)=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

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

\(pt\Leftrightarrow\sqrt{x+1}-2-\left(\sqrt{x-2}-1\right)=0\)

\(\Leftrightarrow\frac{x+1-4}{\sqrt{x+1}+2}-\frac{x-2-1}{\sqrt{x-2}+1}=0\)

\(\Leftrightarrow\frac{x-3}{\sqrt{x+1}+2}-\frac{x-3}{\sqrt{x-2}+1}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x+1}+2}-\frac{1}{\sqrt{x-2}+1}\right)=0\)

Suy ra x=3

Đk:\(-2\le x\le3\)

\(pt\Leftrightarrow\sqrt{x+2}-\left(\frac{1}{3}x+\frac{4}{3}\right)+\sqrt{3-x}-\left(-\frac{1}{3}x+\frac{5}{3}\right)=0\)

\(\Leftrightarrow\frac{x+2-\left(\frac{1}{3}x+\frac{4}{3}\right)^2}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{3-x-\left(-\frac{1}{3}x+\frac{5}{3}\right)^2}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}=0\)

\(\Leftrightarrow\frac{x+2-\frac{x^2+8x+16}{9}}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{3-x-\frac{x^2-10x+25}{9}}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}=0\)

\(\Leftrightarrow\frac{-\frac{x^2-x-2}{9}}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{-\frac{x^2-x-2}{9}}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}=0\)

\(\Leftrightarrow-\frac{x^2-x-2}{9}\left(\frac{1}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{1}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}\right)=0\)

\(\Leftrightarrow-\frac{\left(x+1\right)\left(x-2\right)}{9}\left(\frac{1}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{1}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}\right)=0\)

Suy ra x=-1;x=2

Ta có bất đẳng thức Cauchy với 2 số a,b không âm :\(\frac{a+b}{2}\ge\sqrt{ab}\)

a)Gọi độ dài 2 cạnh liên tiếp của hình chữ nhật là a,b->a+b=k không đổi

->Shcn=ab\(\le\frac{\left(a+b\right)^2}{4}\)=\(\frac{k^2}{4}\)

Dấu "=" xảy ra <=>a=b<=> hình vuông

b)Gọi độ dài 2 cạnh liên tiếp của hình chữ nhật là a,b->ab=k không đổi

Chu Vi HCN=2(a+b)\(\ge\)\(4\sqrt{ab}\)=4\(\sqrt{k}\)

Dấu "=" xảy ra <=> a=b <=>Hình vuông

a ) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)

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

\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

\(\ge\left|x-1+3-x\right|+\left|x-2\right|=\left|x-2\right|+2\ge2\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\\left|x-2\right|=0\end{cases}\Rightarrow x=2}\)(TM)

Vậy \(A_{min}=2\Leftrightarrow x=2\)

b ) \(B=\sqrt{x-2\sqrt{x-1}}-\sqrt{x+2\sqrt{x-1}}\)

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

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

\(=\left|\sqrt{x-1}-1\right|-\left|\sqrt{x-1}+1\right|\)

\(\le\left|\sqrt{x-1}-1-\sqrt{x-1}-1\right|=2\)có GTLN là 2