Theo giả thiết, ta có: \(ab+bc+ca+abc=4\)\(\Leftrightarrow\left(ab+bc+ca\right)+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8\)\(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)

Áp dụng bất đẳng thức Cauchy, ta được: \(P=\frac{4}{\left[\left(a+b\right)^2+4\right]+16}+\frac{4}{\left[\left(b+c\right)^2+4\right]+16}+\frac{4}{\left[\left(c+a\right)^2+4\right]+16}\)\(\le\frac{4}{4\left(a+b\right)+16}+\frac{4}{4\left(b+c\right)+16}+\frac{4}{4\left(c+a\right)+16}\)\(=\frac{1}{\left(a+2\right)+\left(b+2\right)}+\frac{1}{\left(b+2\right)+\left(c+2\right)}+\frac{1}{\left(c+2\right)+\left(a+2\right)}\)\(\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)

Đẳng thức xảy ra khi a = b = c = 1

ĐK: \(x+y\ge0\)

pt (1) \(\Leftrightarrow\sqrt{x+y+1}-\sqrt{3\left(x+y\right)}=4\left(x+y\right)^2-1\)

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

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

\(\Leftrightarrow2x+2y-1=0\)

Từ đó ta có hệ

\(\hept{\begin{cases}2x+2y-1=0\\2x-y=\frac{3}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{6}\end{cases}}}\)

\(ĐK:x\ge\frac{-7}{2}\)

\(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+7x+2x+7\)\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+\left(7x-7\sqrt{2x+7}\right)=0\)\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)         \(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

TH1: \(x-\sqrt{2x+7}=0\left(ĐK:x\ge0\right)\Leftrightarrow x=\sqrt{2x+7}\Leftrightarrow x^2-2x-7=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1+2\sqrt{2}\left(tm\right)\\x=1-2\sqrt{2}\left(L\right)\end{cases}}\)

TH2: \(x+7=\sqrt{2x+7}\Leftrightarrow x^2+12x+42=0\)(vô nghiệm)

Vậy nghiệm duy nhất của phương trình là \(x=1+2\sqrt{2}\)

ĐK: \(x\ge1\)

Áp dụng BĐT Bu-nhi-a-cốp-xki ta có: \(\left(\text{ax}+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)

Dấu '=' xảy ra khi ay=bx

Ta có: \(\left(2.\sqrt{x-1}+x.5\right)^2\le\sqrt{\left(x^2+4\right)\left(x-1+25\right)}=\sqrt{\left(x^2+4\right)\left(x+24\right)}\)

Dấu '=' xảy ra khi \(\sqrt{x-1}.x=2.5\Leftrightarrow x^2\left(x-1\right)=100\Leftrightarrow x^3-x^2-100=0\)

\(\Leftrightarrow x^3-5x^2+4x^2-100=0\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+4x+20\right)=0\Leftrightarrow x=5\)

Vậy pt có nghiệm x=5

Ta có: \(x+y+z=0\); \(x^2+y^2+z^2=a^2\)

\(\Rightarrow\left(x+y+z\right)^2=0\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

\(\Leftrightarrow a^2+2\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow2\left(xy+yz+xz\right)=-a^2\)

\(\Leftrightarrow xy+yz+xz=-\frac{a^2}{2}\)

\(\Rightarrow\left(xy+yz+xz\right)^2=\left(-\frac{a^2}{2}\right)^2\)

\(\Leftrightarrow\left(xy\right)^2+\left(yz\right)^2+\left(xz\right)^2+2\left(x^2y+y^2z+z^2x\right)=\frac{a^4}{4}\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)=\frac{a^4}{4}\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=\frac{a^4}{4}\)( vì \(x+y+z=0\))

Ta có: \(x^2+y^2+z^2=a^2\)

\(\Rightarrow\left(x^2+y^2+z^2\right)^2=\left(a^2\right)^2\)

\(\Leftrightarrow x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)=a^4\)

\(\Leftrightarrow x^4+y^4+z^4+2.\frac{a^4}{4}=a^4\)

\(\Leftrightarrow x^4+y^4+z^4+\frac{a^4}{2}=a^4\)

\(\Leftrightarrow x^4+y^4+z^4=a^4-\frac{a^4}{2}=\frac{a^4}{2}\)

a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

\(=\frac{\left(\sqrt{x}\right)^4-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}.\left(1+2\sqrt{x}\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

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

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

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

b) Ta có: \(B=x-\sqrt{x}+1=x-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\)

Với \(x>0\)\(\Rightarrow\sqrt{x}>0\)\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)\(\Rightarrow B\ge\frac{3}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)( thỏa mãn ĐKXĐ )

Vậy \(minB=\frac{3}{4}\)\(\Leftrightarrow x=\frac{1}{4}\)