Đề bài còn cho x,y >0 nữa

a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )

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

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

\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )

a) Với \(x>0\)và \(x\ne1\)ta có:

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

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

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

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

b) Với \(x>0\)và \(x\ne4\)ta có: 

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

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

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

\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)

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vậy em giải giùm chị nhé

Có \(\sin^2a+\cos^2a=1\)\(\Leftrightarrow\sin^2a=1-\cos^2a=1-\left(\frac{1}{3}\right)^2=\frac{8}{9}\)

\(\Leftrightarrow\sin a=\frac{\sqrt{8}}{3}\)

Xét  \(B=\frac{\sin a-3\cos a}{\sin a+2\cos a}=\frac{\frac{\sqrt{8}}{3}-3\cdot\frac{1}{3}}{\frac{\sqrt{8}}{3}+2\cdot\frac{1}{3}}=\frac{7-5\sqrt{2}}{2}\)

Áp dụng bất đẳng thức AM - GM, ta được: \(2yz+2=x^2+\left(y^2+2yz+z^2\right)=x^2+\left(y+z\right)^2\ge2\sqrt{x^2.\left(y+z\right)^2}=2x\left(y+z\right)\Rightarrow yz+1\ge x\left(y+z\right)\)\(\Rightarrow VT\le\frac{x^2}{x^2+x+x\left(y+z\right)}+\frac{y+z}{x+y+z+1}+\frac{1}{xyz+3}=\frac{x+y+z}{x+y+z+1}+\frac{1}{xyz+3}\)

• Nếu \(x+y+z\le2\)thì \(VT\le1-\frac{1}{x+y+z+1}+\frac{1}{xyz+3}\le1-\frac{1}{3}+\frac{1}{3}=1\)
• Nếu \(x+y+z\ge2\), ta đặt x + y + z = p; xy + yz + zx = q; xyz = r thì áp dụng bất đẳng thức Schur, ta được \(VT\le\frac{p}{p+1}+\frac{1}{\frac{p\left(4q-p^2\right)}{9}+3}=\frac{p}{p+1}+\frac{9}{p^3-4p+27}\)

Khảo sát hàm trên với \(p\in\left[\sqrt{2};2\right]\)ta cũng có \(VT\le1\)

Vậy ta có: \(\frac{x^2}{x^2+yz+x+1}+\frac{y+z}{x+y+z+1}+\frac{1}{xyz+3}\le1\)

Đẳng thức xảy ra khi x = y = 1; z = 0

bài này x,y,z pk không âm

Vì \(-1\le a\le1\Rightarrow a^2\le1\)Tương tự có \(b^2\le1;c^2\le1\)

Suy ra \(P=a^2+2b^2+c^2\le1+2\cdot1+1=4\)hay \(maxP=4\)

   Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\pm1\)