Áp dụng BĐT cô-si, ta có:

\(\frac{1}{\left(x+1\right)}+\frac{1}{\left(y+1\right)}+\frac{1}{\left(z+1\right)}\ge3\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}\ge1-\frac{1}{\left(y+1\right)}+1-\frac{1}{\left(z+1\right)}\)

\(\Leftrightarrow\frac{y}{\left(y+1\right)}+\frac{z}{\left(z+1\right)}\ge3\sqrt{\left(\frac{yz}{\left(y+1\right)\left(z+1\right)}\right)}\)

Ta có:

\(\frac{1}{\left(x+1\right)}\ge3\sqrt{\frac{yz}{\left(x+1\right)\left(y+1\right)}}\)(1)

\(\Leftrightarrow\frac{1}{\left(y+1\right)}\ge3\sqrt{\left(\frac{xy}{\left(x+1\right)\left(z+1\right)}\right)}\)(2)

\(\Leftrightarrow\frac{1}{\left(z+1\right)}\ge3\sqrt{\left(\frac{xy}{\left(x+1\right)\left(y+1\right)}\right)}\)(3)
Từ (1); (2) và (3), ta có:

\(\frac{1}{\left(x+1\right)}+\frac{1}{\left(y+1\right)}+\frac{1}{\left(z+1\right)}\ge8\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Rightarrow xyz\le\frac{1}{8}.\text{ dau }=\text{xay ra khi }x=y=z=\frac{1}{2}\)

\(A+7x^2y-5xy^2-xy=x^2y+8xy^2-5xy\)

\(\Rightarrow A=\left(x^2y+8xy^2-5xy\right)-\left(7x^2y-5xy^2-xy\right)\)

        \(=x^2y+8xy^2-5xy-7x^2y+5xy^2+xy\)          \(=\left(x^2y-7x^2y\right)+\left(8xy^2+5xy^2\right)+\left(-5xy+xy\right)\)

 \(=-6x^2y+13xy^2-4xy\)

A = (x\(^2\)y + 8xy\(^2\)- 5xy) - (7x\(^2\)y - 5xy\(^2\)- xy)

A = x\(^2\)y + 8xy\(^2\)- 5xy - 7x\(^2\)y + 5xy\(^2\)+ xy

A = (x\(^2\)y -7x\(^2\)y) + (8xy\(^2\)+ 5xy\(^2\)) + (-5xy +xy)

A = -6x\(^2\)y + 13xy\(^2\)- 4xy

A = (x\(^2\)y + 8xy\(^2\)- 5xy) - (7x\(^2\)y - 5xy\(^2\)-xy)

A = x\(^2\)y + 8xy\(^2\)- 5xy - 7x\(^2\)y + 5xy\(^2\)+xy

A= (x\(^2\)y -7x\(^2\)y) + (8xy\(^2\)+ 5xy\(^2\)) + (-5xy + xy)

A = -6x\(^2\)y + 13xy\(^2\)- 4xy

cái này mik chịu, mik mới có lớp 7

1. Ta có \(\left(b-a\right)\left(b+a\right)=p^2\)

Mà b+a>b-a ; p là số nguyên tố 

=> \(\hept{\begin{cases}b+a=p^2\\b-a=1\end{cases}}\)

=> \(\hept{\begin{cases}b=\frac{p^2+1}{2}\\a=\frac{p^2-1}{2}\end{cases}}\)

Nhận xét :+Số chính phương chia 8 luôn dư 0 hoặc 1 hoặc 4

Mà p là số nguyên tố 

=> \(p^2\)chia 8 dư 1

=> \(\frac{p^2-1}{2}⋮4\)=> \(a⋮4\)(1)

+Số chính phương chia 3 luôn dư 0 hoặc 1

Mà p là số nguyên tố lớn hơn 3

=> \(p^2\)chia 3 dư 1

=> \(\frac{p^2-1}{2}⋮3\)=> \(a⋮3\)(2)

Từ (1);(2)=> \(a⋮12\)

Ta có \(2\left(p+a+1\right)=2\left(p+\frac{p^2-1}{2}+1\right)=p^2+1+2p=\left(p+1\right)^2\)là số chính phương(ĐPCM)

2,     \(T=\frac{x}{1-yz}+\frac{y}{1-xz}+\frac{z}{1-xy}\)

Áp dụng cosi ta có \(yz\le\frac{y^2+z^2}{2}\)

=> \(\frac{x}{1-yz}\le\frac{x}{1-\frac{y^2+z^2}{2}}=\frac{2x}{2-y^2-z^2}=\frac{2x}{1+x^2}\)

Lại có \(x^2+\frac{1}{3}\ge2x\sqrt{\frac{1}{3}}\)

=> \(\frac{x}{1-yz}\le\frac{2x}{\frac{2}{3}+2x\sqrt{\frac{1}{3}}}=\frac{x}{\frac{1}{3}+x\sqrt{\frac{1}{3}}}\le\frac{x.1}{4}\left(\frac{1}{\frac{1}{3}}+\frac{1}{x\sqrt{\frac{1}{3}}}\right)=\frac{1}{4}.\left(3x+\sqrt{3}\right)\)

Khi đó \(T\le\frac{1}{4}.\left(3x+3y+3z+3\sqrt{3}\right)\)

Mà \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=\sqrt{3}\)

=> \(T\le\frac{6\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}\)

Vậy \(MaxT=\frac{3\sqrt{3}}{2}\)khi \(x=y=z=\frac{1}{\sqrt{3}}\)

\(M=x^2-2xy+4y^2+12xy+22\)

\(M=\left(x^2-2xy+y^2\right)+\left(3y^2+12y+12\right)+10\)

\(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=-2\) 

( Chỗ \(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\) bạn phân tích từng cái đã nhá, mình làm tắt ) 

Lời giải:

Khai triển ta có:

\(M=x^2y^2+\frac{1}{x^2y^2}+2\)

Áp dụng BĐT AM-GM:

\(1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}\)

Tiếp tục áp dụng BĐT AM-GM:

\(M=\left(x^2y^2+\frac{1}{16^2x^2y^2}\right)+\frac{255}{256x^2y^2}+2\geq 2\sqrt{\frac{1}{16^2}}+\frac{255}{256x^2y^2}+2\)

\(\Leftrightarrow M\geq \frac{17}{8}+\frac{255}{256x^2y^2}\) . Mà \(xy\leq \frac{1}{4}\)

\(\Rightarrow M\geq \frac{17}{8}+\frac{255}{256x^2y^2}\geq \frac{17}{8}+\frac{255}{256.\frac{1}{16}}=\frac{289}{16}\)

Vậy \(M_{\min}=\frac{289}{16}\Leftrightarrow x=y=\frac{1}{2}\)

Akai haruma giải dùm mình 5 bài mới đăng đi

a, Xét : 3 - E = 3x^3-3xy-3y^3-x^3-xy-y^2/x^2-xy+y^2

= 2x^2-4xy+2y^2/x^2-xy+y^2

= 2.(x^2-2xy+y^2)/x^2-xy+y^2

= 2.(x-y)^2/x^2-xy+y^2 

>= 0 ( vì x^2-xy+y^2 > 0 )

Dấu "=" xảy ra <=> x-y=0 <=> x=y

Vậy ..........

b, Có : (x+1995)^2 = x^2+3990+1995^2 = (x^2-3990x+1995^2)+7980x

= (x-1995)^2 + 7980x >= 7980x

=> M < = x/7980x = 1/7980 ( vì x > 0 )

Dấu "=" xảy ra <=> x-1995=0 <=> x=1995

Vậy ...............