a)\(M=\frac{y^4+1}{x^2y^4+2y^4+x^2+2}\)

\(M=\frac{y^4+1}{y^4\left(x^2+2\right)+\left(x^2+2\right)}\)

\(M=\frac{y^4+1}{\left(y^4+1\right)\left(x^2+2\right)}\)

\(M=\frac{1}{\left(x^2+2\right)}\left(y^4+1\ne0\right)\)

b) M<1 thì phải~

Ta có: \(x^2\ge0\forall x\Rightarrow x^2+2\ge2\forall x\)

\(\Rightarrow\frac{1}{x^2+2}\le\frac{1}{2}< 1\)

\(\Rightarrow M< 1\)

   đpcm

Ta có: \(M\le\frac{1}{2}\)( ý b)

\(M=\frac{1}{2}\Leftrightarrow x^2+2=2\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Vậy \(M_{max}=\frac{1}{2}\Leftrightarrow x=0\)

Tham khảo nhé~

Với mọi x; y thì phân thức M đều xác định ( vì mẫu lớn hơn 0 )

a) \(M=\frac{y^4+1}{x^2y^4+2y^4+x^2+2}\)

\(M=\frac{y^4+1}{y^4\left(x^2+2\right)+\left(x^2+2\right)}\)

\(M=\frac{y^4+1}{\left(x^2+2\right)\left(y^4+1\right)}\)

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

b) *đề phải là c/m M luôn bé hơn 1*

Dễ thấy \(x^2+2>1\forall x\)

\(\Rightarrow M< 1\forall x;y\) ( vì tử số bé hơn mẫu số )

c) \(M=\frac{1}{x^2+2}\)

Vì \(x^2\ge0\forall x\)

\(\Rightarrow M\le\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Vậy M_max = 1/2 khi và chỉ khi x = 0

\(B=\frac{x+2}{x+3}\cdot\frac{x+3}{x+4}:\frac{x+4}{x+5}\cdot\frac{\left(x+4\right)^2}{x+5}\)

\(B=\frac{x+2}{x+3}\cdot\frac{x+3}{x+4}\cdot\frac{x+5}{x+4}\cdot\frac{\left(x+4\right)^2}{x+5}\)

\(B=\frac{\left(x+2\right)\left(x+3\right)\left(x+5\right)\left(x+4\right)^2}{\left(x+3\right)\left(x+4\right)\left(x+4\right)\left(x+5\right)}\)

\(B=\frac{x+2}{1}\)

\(B=x+2\)

\(B=\frac{x+2}{x+3}.\frac{x+3}{x+4}.\frac{x+4}{x+5}.\frac{\left(x+4\right)^2}{x+5}\)

\(B=\frac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+4\right)^2}{\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+5\right)}\)

\(B=\frac{\left(x+2\right)\left(x+4\right)^2}{\left(x+5\right)^2}\)

Đặt biểu thức cần tìm =A

Theo bài ra ta có:

A=2.1+3.(-1)+...+199(-1)+200.1

 =2-3+4-5+...-199+200

 =(-1)+(-1)+...+(-1)+200(có [(200-2):2=99 số -1]

 =-99+200

 =101

Đặt \(A=2x^2+3x^3+4x^4+....+199x^{199}+200x^{200}\)

\(=\left(2x^2+4x^4+...+200x^{200}\right)+\left(3x^3+5x^5+...+199x^{199}\right)\)

vì \(x=-1\)

\(\Rightarrow A=\left(2+4+...+200\right)-\left(3+5+...+199\right)\)

\(A=10100-9999=101\)

p/s:1cách nữacủa bài này =))

\(4x^3-3x^2+5x-21=4x^3-7x^2+4x^2-7x+12x-21\)

\(=x^2\left(4x-7\right)+x\left(4x-7\right)+3\left(4x-7\right)=\left(4x-7\right)\left(x^2+x+3\right)\)

\(\frac{x^2+y^2+z^2-2xy-2yz+2zx}{x^2-2xy+y^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}=\frac{x-y+z}{x-y-z}\)

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{3x^2\left(x-1\right)-4x\left(x-1\right)+\left(x-1\right)}{2x^2\left(x-1\right)+x\left(x-1\right)-3\left(x-1\right)}\)

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

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

Áp dụng bất dẳng thức Cauchy - Schwartz dạng engel, ta có: 

 \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{a+b+b+c+c+a}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)

Dấu "=" xảy ra khi: \(\frac{a}{a+b}=\frac{b}{b+c}=\frac{c}{c+a}\) 

zzzzzz

\(\frac{\left(2a-1\right)^3+\left(a-1\right)^3}{a^3-\left(a-1\right)^3}=\frac{8a^3-12a^2+6a-1+a^3-3a^2+3a-1}{a^3-\left(a^3-3a^2+3a-1\right)}\)

\(=\frac{9a^3-15a^2+9a-2}{3a^2-3a+1}=\frac{3a\left(3a^2-3a+1\right)-2\left(3a^2-3a+1\right)}{3a^2-3a+1}=3a-2\)