a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

b) b = a - c => b + c = a

\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)

\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)

1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)

\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm 

2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)

tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1

3) kiểm tra lại xem đề đã chuẩn chưa

b. Sử dụng các hằng đẳng thức

 \(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)

và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Do (a - b) + (b - c) + (c - a) =  0 nên áp dụng hđt  \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:

\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)

Bài 1 :

\(b,ax^2+3ax+9=a^2\) 

\(\Leftrightarrow a^2x+3ax+9-a^2=0\) 

\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\) 

\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)

Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\) 

\(\Leftrightarrow ax=a-3\) 

Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\) 

c.Ta có \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{xz}-\frac{2}{xy}+\frac{2}{yz}=1\)

Do x = y + z nên \(\frac{-2}{xz}-\frac{2}{xy}+\frac{2}{yz}=\frac{-2y-2z+2\left(y+z\right)}{\left(y+z\right)zy}=0\)

Vậy nên \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1.\)

a)\(=\frac{ab+a-b-1}{ab-b+a-1}=1\)(Nhân phá ngoặc)

b)\(=\frac{2a+2ab-b-1}{6ab-3b+6a-3}\)(Nhân phá ngoặc)

\(=\frac{2ab+2a-b-1}{3\left(2ab+2a-b-1\right)}=\frac{1}{3}\)