\(\frac{a}{ab+a+2016}+\frac{b}{bc+b+1}+\frac{2016c}{ac+2016c+2016}\)

\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ac+abc^2+abc}\)

\(=\frac{a}{a.\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{abc^2}{ac.\left(1+bc+b\right)}\)

\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)

\(=\frac{1+b+bc}{b+bc+1}=1\)

Bài 1:

a)  ĐKXĐ:  \(x\ne\pm5\)

\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(=\frac{x-5}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{2x+10}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x-5+\left(2x+10\right)-\left(2x+10\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x-5}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+5}\)

b)  \(B=9x^2-42x+49=\left(3x-7\right)^2\)

Tại  \(x=-3\)thì:   \(B=\left[3.\left(-3\right)-7\right]^2=256\)

Bài 2:

a)  ĐKXĐ:  \(x\ne\pm3\)

\(A=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

\(=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)

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

b)  \(A=4\)\(\Rightarrow\)\(\frac{4}{x-3}=4\)

\(\Rightarrow\)\(4\left(x-3\right)=4\)\(\Leftrightarrow\)\(x-3=1\)\(\Leftrightarrow\)\(x=4\)   (t/m ĐKXĐ)

Vậy....

Đặt a=4m+1, b=4n+2(m,n\(\in\)N)

=>ab=(4m+1)(4n+2)

= 16mn+8m+4n+2

Ta thấy 16mn+8m+4n chia hết cho 4

=> ab:14 dư 2

làm a) còn b);c) tương tự

A = (a + b)²  - 2ab = 100 - 8  = 92

a. Có a\(^2\) + b\(^2\) = a\(^2\) + 2ab + b\(^2\) - 2ab

\(\Rightarrow\) a\(^2\) + b\(^2\) = ( a + b ) \(^2\) - 2ab (1)

Thay a + b = 10, ab = 5 vào (1 ) ta có :

a\(^2\) + b\(^2\) = 10\(^2\) - 2 . 5 = 90

KL:.............

b. Có ( a + b ) ( a\(^2\) + b\(^2\) ) = a\(^3\) + ab\(^2\) + a\(^2\)b + b\(^3\) 

\(\Rightarrow\) ( a + b ) ( a\(^2\) + b\(^2\) ) = a\(^3\) + ab ( a + b ) + b\(^3\) ( 2)

Thay a + b = 10, a\(^2\) + b \(^2\) = 90 ( CMa) , ab = (5) vào (2) ta có : 

........................

a) vì a+b=10

=> \(\left(a+b\right)^2=10^2=100\)

\(< =>a^2+2ab+b^2=100\)

\(< =>a^2+b^2+2.4=100\)(vì ab=4)

\(< =>a^2+b^2=100-8\)

\(< =>a^2+b^2=92\)

b) theo câu a ta có \(a^2+b^2=92\)

\(< =>\left(a^2+b^2\right)^2=92^2=8464\)

\(< =a^4+b^4+2a^2b^2=8464\)

\(< =>a^4+b^4+2.\left(ab\right)^2=8464\)

\(< =>a^4+b^4+2.4^2=8464\)

\(< =>a^4+b^4=8464-32\)

\(< =>a^4+b^4=8432\)

Bài 2: 

\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=9\)

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)}{\left(ab\right)^3}\)

\(=\dfrac{5^3-3\cdot5\cdot\left(-2\right)}{\left(-2\right)^3}=\dfrac{125+30}{8}=\dfrac{155}{8}\)

\(a-b=-\sqrt{\left(a+b\right)^2-4ab}=-\sqrt{5^2-4\cdot\left(-2\right)}=-\sqrt{33}\)

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(a^2+b^2=\frac{a^2}{1}+\frac{b^2}{1}\ge\frac{\left(a+b\right)^2}{1+1}=\frac{1^2}{2}=\frac{1}{2}\left(đpcm\right)\)

Đẳng thức xảy ra <=> a = b

úi xin lỗi bài kia thiếu ._. Đẳng thức xảy ra <=> a=b=1/2 nhé

2. Ta có : a³ + b³ + ab = ( a + b )( a² - ab + b² ) + ab

= a² - ab + b² + ac = a² + b² ( do a+b=1 )

Sử dụng kết quả ở bài trước ta có đpcm

Đẳng thức xảy ra <=> a=b=1/2