A =(a+b-2c) -(-a+b+c) -(2a-b-c)

   = a+b-2c+a-b-c-2a+b+c

   = b-2c

B=-(2a-b+c) + (b-2c-3a) -(-5a-3c+b)

  = -2a+b-c+b-2c-3a+5a+3c-b

  = b-c

C=(3a-b-2c)-( 2b+3c-a) +(2a-3b)

  = a-b-2c-2b-3c+a+2a-3b

  = -6b-5c

D=(5a-3b+c) +( 2a-3b+5) -( b-c+a)

   = 5a-3b+c+2a-3b+5-b+c-a

   = 6a-7b+2c

\(A=\left(a+b-2c\right)-\left(-a+b+c\right)-\left(2a-b-c\right)\)

\(=a+b-2c+a-b-c-2a+b+c=b-2c\)

\(B=-\left(2a-b+c\right)+\left(b-2c-3a\right)-\left(-5a-3c+b\right)\)

\(=-2a+b-c+b-2c-3a+5a+3c-b=b\)

\(C=\left(3a-b-2c\right)-\left(2b+3c-a\right)+\left(2a-3b\right)\)

\(=3a-b-2c-2b-3c+a+2a-3b=6a-6b-5c\)

\(D=\left(5a-3b+c\right)+\left(2a-3b+5\right)-\left(b-c+a\right)\)

\(=5a-3b+c+2a-3b+5-b+c-a=6a-7b+2c\)

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)

1 ) Do \(3a-b=5\Rightarrow b=3a-5\)

Ta có : \(A=\frac{5a-b}{2a+5}-\frac{3b-3a}{2b-5}=\frac{5a-3a+5}{2a+5}-\frac{3\left(3a-5\right)-3a}{2\left(3a-5\right)-5}=\frac{2a+5}{2a+5}-\frac{6a-15}{6a-15}=1-1=0\)

Vậy \(A=0\)

2 ) \(P=x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)

Để P là số nguyên tố thì \(Ư\left(P\right)=\left\{1;P\right\}\)

Vì x dương \(\Rightarrow x^2+x+1>x^2-x+1\)

\(\Rightarrow x^2-x+1=1\)

\(\Rightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=1\end{matrix}\right.\)

Vậy x = 1 thì P là số nguyên tố

a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)

\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)

\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)

\(=2a^2-b^2\)

b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)

\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)

\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)

\(=-7ab+b^2\)

c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)

\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)

\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)

\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)

\(=-7bx+3b^2+2x^2\)

d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)

\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)

\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)

\(=-5ax+32x^2-30a^2\)

a: =2ab+8a^2-b^2-4ab+2ab-6a^2

=2a^2-b^2

b: =6a^2-9ab-4ab+6b^2-6a^2+6ab

=-7ab+6b^2

c: =10bx-5b^2-16bx+8b^2+2x^2-xb

=3b^2+2x^2-7xb

d: =2xa+30x^2+5ax+2x^2-30a^2-12ax

=32x^2-30a^2-5ax

a, Vì \(a^2-b^2=4c^2\Rightarrow16a^2-16b^2=64c^2\) (1)

Ta có:\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(5a-3b\right)^2-\left(8c\right)^2\)

\(=25a^2-30ab+9b^2-64c^2\) (2)

Thay (1) vào (2) ta được

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=25a^2-30ab+9b^2-16a^2+16b^2\)

\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2\)

=> đpcm

b, \(M=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2b-b\right)^2\)

\(=4a^2+4b^2+c^2+4b^2+4c^2+a^2+4c^2+4a^2+b^2\)

\(+8ab-4ac-4bc+8bc-4ab-4ac+8ac-4bc-4ab\)

\(=9.\left(a^2+b^2+c^2\right)=9.2017=18153\)

Vậy M=18153

\(P=\frac{3a+7+2a-b-7}{3a+7}-\frac{2b-7+b-2a+7}{2b-7}\)

mà 2a-b=7 hay b-2a=-7 nên ta có

\(P=1+\frac{7-7}{3a+7}-1-\frac{-7+7}{2b-7}=1+0-1-0=0\)

a/ Ta có \(a\left(2a-5c\right)=2a^2-5ac=2bc-5ac=c\left(2b-5a\right)\Rightarrow\frac{c}{2a-5c}=\frac{a}{2b-5a}\)

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