\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

a)...........................

b)\(\Leftrightarrow A=\dfrac{\dfrac{x^2}{4}+x^2y+\dfrac{y}{4}+y^2+x^2y^2+\dfrac{1}{4}+\dfrac{3y}{4}}{x^2y^2+1+y^2-x^2y-y+x^2}\)

\(\Leftrightarrow A=\dfrac{\dfrac{x^2}{4}+\dfrac{1}{4}+y+x^2y+y^2+x^2y^2}{x^2\left(y^2-y+1\right)+\left(y^2-y+1\right)}\)

\(\Leftrightarrow A=\dfrac{\dfrac{\left(x^2+1\right)}{4}+y\left(x^2+1\right)+y^2\left(x^2+1\right)}{\left(y^2-y+1\right)\left(x^2+1\right)}\)

\(\Leftrightarrow A=\dfrac{\left(x^2+1\right)\left(\dfrac{1}{4}+y+y^2\right)}{\left(y^2-y+1\right)\left(x^2+1\right)}=\dfrac{4y^2+4y+1}{4\left(y^2-y+1\right)}\)(không phụ vào x)

\(\Rightarrowđpcm\)

c) Bạn tự làm đi tới đây dễ rồi

a)

\(\begin{array}{l}A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\\A = x - 0,2 - \dfrac{1}{3}x - 2 + 2 - \dfrac{2}{3}x\\ = \left( {x - \dfrac{1}{3}x - \dfrac{2}{3}x} \right) + \left( {\dfrac{{ - 1}}{2} - 2 + 2} \right)\\ =  - \dfrac{1}{2}\end{array}\)

Vậy \(A =  - \dfrac{1}{2}\) không phụ thuộc vào biến x

b)

\(\begin{array}{l}B = \left( {x - 2y} \right)\left( {{x^2} + 2{\rm{x}}y + 4{y^2}} \right) - \left( {{x^3} - 8{y^3} + 10} \right)\\B = \left[ {x - {{\left( {2y} \right)}^3}} \right] - {x^3} + 8{y^3} - 10\\B = {x^3} - 8{y^3} - {x^3} + 8{y^3} - 10 =  - 10\end{array}\)

Vậy B = -10 không phụ thuộc vào biến x, y.

c)

\(\begin{array}{l}C = 4{\left( {x + 1} \right)^2} + {\left( {2{\rm{x}} - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) - 4{\rm{x}}\\{\rm{C = 4}}\left( {{x^2} + 2{\rm{x}} + 1} \right) + \left( {4{{\rm{x}}^2} - 4{\rm{x}} + 1} \right) - 8\left( {{x^2} - 1} \right) - 4{\rm{x}}\\C = 4{{\rm{x}}^2} + 8{\rm{x}} + 4 + 4{{\rm{x}}^2} - 4{\rm{x}} + 1 - 8{{\rm{x}}^2} + 8 - 4{\rm{x}}\\C = \left( {4{{\rm{x}}^2} + 4{{\rm{x}}^2} - 8{{\rm{x}}^2}} \right) + \left( {8{\rm{x}} - 4{\rm{x}} - 4{\rm{x}}} \right) + \left( {4 + 1 + 8} \right)\\C = 13\end{array}\)

Vậy C = 13 không phụ thuộc vào biến x

a: \(B=\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)\)

\(=x^2y+\dfrac{1}{4}x^2+y^2+\dfrac{1}{4}y+x^2y^2+\dfrac{3}{4}y+\dfrac{1}{4}\)

\(=x^2y+x^2y^2+y^2+y+\dfrac{1}{4}x^2+\dfrac{1}{4}\)

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

\(=\left(x^2+1\right)\left(y+\dfrac{1}{2}\right)^2\)

\(C=x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\)

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

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

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

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

=>\(A=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)

b: \(=\dfrac{y^2-y+1+2y-\dfrac{3}{4}}{y^2-y+1}=1+\dfrac{2y-\dfrac{3}{4}}{y^2-y+1}>=1\)

Dấu = xảy ra khi y=3/8

đáp án: a là đúng

A

\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)

\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)

\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)

\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng \(\forall a,b,c\)

a) 3.(ab+bc+ac)≤a²+b²+c²+2ab+2bc+2ac

<=> \(a^2+b^2+c^2-ab-bc-ac\ge0\)

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

<=> (a-b)²+(b-c)²+(a-c)²≥0 ( luon dung voi moi a,b,c)

b) ap dung ket qua tren va vế sau bn xem bài giải của mk ở trên

1: \(MTC=2\left(x-y\right)\left(x+y\right)\)

\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)

\(\dfrac{x+y}{2x^2+4xy+2y^2}\)

\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)

\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)

\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)

2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)