\(x^4+4x^3y+6x^2y^2+4xy^3+y^4-x-y-10\)

\(=\left(x^4+2x^3y+x^2y^2\right)+\left(2x^3y+4x^2y^2+2xy^3\right)+\left(x^2y^2+2xy^3+y^4\right)-\left(x+y\right)-10\)

\(=x^2\left(x^2+2xy+y^2\right)+2xy\left(x^2+2xy+y^2\right)+y^2\left(x^2+2xy+y^2\right)-\left(x+y\right)-10\)

\(=\left(x^2+2xy+y^2\right)\left(x^2+2xy+y^2\right)-\left(x+y\right)-10\)

\(=\left(x+y\right)^2\left(x+y\right)^2-\left(x+y\right)-10\)

\(=\left(x+y\right)^4-\left(x+y\right)-10\)

\(=2^4-2-10\) \(=4\)

Câu 1 :

\(3\left(x-3\right)\left(x+7\right)+\left(1-4\right)\left(x+4\right)+18\)

\(=3\left(x^2+4x-21\right)-3\left(x+4\right)\)

\(=3x^2+12x-63-3x-12=3x^2+9x-75\)

Thay x = 1/2 vào ta được 

\(\dfrac{3.1}{4}+\dfrac{9}{2}-75=-\dfrac{279}{4}\)

Câu 2 : 

\(5x^2+5xy+5x=5x\left(x+y+1\right)\)

Thay x = 60 ; y = 50 ta được 

\(300\left(60+50+1\right)=33300\)

Câu 3 : 

\(4x^2y^2+2xy^2+6x^2y=2xy\left(2xy+y+3x\right)\)

Thay x = 10 ; y  = 1/2 ta được 

\(\dfrac{2.10.1}{2}\left(\dfrac{2.10.1}{2}+\dfrac{1}{2}+30\right)=405\)

1: \(=3\left(x^2+4x-21\right)+x^2-16+18\)

\(=3x^2+12x-63+x^2+2\)

\(=4x^2+12x-61\)

\(=4\cdot\dfrac{1}{4}+12\cdot\dfrac{1}{2}-61=1-61+6=-54\)

2: \(=5\cdot60^2+5\cdot60\cdot50+5\cdot60=33300\)

3: \(=4\cdot10^2\cdot\dfrac{1}{4}+2\cdot10\cdot\dfrac{1}{4}+6\cdot100\cdot\dfrac{1}{2}=405\)

\(P=\left(x+2y\right)^2-2\left(x+2y\right)\left(y-1\right)+\left(y-1\right)^2\\ P=\left(x+2y-y+1\right)^2=\left(x+y+1\right)^2\\ Q.sai.đề\\ M=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\\ M=1^3-3xy\left(x+y-1\right)=1-3xy\left(1-1\right)=1-0=1\\ x+y=2\Leftrightarrow\left(x+y\right)^2=4\\ \Leftrightarrow x^2+y^2+2xy=4\\ \Leftrightarrow2xy=4-10=-6\\ \Leftrightarrow xy=-3\\ N=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\\ N=2\left(10+3\right)=2\cdot13=26\)

Thay \(x=\dfrac{1}{2};y=-1\) vào B, ta được:

\(B=\left[\left(\dfrac{1}{2}\right)^3-4\cdot\left(\dfrac{1}{2}\right)^2\cdot\left(-1\right)+3\cdot\left(-1\right)^2-4\right]:\left[3\cdot\left(\dfrac{1}{2}\right)^3-3\cdot\left(-1\right)^2-3\cdot\left(-1\right)\right]\)

\(=\left(\dfrac{1}{8}+4\cdot\dfrac{1}{4}+3\cdot1-4\right):\left(3\cdot\dfrac{1}{8}-3\cdot1+3\right)\)

\(=\left(\dfrac{1}{8}+1+3-4\right):\left(\dfrac{3}{8}-3+3\right)\)

\(=\dfrac{1}{8}\cdot\dfrac{8}{3}=\dfrac{1}{3}\)

a, \(A=x^2-y^2-4x\)

\(=\left(x^2-y^2\right)-4x\)

\(=\left(x+y\right)\left(x-y\right)-4x\)

\(=2\left(x-y\right)-2.2x\)

\(=2\left(x-y-2x\right)\)

\(=2\left(-x-y\right)\)

\(=2\left[-\left(x+y\right)\right]\)

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

\(=-2.2=-4\)

Vậy \(A=-4\)

b, \(B=x^2+y^2+2xy-4x-4y-3\)

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

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

\(=4^2-4.4-3\)

\(=-3\)

Vậy \(B=-3\)

c, Phần này hình như đề bài sai, bạn xem lại đề hộ mk cái nhé ;)

A =(x+y)(x-y) -4x = 2(x-y) -4x = 2x -2y - 4x = - 2(x+y) = -4

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