Lời giải:

Ta có:

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

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

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

\(=x^3(x+y)+3xy[x(x+y)+y(x+y)]+y^3(x+y)-(x+y)-10\)

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

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

\(=2[x^3+3xy(x+y)+y^3]-12\)

\(=2[x^2(x+y)+y^2(x+y)+2xy(x+y)]-12\)

\(=2(x+y)(x^2+y^2+2xy)-12=2(x+y)(x+y)^2-12\)

\(=2(x+y)^3-12=2.2^3-12=4\)

Nếu bạn đã biết hằng đẳng thức thì:

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

\(=(x+y)^4-(x+y)-10=2^4-2-10=4\)

Lời giải:

Ta có:

x4+4x3y+6x2y2+4xy3+y4−x−y−10

=(x4+x3y)+3(x3y+2x2y2+xy3)+(xy3+y4)−(x+y)−10

=x3(x+y)+3xy(x2+2xy+y2)+y3(x+y)−(x+y)−10

=x3(x+y)+3xy[x(x+y)+y(x+y)]+y3(x+y)−(x+y)−10

=x3(x+y)+3xy(x+y)2+y3(x+y)−(x+y)−10

=2x3+6xy(x+y)+2y3−2−10

=2[x3+3xy(x+y)+y3]−12

=2[x2(x+y)+y2(x+y)+2xy(x+y)]−12

=2(x+y)(x2+y2+2xy)−12=2(x+y)(x+y)2−12

=2(x+y)3−12=2.23−12=4

Nếu bạn đã biết hằng đẳng thức thì:

x4+4x3y+6x2y2+4xy3+y4−x−y−10

=(x+y)4−(x+y)−10=24−2−10=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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