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Bài 2:
\(A=x^2+4y^2-2x+10-4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 vào biểu thức A ta được: \(A=5^2-2.5+10=25\)
\(B=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)\left(y-1\right)+y^2-2y+1\)
\(=x^2+4xy+4y^2-2xy+2x-4y^2+4y+y^2-2y+1\)
\(=x^2+2xy+y^2+2x+2y+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1\)
Thay x + y = 5 vào biểu thức B ta được: \(B=5^2+2.5+1=25+10+1=36\)
\(C=x^2-y^2-4x=\left(x^2-4x+4\right)-y^2-4\)
\(=\left(x-2\right)^2-y^2-4\) \(=\left(x-y-2\right)\left(x-2+y\right)-4\)
Thay x + y = 2 vào C ta được: \(C=\left(x-2-y\right)\left(2-2\right)-4=0-4=-4\)
\(D=x^2+y^2+2xy-4x-4y-3\)
\(=\left(x+y\right)^2-4\left(x+y\right)-3\) Thay x + y = 4 vào D ta được:
\(D=4^2-4.4-3=16-16-3=-3\)
Bài 3:
a) \(N=-9x^2+12x-5=-\left(9x^2-12x+4\right)-1\)
\(=-\left(3x-2\right)^2-1\)
Do \(\left(3x-2\right)^2\ge0\) nên \(-\left(3x-2\right)^2-1< 0\)
Vậy N < 0
b) ghi đề cẩn thận lại đi, mk k hiểu
\(\text{a) }\left(x-1\right)\left(x^2+y\right)-\left(x^2-y\right)\left(x-2\right)-x\left(x+2y\right)+3\left(y-5\right)\)
\(=\left(x^3+xy-x^2-y\right)-\left(x^3-2x^2-xy+2y\right)-\left(x^2+2xy\right)+\left(3y-15\right)\)
\(=x^3+xy-x^2-y-x^3+2x^2+xy-2y-x^2-2xy+3y-15\)
\(=\left(x^3+x^3\right)+\left(-x^2+2x^2-x^2\right)+\left(xy+xy-2xy\right)+\left(-y-2y+3y\right)-15\)
\(=0+0+0+0-15\)
\(=-15\)
\(\text{b) }6\left(x^3y+x-3\right)-6x\left(2xy^3+1\right)-3x^2y\left(2x-4y^2\right)\)
\(=\left(6x^3y+6x-18\right)-\left(12x^2y^3+6x\right)-\left(6x^3y-12x^2y^3\right)\)
\(=6x^3y+6x-18-12x^2y^3-6x-6x^3y+12x^2y^3\)
\(=\left(6x^3y-6x^3y\right)+\left(6x-6x\right)+\left(-12x^2y^3+12x^2y^3\right)-18\)
\(=0+0+0-18\)
\(=-18\)
\(\text{c) }\left(x^2+2xy+4y^2\right)\left(x-2y\right)-6\left(\frac{1}{2}-\frac{4}{3}y^3\right)\)
\(=\left(x^3-2x^2y+2x^2y-4xy^2+4xy^2-8y^3\right)-\left(3-8y^3\right)\)
\(=\left(x^3-8y^3\right)-\left(3-8y^3\right)\)
\(=x^3-8y^3-3+8y^3\)
\(=x^3-3\)
Câu 2:
\(B=x^2+2x+y^2-2x-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2\cdot7+37=49+37+14=100\)
Câu 3:
\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2\cdot5+10=25\)
A=\(\left(x-y\right)^2+\left(x+y\right)^2=x^2-2xy+y^2+x^2+2xy+y^2=2x^2+2y^2\)
B=\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=\left(2y\right).\left(2x\right)\)
C=\(\left(2a+b\right)^2-\left(2a-b\right)^2=\left(2a+b-2a+b\right)\left(2a+b+2a-b\right)=\left(2b\right).\left(4a\right)\)
D=\(\left(2x-1\right)^2-2\left(2x-3\right)^2+4=4x^2-4x+1-4x+6+4=4x^2-8x+11\)
E=\(\left(x+3y\right)^2-\left(x-3y\right)^2=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)=\left(6y\right).\left(2x\right)\)
F=\(\left(2x+y\right)^2-\left(2x-y\right)^2=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)=\left(2y\right).\left(4x\right)\)
G=\(\left(x-2y\right)^2+4\left(x-2y\right)y+4y^2=x^2-4xy+4y^2+4xy-8y^2+4y^2=x^2\)
H=\(\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y^{ }\right)^2=x^2-2xy+y^2-4\left(x^2+2xy-xy-2y^2\right)+4x+8y=x^2-2xy+y^2-4x^2-8xy+4xy+8y^2+4x+8y=3x^2+12xy-9y^2+4x+8y\)
Ta có:
a) A= (x-y)^2 + (x+y)^2
A= x^2 -2xy + y^2 + x^2 + 2xy + y^2
A= 2x^2+ 2y^2
b) B= (x+y)^2 -( x-y)^2
B= (x+y-x+y)(x+y+x-y)
B= 2y.2x= 4xy
c) C= (2a+b)^2 -( 2a-b)^2
C= (2a+b-2a+b)(2a+b+2a-b)
C= 2b.4a
C= 8ab
d) D= (2x-1)^2 -2(2x-3)^2+4
D= 4x^2 -4x+1 -2( 4x^2 -12x + 9) +4
D= 4x^2 -4x+1 -8x^2 + 24x -18 +4
D= -4x^2 + 20x-13
e) E= (x+3y)^2-(x-3y)^2
E= (x+3y-x+3y)(x+3y+x-3y)
E= 6y.2x= 12xy
f) F= (2x+y)^2-(2x-y)^2
F=(2x+y-2x+y)(2x+y+2x-y)
F= 2y.4x= 8xy
g) G= (x-2y)^2 + 4(x-2y)y + 4y^2
G= (x-2y)^2 + 2(x-2y)2y + (2y)^2
G= (x-2y+2y)^2
G= x^2
h) H= (x-y)^2 -4(x-y)(x+2y)+ 4(x+2y)^2
H= (x-y)^2 - 2(x-y)2(x+2y) + [2(x+2y)]^2
H= (x-y- 2x-4y)^2
H= (-x-5y)^2
Lưu ý (-A-B)^2 = ( A+ B)^2
=> H= (x+5y)^2
a) Ta có: \(\frac{\left[2\left(y-x\right)^3-2\left(y-x\right)^2+\left(x-y\right)\right]}{y-x}\)
\(=\frac{2\left(y-x\right)^3}{y-x}-\frac{2\left(y-x\right)^2}{y-x}+\frac{x-y}{y-x}\)
\(=2\left(y-x\right)^2-2\left(y-x\right)-1\)
\(=2y^2-4yx+2x^2-2y+2x-1\)
cảm ơn bạn nha