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a)Tại \(x=87;y=13\) thì
\(A=x^2-y^2=\left(x-y\right)\left(x+y\right)\)
\(=\left(87-13\right)\left(87+13\right)=74\cdot100=7400\)
b)Tại \(x=\dfrac{1}{3}\) thì
\(B=9x^2-6x+1=\left(x-\dfrac{1}{3}\right)^2\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)^2=0^2=0\)
c)Tại \(x=1;y=2\) thì
\(C=4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
\(=\left(2\cdot1-3\cdot2\right)^2=\left(-4\right)^2=16\)
a, Ta có:
\(A=x^2-y^2=\left(x-y\right)\left(x+y\right)\)
Thay \(x=87;y=13\) vào A ta được:
\(\left(87-13\right)\left(87+13\right)=74.100=7400\)
b, Ta có:
\(B=9x^2-6x+1=9x^2-3x-3x+1\)
\(=3x\left(3x-1\right)-\left(3x-1\right)\)
\(=\left(3x-1\right)^2\)
Thay \(x=\dfrac{1}{3}\) vào B ta được:
\(\left(3.\dfrac{1}{3}-1\right)^2=0\)
c, Ta có:
\(C=4x^2-12xy+9y^2=4x^2-6xy-6xy+9y^2\)
\(=2x\left(2x-3y\right)-3y\left(2x-3y\right)\)
\(=\left(2x-3y\right)^2\)
Thay \(x=1;y=2\) vào biểu thức C ta được:
\(\left(2.1-3.2\right)^2=\left(2-6\right)^2=\left(-4\right)^2=16\)
Chúc bạn học tốt!!!

8) \(y^2-y-30=y^2+5y-6y-30=y\left(y+5\right)-6\left(y+5\right)=\left(y-6\right)\left(y+5\right)\)
9) \(y^2-8y+15=y^2-3y-5y+15=y\left(y-3\right)-5\left(y-3\right)=\left(y-5\right)\left(y-3\right)\)
10) \(y^2+y-6=y^2-2y+3y-6=y\left(y-2\right)+3\left(y-2\right)=\left(y+3\right)\left(y-2\right)\)
11) \(y^2-y-12=y^2+3y-4y-12=y\left(y+3\right)-4\left(y+3\right)=\left(y-4\right)\left(y+3\right)\)
12) \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)
13) \(u^2+u-42=u^2+7u-6u-42=u\left(u+7\right)-6\left(u+7\right)=\left(u-6\right)\left(u+7\right)\)
14) \(2x^2+x-6=2x^2+4x-3x-6=2x\left(x+2\right)-3\left(x+2\right)=\left(2x-3\right)\left(x+2\right)\)
15) \(7x^2+50x+7=7x^2+49x+x+7=7x\left(x+7\right)+\left(x+7\right)=\left(7x+1\right)\left(x+7\right)\)
16) \(12x^2+7x-12=12x^2+16x-9x-12=4x\left(3x+4\right)-3\left(3x+4\right)=\left(4x-3\right)\left(3x+4\right)\)
17) \(15x^2+7x-2=15x^2-3x+10x-2=3x\left(5x-1\right)+2\left(5x-1\right)=\left(3x+2\right)\left(5x-1\right)\)
18) \(2x^2-y^2+xy=2x^2+2xy-xy-y^2=2x\left(x+y\right)-y\left(x+y\right)=\left(2x-y\right)\left(x+y\right)\)
19) \(x^2-3xy+2y^2=x^2-xy-2xy+2y^2=x\left(x-y\right)-2y\left(x-y\right)=\left(x-2y\right)\left(x-y\right)\)

1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3

\(a,x^2-y^2=\left(x+y\right)\left(x-y\right)=\left(87+13\right)\left(87-13\right)=100.74=7400\)\(b,x^3-3x^2+3x-1=\left(x-1\right)^3=\left(101-1\right)^3=100^3=1000000\)c,\(x^3+9x^2+27x+27=\left(x+3\right)^3=\left(97+3\right)^3=1000000\)
a) x2 - y2 = (x+y)(x-y)
Thay x=87; y=13 có:
(87+13)(87-13) = 100.74 = 7400
b)x3-3x2+3x-1 = x3 - 3x2.1+ 3x .12 -13 = (x-1)3
Thay x=101 có:
(101-1)3 =1003 =1000000
c)x3+9x2+27x+27= x3 +3x2.1+3x.12+33= (x+3)3
Thay x=97 có:
(97+3)3= 1003=1000000

a) x2 - y2 = ( x+y )( x-y )
Thay x = 87 và y = 13 vào biểu thức a) ta có :
( 87+13 )( 87-13 ) = 100.74 = 7400

Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
\(x^2-2\)
\(=x^2-\left(\sqrt{2}\right)^2\)
\(=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
\(y^2-13\)
\(=y^2-\left(\sqrt{13}\right)^2\)
\(=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)