Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Trả lời:
a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)
b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)
\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)
\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)
a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)
b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)
a) \(73^2-27^2=\left(73+27\right)\left(73-27\right)=100.46=4600\)
b) \(55^2+20^2-25^2+40.45=\left(55^2-25^2\right)+\left(20^2+40.45\right)\)
\(=\left(55-25\right)\left(55+25\right)+\left(40.10+40.45\right)=30.80+40.55\)
\(=40\left(60+55\right)=40.115=4600\)
Trả lời:
a) x2 + 4y2 + 4xy = x2 + 2.x.2y + (2y)2 = ( x + 2y )2
b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\)
c) x3 - 6x2 + 12x - 8 = x3 - 3.x2.2 + 3.x.22 - 23 = ( x - 2 )3
d) x2 - x - y2 - y = ( x2 - y2 ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 )
e) 5x - 5y + ax - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a )
Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
\(=\dfrac{a+b+a-b}{a^2-b^2}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{2a^3+2a^2b^2+2a^3-2ab^2}{a^4-b^4}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{4a^7+4a^3b^4+4a^7-4a^3b^4}{a^8-b^8}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{8a^7}{a^8-b^8}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{8a^{15}+8a^7b^8+8a^{15}-8a^7b^8}{a^{16}-b^{16}}=\dfrac{16a^{15}}{a^{16}-b^{16}}\)
Bài 62: 25x2y6-60xy4z2+36y2z4=(5xy3)2-2.5xy3.(6yz2)2
Bài 63: 1/9u4v6-1/3u5v4+(1/2u3v)=(1/3u2v3)-2.1/3u2v3.1/2u2v3+(1/2u3v)
Ta có: a + b + c = 0
<=> a2 + b2 + c2 + 2(ab + bc + ac) = 0
<=> a2 + b2 + c2 = -2(ab + bc + ac)
<=> a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2 = 4[a2b2 + b2c2 + a2c2 + 2abc(a + b + c)] (vì a + b + c= 0)
<=> a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2) = 4(a2b2 + b2c2 + a2c2)
<=> a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2) (đpcm)
b) Từ a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2)
<=> (a4 + b4 + c4)/2 = a2b2 + b2c2 + a2c2 + 2abc(a + b + c) (vì a + b + c) = 0
<=> (a4 + b4 + c4)/2 = (ab + bc + ac)2
<=> a4 + b4 + c4 = 2(ab + bc + ac)2 (đpcm)
c) Từ a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2)
<=> 2(a4 + b4 + c4) = a4+ b4 + c4 + 2(a2b2 + b2c2 + a2c2)
<=> 2(a4 + b4 + c4) = (a2 + b2 + c2)2
<=> a4 + b4 + c4 = (a2 + b2 + c2)2/2 (đpcm)
a) Áp dụng hằng đẳng thức thứ hai ta có
\(x^2 - 2xy + y^2\) \(<=>\) \((x +y )^2\)
b) Ta có :
\(x^2 - 2xy - 4z^2 + y^2 <=> (x^2 - 2xy + y^2) - (2z)^2 <=> ( x-y)^2 - (2z)^2 <=> ( x-y+2z) (x-y-2z)\)