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\(=\dfrac{a+b+a-b}{a^2-b^2}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{2a^3+2a^2b^2+2a^3-2ab^2}{a^4-b^4}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{4a^7+4a^3b^4+4a^7-4a^3b^4}{a^8-b^8}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{8a^7}{a^8-b^8}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{8a^{15}+8a^7b^8+8a^{15}-8a^7b^8}{a^{16}-b^{16}}=\dfrac{16a^{15}}{a^{16}-b^{16}}\)
\(x^2-5y+y^2-2xy+5x=\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
a/ x2 – 5y + y2 -2xy + 5x = ( x2 - 2xy + y2 ) - 5( y - x ) = ( x - y )2 - 5( y - x ) = ( y - x )2 - 5( y - x ) = ( y - x )( y - x - 5 )
b/ 4x2 – 81(y – 2)2 = 4x2 - 92(y – 2)2= 4x2 – ( 9y – 18)2 = ( 2x -9y -18 )( 2x + 9y + 18 )
c/ x2z – y2z + 2yz – z = ( x2z + yz ) - ( y2z - yz ) - z = z( x2 + y ) - z( y2 - y ) -z = z( x2 + y - y2 +y - 1 ) = z( x2 + 2y - y2 - 1 ) \(=z[x^2-\left(y^2-2y+1\right)]=z[x^2-\left(y-1\right)^2=z\left(x-y+1\right)\left(x+y-1\right)\)
d/ x3 – 8y3 + x2 + 2xy + 4y2 = ( x3 – 8y3 ) + x2 + 2xy + 4y2 = ( x -2y )( x2 + 2xy + 4y2 ) + ( x2 + 2xy + 4y2 0 = ( x2 + 2xy + 4y2)( x -2y +1)
e/ 7x2 – 11x + 4 = 7x2 -7x -4x +4 = 7x( x-1 ) - 4( x - 1 ) = ( x - 1 )( 7x - 4 )
g/ 13x2 + 2xy – 15y2 = 13x2 - 13xy + 15xy - 15y2 = 13x( x - y ) + 15y( x - y ) = ( x - y )( 13x + 15y )
h/ x3 + 3x2 + 3x + 2 = x3 +2x2 + x2 +2x + x + 2 = x2( x + 2 ) + x( x + 2 ) + ( x + 2 ) = ( x + 2 )( x2 + x + 1 )
i/ x3 – 3x2 + 3x – 2 + xy – 2y = x3 - 2x2 - x2 + 2x + x - 2 +xy - 2y = x2( x - 2 ) - x( x - 2 ) + ( x - 2 ) + y( x - 2 ) = ( x - 2 )( x2 - x +1 + y )
Trả lời:
a, \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)
Vậy x = 0; x = - 1/5 là nghiệm của pt.
b, \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy x = 5 là nghiệm của pt.
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.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)
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, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
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 )