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Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)
Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)
\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=24-11x\)
b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)
\(=8x^2y-6y^2-9x^2y+12y^2\)
\(=6y^2-x^2y\)
c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)
\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)
\(=4y^3+y^2+6xy^2\)
b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)
\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)
\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+16\)
\(=-11x+16\)
b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(=8x^2y-6y^2-\left(9x^2y-12y^2\right)\)
\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)
c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2.3y-y+y^2+y^3+y\)
\(=9y^3+y^2+y^3=10y^3+y^2\)
Chúc bạn học tốt!!!
a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+16\)
\(=-11x+16\)
b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(=8x^2y-6y^2-9x^2y+12y^2\)
\(=-x^2y+6y^2\)
c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2.3y-y\left(1-y-y^2-1\right)\)
\(=9y^3-y\left(-y-y^2\right)\)
\(=9y^3+y^2+y^3=10y^3+y^2\)
a) Đề sai nha bạn :) mấy dấu cộng bạn phỉa chuyển thành dấu nhân nhé
\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(A=2^{512}-1+1\)
\(A=2^{512}\)
b . ( 5x - 3y + 4z )( 5x - 3y - 4z ) = ( 5x - 3y )^2 - ( 4z )^2 = 25x^2 - 30xy + 9y^2 - 16z^2 = 25( y^2 + z^2 ) - 30xy + 9y^2 - 16z^2 = 9z^2 + 34y^2 - 30xy ( 1 )
( 3x - 5y )^2 = 9x^2 - 30xy + 25y^2 = 9( y^2 + z^2 ) - 30xy + 25y^2 = 34y^2 + 9z^2 - 30xy ( 2 )
Tu ( 1 ) va ( 2 ) => dpcm