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\(A=x^2-2xy-4z^2+y^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y+2z\right)\left(x-y-2z\right)\)
\(=\left(6+4+45\right)\left(6+4-45\right)\)
\(=-1925\)
a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)
a: Ta có: \(x^2-2xy+y^2-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\)
\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)
\(=-8000\)
b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)
\(=3x^2+12x-63+x^2-8x+16+48\)
\(=2x^2+4x+1\)
\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)
\(=\dfrac{7}{2}\)
a)x2-2xy-4x2+y2
= (x2-2xy+y2)-(2x)2
= (x-y)2-(2x)2 = (x-y-2x)(x-y+2x)(1)
Thay x=6; y=-4; z=45 ta được:
(1)<=>(6+4-90)(6+4+90)= (10-90).(10+90)=-80.100= -8000
C=\(\left(x-1\right)x^2-4x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2x-2x+4\right)\)
C= \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\)
bạn thay x vào rồi tính là được
B=\(x\left(2x-y\right)-z\left(y-2x\right)=x\left(2x-y\right)+z\left(2x-y\right)=\left(2x-y\right)\left(x+z\right)\)
bạn thay x,y,z tính là ok
Bài a mình k chắc lắm nhưng nghĩ là thay vào rồi tính
\(x^2-2xy-4z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\) ( 1 )
Thay vào bấm máy tính ta được ( 1 )=19
b) \(3\left(x-3\right)\left(x+7\right)-\left(x-4\right)^2\)
\(=\left(3x-9\right)\left(x+7\right)-\left(x^2-8x+16\right)\)
\(=3x^2+12-63-x^2+8x-16\)
\(=2x^2+20x-79\)
\(=2x^2+20x+50-129\)
\(=2\left(x+5\right)^2-129\)
Thay x vào
`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`
`= 2xy`.
Thay `x = 2/3; y = -3/4` vào BT:
`2 . 2/3 . -3/4 = -1.`
`b, x(x-2y) - y(y^2-2x)`
`= x^2 - 2xy - y^3 + 2xy`
`= x^2 - y^3`
Thay `x = 5; y =3` vào BT:
`= 5^2 - 3^3 = 25 - 27 = -2`
a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)
\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)
\(=2xy\)
Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:
\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)
b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)
\(=x^2-2xy-y^3+2xy\)
\(=x^2-y^3\)
Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)
a, \(A=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(9x^2-4\right)\)
\(=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(3x-2\right)\left(3x+2\right)\)
\(=\left(3x-2+3x+2\right)^2\)
\(=36x^2=36.\left(-\frac{1}{3}\right)^2=4\)
b, \(B=\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left[\left(x+y-7\right)-\left(y-6\right)\right]^2\)
\(=\left(x-1\right)^2\)
\(=\left(101-1\right)^2=10000\)
c, \(C=4x^2-20x+27\)
\(=\left(2x\right)^2-2.2x.5+5^2+2\)
\(=\left(2x-5\right)^2+2\)
\(=\left(52,5.2-5\right)^2+2\)
\(=100^2+2=10002\)
Bài này dễ mà chỉ dùng hằng đẳng thức thôi. Chúc bạn học tốt.
a) \(x^2-2xy-4z^2+y^2\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(2z\right)^2\)
\(\Leftrightarrow\left(x-y\right)^2-\left(2z\right)^2\)
\(\Leftrightarrow\left[\left(x-y\right)+2z\right]\left[\left(x-y\right)-2z\right]\)
\(\Leftrightarrow\left(x-y+2z\right)\left(x-y-2z\right)\)
Tại x=6, y=-4, z=45
\(\left[6-\left(-4\right)+2.45\right]\left[6-\left(-4\right)-2.45\right]=100.\left(-80\right)=-8000\)
b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(\Leftrightarrow3\left(x^2+7x-3x-21\right)+\left(x^2-4x+4\right)+48\)
\(\Leftrightarrow3x^2+21x-9x-63+x^2-4x+4+48\)
\(\Leftrightarrow4x^2+8x-11\)
Tại x=0,5 ta có:
\(4.\left(0,5\right)^2+8.0,5-11=-6\)
a)Đặt \(A=x^2-2xy-4z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-\left(2z\right)^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\)
Thay \(x=6;y=-4;z=45\) vào A, ta có:
\(A=\left[6-\left(-4\right)-2\cdot45\right]\left[6-\left(-4\right)+2\cdot45\right]\)
\(=100\cdot\left(-80\right)\)
\(=-8000\)
Vậy \(A=-8000\)
b) Đặt \(B=3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2+7x-3x-21\right)+x^2-4x+4+48\)
\(=3x^2+12x-63+x^2-4x+52\)
\(=4x^2+8x-11\)
Thay \(x=0,5\) vào B, ta có:
\(B=4\cdot\left(0,5\right)^2+8\cdot0,5-11\)
\(=1\cdot4-11\)
\(=-6\)
Vậy \(B=-6\)