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a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) Ta có : x2 + 7x + 12
= x2 + 3x + 4x + 12
= (x2 + 3x) + (4x + 12)
= x(x + 3) + 4(x + 3)
= (x + 4)(x + 3)
Bạn ơi mk nhầm đề rồi số 30 thay bằng số 60 còn 36 thay bằng 72 và 39 thay bằng 75 nha
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
A) 7X2 - 7Y2 - 14X + 14Y
= ( 7X2 - 7Y2 ) - ( 14X - 14Y )
= 7( X2 - Y2 ) - 14( X - Y )
= 7( X - Y )( X + Y ) - 14( X - Y )
= 7( X - Y )( X + Y - 14 )
B) X2 - Y2 + 14X + 49
= ( X2 + 14X + 49 ) - Y2
= ( X + 7 )2 - Y2
= ( X - Y + 7 )( X + Y + 7 )
C) X2 - Y2 - X + Y
= ( X2 - Y2 ) - ( X - Y )
= ( X - Y )( X + Y ) - ( X - Y )
= ( X - Y )( X + Y - 1 )
D) X2 + 12Y - Y2 - 36
= X2 - ( Y2 - 12Y + 36 )
= X2 - ( Y - 6 )2
= ( X - Y + 6 )( X + Y - 6 )
E) X3 + X2 - 9X - 9
= ( X3 + X2 ) - ( 9X + 9 )
= X2( X + 1 ) - 9( X + 1 )
= ( X + 1 )( X2 - 9 )
= ( X + 1 )( X - 3 )( X + 3 )
x2-2xy+y2-z2+2zt-t2
=(x2-2xy+y2)-(z2-2zt+t2)
=(x-y)2-(z-t)2
=(x-y+z-t)(x-y-z+t)
x2 – 2xy + y2 – z2 + 2zt – t2 = (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2
= [(x – y) – (z – t)] . [(x – y) + (z – t)]
= (x – y – z + t)(x – y + z – t)
Bài làm:
a) \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(\left(x-y\right)\left(x-y-z\right)\)
a/ \(x^2-2xy+y^2-zx+yz.\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c/ \(x^2-y^2-2x-2y.\)
\(=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
Đặt x = 0,5 (*)
\(3.\left(x-2\right).\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2-7x-3x-21\right)+x^2-2.x.4+4^2\)
\(=3x^2-21x-9x-63+x^2-8x+8+48\)
\(=\left(3x^2+x^2\right)+\left(-21x-9x-8x\right)+\left(-63+8+48\right)\)
\(=4x^2+4x+1\)
\(=\left(2x+1\right)^2\)
Thay (*) vào biểu thức trên ta có :
\(=\left(2\times0,5+1\right)^2=4\)
CHÚC BẠN HỌC TỐT !!!
ta có: 3 ( ( x - 3 ) ( x + 7 ) ) + x2 - 2x4 + 42 + 48
= 3 ( x2 - 7x - 3x - 21 ) + x2 - 2x4 + 42 + 48
= 3x2 - 21x - 9x - 63 + x2 - 8x +8 + 48
= ( 3x2 + x2 ) + ( - 21x - 9x - 8x ) + ( - 63 + 8+ 48)
= 4x2 + 4x + 1
= ( 2x )2 + 2*2*x 12
= ( 2x + 1)2
Thay x = 0.5 vào biểu thức trên ta được:
=(2*0.5 + 1)2
= 22 = 4
\(x^3-x^2+7x-7=x^2\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(x^2+7\right)\)