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Bài 1:
Theo bài ra ta có:
\(\left(x-y\right)^2=x^2-2xy+y^2\)
\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)
\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)
\(=25-10y+y^2+25-10x+x^2-4\)
\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)
\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)
\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)
\(=50-50+5^2-4-4\)
\(=25-8=17\)
Vậy giá trị của \(\left(x-y\right)^2\)là 17
a) \(\left(x+y-z\right)^2=\left[\left(x+y\right)-z\right]^2\)
\(=\left(x+y\right)^2-2\left(x+y\right)z+z^2\)
\(=x^2+2xy+y^2-2zx-2yz+z^2\)
\(=x^2+y^2+z^2+2xy-2yz-2zx\)
b) \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=x^4-y^4\)
c) \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5\)
I,
\(B=\left(3x-1\right)^2-\left(x+7\right)^2-2\left(2x-5\right)\left(2x+5\right)\\ =9x^2-6x+1-x^2-14x-49-2\left(4x^2-25\right)\\ =8x^2-20x-48-8x^2+50\\ =-20x+2\)
II,
\(a,2x^2+6xy-10.Thayx=-4,y=3,tacó: 2\cdot\left(-4\right)^2+6\cdot\left(-4\right)\cdot3-10=-50\)
\(b,x\left(x+y\right)+y\left(x+y\right)=\left(x+y\right)\left(x+y\right)=\left(x+y\right)^2\\ Thayx=19,6;y=0,4tacó:\\ \left(19,6+0,4\right)^2=400\)
\(c,x\left(x-3\right)-y\left(3-x\right)=x\left(x-3\right)+y\left(x-3\right)=\left(x+y\right)\left(x-3\right)\\ Thayx=\frac{1}{3};y=\frac{8}{3},tacó:\\ \left(\frac{1}{3}+\frac{8}{3}\right)\left(\frac{1}{3}-3\right)=-8\)
\(d,2x^2\left(x^2+y^2\right)+2y^2\left(x^2+y^2\right)+5\left(x^2+y^2\right)\\ =\left(x^2+y^2\right)\left[2\left(x^2+y^2\right)+5\right]\\ Thayx^2+y^2=1,tacó:\\ 1\cdot\left(2\cdot1+5\right)=7\)
1, 5a2xy-10a3x-15ay = 5a( axy - 2a\(^2\)x - 3y )
2, mxy-m2x+my = m( xy - mx + y )
3, 2mx-4m2xy+6mx = 2mx( 1 - 2my + 3 ) = 2mx( 4 - 2my )
4, a2b-2ab2+ab = ab( a - 2b + 1 )
5, 5a2b-2ab2+ab = ab( 5a - 2b +1 )
6, -3x2y3-6x3y2-x2y2 = -3x\(^2\)y\(^2\) ( y + 2x + 1 )
7, 5x2y4-10x4y2+5x2y2 = 5x\(^{^2y^2}\)( y\(^2\) - 2x\(^2\) + 1 )
8, -2x3y4-4x4y3+2x3y3 = 2\(x^3y^3\) ( -y - 2x + 1 )
9, 4x3y2-8x3y+16xy2-24 = 4( x\(^3\)y\(^2\) - 2x\(^3\)y + 4 xy\(^2\) - 6 )
10, 12x3y-6xy+3x = 3x( 4x\(^2\)y - 2y + 1 )
11, 2(x-y)-a(x-y) = ( 2 - a ) ( x - y )
12, a(x-y)+b(x-y)= ( a + b ) ( x - y )
13, m(x+y)-n(x+y) = ( m - n ) ( x + y )
14, 2a(x+y)-4(x+y) = ( 2a - 4 )( x + y ) = 2( a - 2 ) ( x + y )
15, 3a(x+y)-6ab(x+y) = ( 3a - 6ab )( x + y ) = 3a( 1 - 2b ) ( x + y )
16, 5a2(x-y)+10a(x-y) = ( 5a\(^2\)+10a )( x - y ) = 5a( a + 2 ) ( x - y )
17, -2ab(x-y)-4a(x-y) = ( -2ab - 4a )( x - y ) = -2a( b + 2 )( x - y )
18, 3a(x-y)+2(x-y) = ( 3a + 2 ) ( x - y )
19, m(a-b)-m2(a-b) = ( m - m\(^2\) ) ( a - b ) = m( 1 - m ) ( a - b )
20, mx(a+b)-m(a+b) = ( mx - m ) ( a + b ) = m( x - 1 )( a + b )
21, x(a-b)-y(b-a) = x( a - b ) + y( a - b ) = ( x + y ) ( a - b )
22, ab(x-5)-a2(5-x) = ab( x - 5 ) + a\(^2\)( x - 5 ) = ( ab + a\(^2\) ) ( x - 5 ) = a( b + a )( x - 5 )
23, 2a2(x-y)-4a(y-x)= 2a\(^2\)( x - y ) + 4a( x - y )=( 2a\(^2\) + 4a ) ( x - y )= 2a( a + 2 )( x - y )
Đăng ít thôi =))
a. \(5a^2xy-10a^3x-15ay=5a\left(axy-2a^2x-3y\right)\)
b. \(mxy-m^2x+my=m\left(xy-mx+y\right)\)
c. \(2mx-4m^2xy+6mx=2mx\left(1-2my+3\right)=2mx\left(-2my+4\right)\)
d. \(a^2b-2ab^2+ab=ab\left(a-2b+1\right)\)
e. \(5a^2b-2ab^2+ab=ab\left(5a-2b+1\right)\)
g.
a) (3x-1)(9x2+3x+1)=27x3-1
27x3-1=27x3-1
27x3-1-(27x3-1)=0
27x3-1-27x3+1=0
⇒x=0
b)(x2-5x+25)(x+5)=x3+125
(x+5)(x2-x.5+52)=x3+125
x3+125-(x3+125)=0
x3+125-x3-125=0
⇒x=0
c)(x-3)(x2-6x+9)=(x-3)3
x3-33-(x-3)3=0
x3-27-x3+27=0
⇒x=0
d) Đề phải là thế này chứ \(\left(x-y+4\right).\left(x-y-4\right)\)
\(=\left(x-y\right)^2-4^2\)
\(=\left(x-y\right)^2-16\)
\(=x^2-2.x.y+y^2-16\)
\(=x^2-2xy+y^2-16.\)
Chúc bạn học tốt!
1. \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3\)
\(=\left(5x+y^2\right)\left[\left(5x\right)^2-5x.y^2+\left(y^2\right)^2\right]\)
\(=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
2. \(4x\left(x-2y\right)+8y\left(2y-x\right)\)
\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(4x-8y\right)\)
3. \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)
\(=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
4. \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
5. \(a^3x-ab+b-x\)
\(=a^3x-x-ab+b\)
\(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left[x\left(a^2+a+1\right)-b\right]\)
6. \(x^3-64=x^3-4^3\)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
7. \(0,125\left(a+1\right)^3-1\)
\(=\left[0,5\left(a+1\right)\right]^3-1^3\)
\(=\left[0,5\left(a+1\right)-1\right]\left\{\left[0,5\left(a+1\right)\right]^2+\left[0,5\left(a+1\right).1\right]+1^2\right\}\)
\(=\left[0,5\left(a+1-2\right)\right]\left[0,25a^2+0,5a+0,25+0,5a+0,5+1\right]\)
\(=\left[0,5\left(a-1\right)\right]\left(0,25a^2+a+1,75\right)\)
8. \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\)
\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)
\(=\left(2x+22\right)\left(4x+8\right)\)
9. \(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
10. \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
11. \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
12. \(x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
a) Ta có: \(VP=x^2+y^2+z^2-2xy+2yz-2zx\)
\(=\left(x^2-xy-xz\right)+\left(y^2-xy+yz\right)+\left(z^2-yz-zx\right)\)
\(=x\left(x-y-z\right)+y\left(y-x+z\right)+z\left(z-y-x\right)\)
\(=x\left(x-y-z\right)-y\left(x-y-z\right)-z\left(x-y-z\right)\)
\(=\left(x-y-z\right)\left(x-y-z\right)\)
\(=\left(x-y-z\right)^2=VT\)(đpcm)
b) Ta có: \(VP=x^2+y^2+z^2+2xy-2yz-2zx\)
\(=\left(x^2+xy-zx\right)+\left(y^2+xy-2yz\right)+\left(z^2-yz-zx\right)\)
\(=x\left(x+y-z\right)+y\left(x+y-z\right)+z\left(z-y-x\right)\)
\(=\left(x+y-z\right)\left(x+y\right)-z\left(x+y-z\right)\)
\(=\left(x+y-z\right)\left(x+y-z\right)\)
\(=\left(x+y-z\right)^2=VT\)(đpcm)
c) Ta có: \(VP=x^4-y^4\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)=VT\)(đpcm)
d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5=VP\)(đpcm)
a: =2(x-y)^3/(x-y)-7(x-y)^2/(x-y)+(x-y)/(x-y)
=2(x-y)^2-7(x-y)+1
b: =3(x-y)^5/5(x-y)^2-2(x-y)^4/5(x-y)^2+3(x-y)^2/5(x-y)^2
=3/5(x-y)^3-2/5(x-y)^2+3/5
\(a,\)
\(\left[2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right]:\left(x-y\right)\)
\(=\left[2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right]:\left(x-y\right)\)
\(=\left\{\left(x-y\right)\left[2\left(x-y\right)^2-7\left(x-y\right)+1\right]\right\}:\left(x-y\right)\)
\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)
\(b,\)
\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:\left[5\left(x-y\right)^2\right]\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)