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a) (a - 2b)x(a + 2b)
b) x2-(y-3)2
=> (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))2 - (5(x + y))2
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)2
f) (5x - 2y)2
h) (x - 5)(x2 + 5x + 25)
k) (x + 5)3
\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)
\(a,x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\\ b,2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\left(x+1\right)^2\\ c,3x^3y-12x^2y+12xy=2xy\left(x^2-4x+4\right)=2xy\left(x-2\right)^2\\ d,6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\left(x+y\right)^2\\ e,x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x+y\right)\\ f,9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(9x^2-4y^2\right)\left(x-2\right)=\left(3x-2y\right)\left(3x+2y\right)\left(x-2\right)\)
Tick plz
a: \(x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\)
b: \(2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\cdot\left(x+1\right)^2\)
c: \(3x^3y-12x^2y+12xy=3xy\left(x^2-4x+4\right)=3xy\cdot\left(x-2\right)^2\)
d: \(6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\cdot\left(x+y\right)^2\)
e: \(x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
f: \(9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(x-2\right)\left(3x-2y\right)\left(3x+2y\right)\)
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
a,x2-y2-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x2-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a2-6ab+3b2-12c2
= 3(a2 - 2ab + b2 - 4c2)
= 3[(a-b)2 - 4c2)
= 3(a-b-2c)(a-b+2c)
d,x2-25+y2+2xy
= (x+y)2 - 25
= (x+y+5)(x+y-5)
e) a2+2ab+b2-ac-bc
= (a+b)2-c(a+b)
= (a+b)( a+b-c)
f) x2-2x-4x2-4y
= -3x2-2x-4y
= -(3x2+2x+4y)
g)x2y-x3-9y+9x
= x2(y-x)-9(y-x)
= (y-x)(x2-9)
h) x2(x-1)+16(1-x)
= x2(x-1)-16(x-1)
= (x-1)(x2-16)
= (x-1)(x-4)(x+4)
n) 81x2-6yz-9y2-z2
= (9x)2-[(3y)2+6yz+z2]
=(9x)2-(3y+z)2
=(9x+3y+z)(9x-3y-z)
m) xz- yz-x2+2xy-y2
= z(x-y)-(x2-2xy+y2)
= z(x-y)-(x-y)2
= (x-y)(z-x+y)
p) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x+3) + 5(x+3)
= (x+3)(x+5)
k) x2 - x - 12
= x2 + 3x - 4x - 12
= x(x+3) - 4(x+3)
= (x+3)(x-4)
Bài 1:
a) \(x^2-10x+26+y^2+2y\)
\(=x^2-2.x.5+25+y^2+2y+1\)
\(=\left(x-5\right)^2+\left(y+1\right)^2\)
b) Sửa đề \(z^2-6z+5-t^2-4t\)
\(=z^2-2.z.3+9-4-t^2-4t\)
\(=\left(z-3\right)^2-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
c) \(\left(x+y-4\right)\left(x+y+4\right)\)
\(=\left(x+y\right)^2-4^2\)
d) \(a^2-b^2+c^2-2ac-d^2+2bd\)
\(=\left(a^2-2ac+c^2\right)-\left(b^2-2bd+d^2\right)\)
\(=\left(a-c\right)^2-\left(b-d\right)^2\)
e) \(\left(a-b-c\right)\left(a+b-c\right)\)
\(=\left(a-c-b\right)\left(a-c+b\right)\)
\(=\left(a-c\right)^2-b^2\)
f) \(4a^2+2b^2-4ab-2b+1\)
\(=\left(2a\right)^2-2.2a.b+b^2+b^2-2b+1\)
\(=\left(2a-b\right)^2+\left(b-1\right)^2\)
Bài 2:
a) Sửa đề \(4x^2-4xy+y^2\)
\(=\left(2x\right)^2-2.2x.y+y^2\)
\(=\left(2x-y\right)^2\)
b) \(y^2-6y+9\)
\(=y^2-2.y.3+3^2\)
\(=\left(y-3\right)^2\)
c) \(a^2+a+\dfrac{1}{4}\)
\(=a^2+2a.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(a+\dfrac{1}{2}\right)^2\)
d) \(a^2-12a+36\)
\(=a^2-2.a.6+6^2\)
\(=\left(a-6\right)^2\)
i) \(x^2-xy+\dfrac{1}{4}y^2\)
\(=x^2-2.x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\)
\(=\left(x-\dfrac{1}{2}y\right)^2\)
e) \(9x^2-24x+16\)
\(=\left(3x\right)^2-2.3x.4+4^2\)
\(=\left(3x-4\right)^2\)
f) \(x^2-3x+\dfrac{9}{4}\)
\(=x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\)
\(=\left(x-\dfrac{3}{2}\right)^2\)
g) \(1-2xy^2+x^2y^4\)
\(=1-2xy^2+\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)^2\)
h) \(\left(2a-b\right)^2+2\left(2a-b\right)+1\)
\(=\left(2a-b+1\right)^2\)
Bài 3:
a) \(A=\dfrac{1}{4}x^2-xy+y^2\)
\(A=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.y+y^2\)
\(A=\left(\dfrac{1}{2}x-y\right)^2\)
Thay x = 2012 và y = 1004 vào A ta được
\(A=\left(\dfrac{1}{2}.2012-1004\right)^2\)
\(A=\left(1006-1004\right)^2\)
\(A=2^2=4\)
b) \(B=9x^2-3xy+\dfrac{1}{4}y^2\)
\(B=\left(3x\right)^2-2.3x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\)
\(B=\left(3x-\dfrac{1}{2}y\right)^2\)
Thay x = 231 và y = 1384 vào B ta được
\(B=\left(3.231-\dfrac{1}{2}.1384\right)^2\)
\(B=\left(693-692\right)^2\)
\(B=1^2=1\)
thank you bạn nha