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a, (x+2)^2
b, (x-3)^2
c, (2x+3)^2
d, (3x-1)^2
e, (x+5)^2
g, (4x-1)^2
a) x2 + 4x + 4 = ( x + 2 )2
b) x2 - 6x + 9 = (x-3)2
c) 4x2 + 12x + 9 = (2x)2 + 2.2x.3 + 3^2 = (2x + 3)2
d) 9x2 - 6x + 1 = (3x)2 - 2.3x.1 + 1^2 = (3x-1)2
e) x2 + 25 +10x = x2 + 2.x.5 + 52 = (x+5)2
g) 16x2 +1 - 8x = (4x)2 - 2.4x.1 + 1^2 = (4x-1)2
a) \(\frac{1}{9}x^4-2x^2y+9y^2=\left(\frac{1}{3}\right)^2\left(x^2\right)^2-2x^2y+\left(3y\right)^2\)
\(=\left(\frac{1}{3}x^2\right)^2-2\frac{1}{3}x^23y+\left(3y\right)^2\)
\(=\left(\frac{1}{3}x^2-3y\right)^2\)
b) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
\(\frac{1}{9}x^4-2x^2y+9y^2\)
\(=\left(\frac{1}{3}x^2\right)^2-2\times\frac{1}{3}x^2\times3y+\left(3y\right)^2\)
\(=\left(\frac{1}{3}x^2-3y\right)^2\)
\(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2\times5x\times2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
a, \(\dfrac{1}{9}x^4-2x^2y+9y^2=\left(\dfrac{1}{3}x^2\right)^2-\left(2.\dfrac{1}{3}x^2.3y\right)^2+\left(3y\right)^2\)
\(=\left(\dfrac{1}{3}x^2-3y\right)^2\)
b, \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2=\left(5x-2y\right)^2\)
*Trả lời:
a) Có vẻ như đề sai nên mình sửa lại:
\(2x^2y+2xy^2-x-y=\left(2x^2y+2xy^2\right)-\left(x+y\right)=2xy\cdot\left(x+y\right)-\left(x+y\right)=\left(2xy-1\right)\left(x+y\right)\)
b) \(8x^3-12x^2+6x-1=\left(2x\right)^3-3\cdot4x^2+3.2x-1=\left(2x-1\right)^3\)
c)\(4x^2-4xy+y^2-9=\left(4x^2-4xy+y^2\right)-9=\left(2x-y\right)^2-3^2=\left(2x-y-3\right)\left(2x-y+3\right)\)
e)\(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2y+y^2=\left(5x^2-y\right)^2\)
h)\(x^2-7xy+10y^2=x^2-2xy-5xy+10y^2=\left(x^2-2xy\right)-\left(5xy-10y^2\right)=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-5y\right)\left(x-2y\right)\)
\(a,2x^2+7x+100=2\left(x+\frac{7}{4}\right)^2+\frac{751}{8}\ge\frac{751}{8}\)
Dấu " =" xảy ra khi
\(x=\frac{-7}{4}\)
Vậy..............................
\(b,4x^2-25x+9=4\left(x^2-\frac{25}{4}x+\frac{9}{4}\right)\)
\(=4\left(x-\frac{25}{8}\right)^2-\frac{481}{16}\ge\frac{-481}{16}\)
Dấu "=" xảy ra khi \(x=\frac{25}{8}\)
Vậy............................................
A= 2.(x2+2.x.7/4+49/16)2+751/8
= 2.(x+7/4)2+751/8
Lại có (x+7/4)2\(\ge\)0
=> A \(\ge\)751/8
Vậy Min A = 751/8 <=> x= -7/4
b,B= (2x)2-2.2x.25/4+625/16 -481/16
= (2x-25/4)2-481/16
Lại có (2x-25/4)2\(\ge\)0
=> B \(\ge\)-481/16
Vậy min B = -481/16 <=> x= 25/8
(Máy mình hỏng từ đây mình làm tắt một chút)
c, C= (3x)2-24x+16+40= (3x-4)2+40
Lại có (3x-4)2\(\ge\)0
=> C \(\ge\)40
Vậy Min C = 40 <=> 3x-4 =0 <=> x= 4/3
d, D= (2x)2+4x+1+10= (2x+1)2+10
Lại có (2x+1)\(\ge\)0
=> D\(\ge\)10
Vậy min D = 10 <=> x= -1/2
e,E= x^2-2x+1+y2 -4y+4+2
= (x-1)2+(y-2)2+2
Lại có (x-1)2+(y-2)2\(\ge\)0
=> E \(\ge\)2
Vậy Min E = 2 <=> x= 1; y=2
a) \(12x-9-4x^2\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left(2x-3\right)^2\)
b)\(1-9x+27x^2-27x^3\)
\(=\left(1-3x\right)^{^3}\)
c)\(\frac{x^2}{4}+2xy+4y^2-25\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
d)\(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)^2-3\left(x^2-4x\right)-5\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)\left(x^2-4x-3\right)-5\left(x^2-4x-3\right)\)
\(=\left(x^2-4x-5\right)\left(x^2-4x-3\right)\)
\(=\left(x^2+x-5x-5\right)\left(x^2-4x-3\right)\)
\(=\left[x\left(x+1\right)-5\left(x+1\right)\right]\left(x^2-4x-3\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x^2-4x-3\right)\)
Chúc bạn học tốt !
a) \(a^2-2a+1=a^2-2\times a\times1+1^2=\left(a-1\right)^2\)
b) sửa lại đề thành \(1-4x+4x^2\)
\(1-4x+4x^2=1^2-2\times2x\times1+\left(2x\right)^2=\left(1-2x\right)^2\)
c) bạn xem lại đề câu C nha mik làm lun
d) \(25x^2-20xy+4y^2=\left(5x\right)^2-2\times5x\times2y+\left(2y\right)^2=\left(5x-2y\right)^2\)
a) \(a^2-2a+1\)
\(=\left(a-1\right)^2\)
b) Sửa đề \(1-4x+4x^2\)
\(=\left(2x\right)^2-2.2x+1\)
\(=\left(2x-1\right)^2\)
c) \(a^2+9-6x\)
\(=a^2-2.x.3+3^2\)
\(=\left(a-3\right)^2\)
d) \(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)