Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) \(-4x^2-4x-1\)
\(=-\left(4x^2+4x+1\right)\)
\(=-\left(2x+1\right)^2\)
c) \(\frac{4}{9}x^2-25y^2\)
\(=\left(\frac{2}{3}x+5y\right)\left(\frac{2}{3}x-5y\right)\)
d) \(\frac{1}{27}x^3-8\)
\(=\left(\frac{1}{3}x-2\right)\left(\frac{1}{9}x+\frac{2}{3}x+4\right)\)
a, \(x^2+10x+25=x^2+5x+5x+25\)
\(=\left(x+5\right)^2\)
b, \(x^2-12x+36=x^2-6x-6x+36\)
\(=\left(x-6\right)^2\)
c, \(9x^2+4+12x=9x^2+6x+6x+4\)
\(=3x\left(3x+2\right)+2\left(3x+2\right)=\left(3x+2\right)^2\)
d, \(x^2+49-14x=x^2-7x-7x+49\)
\(=\left(x-7\right)^2\)
e, \(9x^4+24x^2+16=9x^4+12x^2+12x^2+16\)
\(=3x^2\left(3x^2+4\right)+4\left(3x^2+4\right)=\left(3x^2+4\right)^2\)
g,\(4x^2-12xy+9y^2=4x^2-6xy-6xy+9y^2\)
\(=2x\left(2x-3y\right)-3y\left(2x-3y\right)=\left(2x-3y\right)^2\)
Chúc bạn học tốt!!!
\(a,2x^2+7x+100=2\left(x+\frac{7}{4}\right)^2+\frac{751}{8}\ge\frac{751}{8}\)
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, (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)Tại \(x=87;y=13\) thì
\(A=x^2-y^2=\left(x-y\right)\left(x+y\right)\)
\(=\left(87-13\right)\left(87+13\right)=74\cdot100=7400\)
b)Tại \(x=\dfrac{1}{3}\) thì
\(B=9x^2-6x+1=\left(x-\dfrac{1}{3}\right)^2\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)^2=0^2=0\)
c)Tại \(x=1;y=2\) thì
\(C=4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
\(=\left(2\cdot1-3\cdot2\right)^2=\left(-4\right)^2=16\)
a, Ta có:
\(A=x^2-y^2=\left(x-y\right)\left(x+y\right)\)
Thay \(x=87;y=13\) vào A ta được:
\(\left(87-13\right)\left(87+13\right)=74.100=7400\)
b, Ta có:
\(B=9x^2-6x+1=9x^2-3x-3x+1\)
\(=3x\left(3x-1\right)-\left(3x-1\right)\)
\(=\left(3x-1\right)^2\)
Thay \(x=\dfrac{1}{3}\) vào B ta được:
\(\left(3.\dfrac{1}{3}-1\right)^2=0\)
c, Ta có:
\(C=4x^2-12xy+9y^2=4x^2-6xy-6xy+9y^2\)
\(=2x\left(2x-3y\right)-3y\left(2x-3y\right)\)
\(=\left(2x-3y\right)^2\)
Thay \(x=1;y=2\) vào biểu thức C ta được:
\(\left(2.1-3.2\right)^2=\left(2-6\right)^2=\left(-4\right)^2=16\)
Chúc bạn học tốt!!!
\(a,\)\(100a^2+20a+1\)
\(=\left(10a\right)^2+2.10a.1+1^2\)
\(=\left(10a+1\right)^2\)
Bài làm:
a) Sửa đề: \(4x^2-y^2\)
\(=\left(2x\right)^2-y^2\)
\(=\left(2x-y\right)\left(2x+y\right)\)
b) \(a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
c) \(x^2-2xy+y^2\)
\(=\left(x-y\right)^2\)
d) \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
b) \(a^2+2ab+b^2=\left(a+b\right)^2.\)
c) \(x^2-2xy+y^2=\left(x-y\right)^2.\)
d) \(x^2+4xy+4y=\left(x+2y\right)^2\)
câu a chịu
a) \(x^2-9\)
= \(x^2-3^2\)
\(=\left(x-3\right)\left(x+3\right)\)
b) \(4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
c) \(9x^2-y^2\)
\(=\left(3x\right)^2-y^2\)
\(=\left(3x-y\right)\left(3x+y\right)\)
d) \(25x^2-9y^2\)
\(=\left(5x\right)^2-\left(3y\right)^2\)
\(=\left(5x-3y\right)\left(5x+3y\right)\)
e) \(36x^2-25y^2\)
\(=\left(6x\right)^2-\left(5y\right)^2\)
\(=\left(6x-5y\right)\left(6x+5y\right)\)
a) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
b) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x-1\right)\left(2x+1\right)\)
tương tự