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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 , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
\(\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\)
\(\left(x-2y\right)^2-4\left(x-2y\right)+4=\left(x-2y-2\right)^2\)
\(\left(a^2+1\right)^2-6\left(a^2+1\right)+9=\left(a^2+1-3\right)^2=\left(a^2-2\right)^2\)
\(\left(x+y\right)^2+\left(x+y\right)x+\frac{1}{4}x^2=\left(x+y+\frac{1}{2}x\right)^2=\left(\frac{3}{2}x+y\right)^2\)
\(16x^4-9x^2=x^2\left(16x^2-9\right)=x^2\left(4x-4\right)\left(4x+3\right)\)
\(a^2-b^4=\left(a-b^2\right)\left(a+b^2\right)\)
(x + 2y)2 - 16
= (x + 2y)2 - 42
= (x + 2y - 4).(x + 2y + 4)
(x - 2y)2 - 4.(x - 2y) + 4
= (x - 2y)2 - 2.(x - 2y).2 + 22
= (x - 2y - 2)2
(a2 + 1)2 - 6.(a2 + 1) + 9
= (a2 + 1)2 - 2.(a2 + 1).3 + 32
= (a2 + 1 - 3)2
= (a2 - 2)2
(x + y)2 + (x + y).x + 1/4.x2
= (x + y)2 + 2.(x + y).1/2.x + (1/2.x)2
= (x + y + 1/2.x)2
= (3/2.x + y)2
16x4 - 9x2
= (4x2)2 - (3x)2
= (4x2 - 3x).(4x2 + 3x)
a2 - b4
= a2 - (b2)2
= (a - b2).(a + b2)
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a, b, c, dung hang dang thuc A2-B2 =9A-B) (A+B)
d) a4 - 16 =(a2)2 -42= (a2-4) (a2+4) =(a-2)(a+2) (a2+4)
\(a..9-4x^2=3^2-\left(2x\right)^2=\left(3-2x\right)\left(3+2x\right)\)
\(b..16x^2-25=\left(4x\right)^2-5^2=\left(4x-5\right)\left(4x+5\right)\)
\(c..a^4-16=\left(a^2\right)^2-4^2=\left(a^2-4\right)\left(a^2+4\right)\)
\(d..\left(a+b\right)^2-1=\left(a+b\right)^2-1^2=\left(a+b-1\right)\left(a+b+1\right)\)
k mik đi
1)
\(=x^2-4x+4+y^2+2y+1\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
2)
\(=a^2+2ab+b^2+a^2-2ax+x^2\)
\(=\left(a+b\right)^2+\left(a-x\right)^2\)
3)
\(=x^2-2x+1+y^2+6y+9\)
\(=\left(x-1\right)^2+\left(y+3\right)^2\)
4)
\(=x^2-2xy+y^2+x^2+10x+25\)
\(=\left(x-y\right)^2+\left(x+5\right)^2\)
5)
\(=a^2+2ab+b^2+4b^2+4b+1\)
\(=\left(a+b\right)^2+\left(2b+1\right)^2\)
1/ x2 - 4x + 5 + y2 + 2y
= ( x2 - 4x + 4 ) + ( y2 + 2y + 1 )
= ( x - 2 )2 + ( y + 1 )2
2/ 2a2 + 2ab - 2ax + x2 + b2
= ( a2 + 2ab + b2 ) + ( x2 - 2ax + a2 )
= ( a + b )2 + ( x - a )2
3/ x2 - 2x + y2 + 6y + 10
= ( x2 - 2x + 1 ) + ( y2 + 6y + 9 )
= ( x - 1 )2 + ( y + 3 )2
4/ 2x2 + y2 - 2xy + 10x + 25
= ( x2 - 2xy + y2 ) + ( x2 + 10x + 25 )
= ( x - y )2 + ( x + 5 )2
5/ a2 + 2ab + 5b2 + 4b + 1
= ( a2 + 2ab + b2 ) + ( 4b2 + 4b + 1 )
= ( a + b )2 + ( 2b + 1 )2
oh
no way bro
⇒ \(x^{2} + 14 x + 49 = \left(\right. x + 7 \left.\right)^{2} .\)
mà :\(4 x^{2} + 20 x + 25 = \left(\right. 2 x \left.\right)^{2} + 2 \cdot \left(\right. 2 x \left.\right) \cdot 5 + 5^{2}\)
⇒ \(4 x^{2} + 20 x + 25 = \left(\right. 2 x + 5 \left.\right)^{2} .\)
Do đó: \(0 , 25 x^{2} + 3 x + 9 = \left(\left(\right. \frac{x}{2} \left.\right)\right)^{2} + 2 \cdot \frac{x}{2} \cdot 3 + 3^{2}\)
⇒ \(0 , 25 x^{2} + 3 x + 9 = \left(\left(\right. \frac{x}{2} + 3 \left.\right)\right)^{2} .\)
Do đó: \(16 x^{2} + 4 x + \frac{1}{4} = \left(\right. 4 x \left.\right)^{2} + 2 \cdot 4 x \cdot \frac{1}{2} + \left(\left(\right. \frac{1}{2} \left.\right)\right)^{2}\)
⇒ \(16 x^{2} + 4 x + \frac{1}{4} = \left(\left(\right. 4 x + \frac{1}{2} \left.\right)\right)^{2} .\)
help me pls
hay
Đa tạ bn Vũ Thư Anh
1. \(x^2+14x+49\)
=\(x^2+2\cdot x\cdot7+7^2\)
=\(\left(x+7\right)^2\)
2. \(20x+4x^2+25\)
\(=\left(2x\right)^2+2\cdot2x\cdot5+5^2\)
=\(\left(2x+5\right)^2\)
3. \(0,25x^2+3x+9\)
\(=\left(0,5x\right)^2+2\cdot0,5x\cdot3+3^2\)
= \(\left(0,5x+3\right)^2\)
4.\(16x^2+4x+\frac14\)
\(=\left(4x\right)^2+2\cdot4x\cdot\frac12+\left(\frac12\right)^2\)
\(=\left(4x+\frac12\right)^2\)
t kbt
Câu 1. x² + 14x + 49 = (x + 7)², vì x² là bình phương của x, 49 là bình phương của 7, 14x = 2.x.7 nên có dạng bình phương của một tổng
Câu 2. 20x + 4x² + 25 = 4x² + 20x + 25 = (2x + 5)², vì 4x² là bình phương của 2x, 25 là bình phương của 5, 20x = 2.2x.5 nên có dạng bình phương của một tổng