Viết các biểu thức sau dưới dạng bình phương của một tổng hoặc hiệu
a)-8x+16+x^2
b)
Chỉ rõ cho mình cách giải nha cảm ơn mọi người
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\(a,=\left(x-4\right)^2\\ b,=\left(\dfrac{1}{2}xy^2+1\right)^2\)
a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
a) ( x + 1 ) 2 . b) ( x – 4 ) 2 .
c) x 2 4 + x + 1 ; d) ( 2 x – 2 y ) 2 .
\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)
`a)x^2+20x+100=(x+10)^2`
`b)16x^2+24xy+9y^2=(4x+3y)^2`
`c)y^2-14y+49=(y-7)^2`
`d)9x^2-42xy+49y^2=(3x-7y)^2`
a, \(x^2+2x.10+100=\left(x+10\right)^2\)
\(b,16x^2+2.4x.3y+9y^2=\left(4x+3y\right)^2\)
\(c,y^2-14y+49=\left(y-7\right)^2\)
\(d,9x^2-2.3x.7x+49y^2=\left(3x-7y\right)^2\)
\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)
\(a,\)
với \(a=100\)
\(=>9x^2+30x+25=\left(3x\right)^2+2.3.5x+5^2=\left(3x_{ }+5\right)^2\)
\(b,\)
với \(a=\dfrac{1}{25}\)
\(25x^2-2x+\dfrac{1}{25}=\left(5x\right)^2-2.5.x.\dfrac{1}{5}+\left(\dfrac{1}{5}\right)^2=\left(5x-\dfrac{1}{5}\right)^2\)
\(c,\)
với \(a=6\)
\(=>x^2+2.3.x+3^2=\left(x+3\right)^2\)
\(d.\)
với \(a=\dfrac{4}{3}\)
\(=>\left(2x\right)^2-2.2.\dfrac{1}{3}x+\left(\dfrac{1}{3}\right)^2=\left(2x-\dfrac{1}{3}\right)^2\)
\(a,=\left(x+\dfrac{5}{2}\right)^2\\ b,=\left(2x+3y\right)^2\\ c,=a^2+b^2+c^2+2ab-2bc-2ac\\ d,=\left(4x-1\right)^2\\ e,=a^2+b^2+c^2+2ab+2bc+2ac\\ f,=a^2+b^2+c^2-2ab+2bc-2ac\)
1, \(x^2+2xy+y^2=\left(x+y\right)^2\)
2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)
5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
1: =(x+y)^2
2: =(2x+3)^2
3: =(x+5/2)^2
4: =(4x-1)^2
5: =(x+1/2)^2
6: =(x-3/2)^2
7: =(x+1)^3
8: =(1/2x+1)^2
9: =(3y-1/3)^3
10: =(2x+y)^3
Lời giải:
a. $-8x+16+x^2=x^2-2.x.4+4^2=(x-4)^2$
b. $xy^2+\frac{1}{4}x^2y^4+1=(\frac{1}{2}xy^2)^2+2.\frac{1}{2}xy^2.1+1^2$
$=(\frac{1}{2}xy^2+1)^2$
a: \(x^2-8x+16=\left(x-4\right)^2\)
b: \(\dfrac{1}{4}x^2y^4+xy^2+1=\left(\dfrac{1}{2}xy^2+1\right)^2\)