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Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
b)(y-2)^3=y^3-8+12y-6y^2
c)8x^3+y^3=(2x+y)(4x^2+y^2-4xy)
2)
=(xy+2/3)^2
a) 4x2+4xy+y2
=(2x)2+2(2x)(y)+y2
=(2x+y)2
b)9m2+n2-6mn
=(3m)2-2(3m)n+n2
=(3m-n)2
c)16a2+25b2+40ab
=(4a)2+(5b)2+2(4a)(5b)
=(4a+5b)2
d)x2-x+\(\frac{1}{4}\)
=x2-2x.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)
=\(\left(x-\frac{1}{2}\right)^2\)
\(a,x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2 \)
b,\(a^2-\frac{4}{3}ba+\frac{4}{9}b^2\)=\(\left(a-\frac{2}{3}b\right)^2\)
c,\(a^2+10ab+25b^2\)=\(\left(a+5b\right)^2\)
d,\(x^2-6xy^2+9y^4\) =(\(x-3y^2\))\(^2\)
e,\(a^2+a+\frac{1}{4}\)=\(\left(a+\frac{1}{2}\right)^2\)
Cho mik điểm SP đi ak <3
a) \(x^2+6x+9=\left(x+3\right)^2\)
b) \(2xy^2+x^2y^4+1=x^2y^4+2xy^2+1=\left(xy^2+1\right)^2\)
c) \(x^2+x+\frac{1}{4}=x^2.2.\frac{1}{2}x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
a.
\(\frac{x^2}{4}+x+3=\frac{x^2}{4}+x+1+2=\left(\frac{x}{2}+1\right)^2+2>0;\forall x\)
b.
\(A=-3x^2+2x-5=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}\right)-\frac{14}{3}=-3\left(x-\frac{1}{3}\right)^2-\frac{14}{3}\le-\frac{14}{3}\)
\(A_{max}=-\frac{14}{3}\) khi \(x=\frac{1}{3}\)
c.
Đề thiếu (để ý 2 số hạng cuối)
\(A=x^4-2x^3+x^2+3x^2-6x+3-1\)
\(=\left(x^2-x\right)^2+3\left(x-1\right)^2-1\ge-1\)
\(A_{min}=-1\) khi \(x=1\)
d.
\(27x^2-\frac{9}{2}x+\frac{3}{16}=3\left(9x^2-\frac{3}{2}x+\frac{1}{16}\right)=3\left(3x-\frac{1}{4}\right)^2\)
e.
\(=\left[\left(b+c\right)+a\right]^2+\left[\left(b+c\right)-a\right]^2+\left[a-\left(b-c\right)\right]^2+\left[a+\left(b-c\right)\right]^2\)
\(=2\left(b+c\right)^2+2a^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2b^2+4bc+2c^2+2b^2-4bc+2c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
f.
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+b^2d^2+2ac.bd\right)+\left(a^2d^2+b^2c^2-2ad.bc\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
a)x2+2x+1=x2+2x.1+12=(x+1)2
b)x2-x+\(\frac{1}{4}\)=x2-2.x.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)=\(\left(x-\frac{1}{2}\right)^2\)
1) viết biểu thức dưới dạng bình phương của 1 tổng hoặc 1 hiệu
a) x2 - 2x +1
= x2 - 2 . x . 1 + 12
= ( x + 1 )2
b) 81x2 + 1 +18x
= 81x2 + 18x + 1
= (9x)2 + 2 . 9x . 1 + 12
= (9x + 1)2
c)1/4 + x2 + x
= x2 + 2 . x . 1/2 + (1/2)2
= (x + 1/2)2
~.~
a) (a - 2b)2 = a2 - 2.a.2b + 4b2
= a2 - 4ab + 4b2
b) m2 - 4n2 = m2 - (2n)2 = (m - 2n)(m + 2n)
. Bài 1:
a; 9m^2 + n^2 - 6mn
= (3m)^2 - 2.3m.n + (n)^2
= ( 3m-n )^2
b; x^2-x+1/4
= x^2-2.(x).1/2+(1/2)^2
= (x-1/2)^2