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a) x2+6xy+3y2=(x+3y)2
b) (a-2b2)2=a2-2.a.2b2+4b4
c) (m+1/2)2=m2+m+1/4
d)m2-4n4=(m+2n2)(m-2n)2
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a)x2+6xy+9y2=(x+3y)2
b)(a-2b)2=a2-4ab+4b2
c)(m+\(\dfrac{1}{2}\))2=m2+m+\(\dfrac{1}{4}\)
d)m2-4n2=(m-2n)(m+2n)
Bài giải:
a) x2 + 2 . x . 3y + … = (…+3y)2
x2 + 2 . x . 3y + (3y)2 = (x + 3y)2
Vậy: x2 + 6xy +9y2 = (x + 3y)2
b) …-2 . x . 5y + (5y)2 = (… - …)2;
x2 – 2 . x . 5y + (5y)2 = (x – 5y)2
Vậy: x2 – 10xy + 25y2 = (x – 5y)2
Bài giải:
a) x2 + 2 . x . 3y + … = (…+3y)2
x2 + 2 . x . 3y + (3y)2 = (x + 3y)2
Vậy: x2 + 6xy +9y2 = (x + 3y)2
b) …-2 . x . 5y + (5y)2 = (… - …)2;
x2 – 2 . x . 5y + (5y)2 = (x – 5y)2
Vậy: x2 – 10xy + 25y2 = (x – 5y)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
c)\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
d)\(\dfrac{a^2}{4}-2a+4=\left(\dfrac{a}{2}-2\right)^2\)
e) \(4y^2-9x^2=\left(2y-3x\right)\left(2y+3x\right)\)
f)\(9y^2-\dfrac{1}{4}=\left(3y-\dfrac{1}{2}\right)\left(3y+\dfrac{1}{2}\right)\)
g)\(8x^3+8a^3=\left(2x+2a\right)\left(4x^2-4xa+4a^2\right)\)
1,
\(\left(\frac{2}{3}x+y\right)^2=\left(\frac{2}{3}x\right)^2+2.\frac{2}{3}x.y+\left(y\right)^2=\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
\(\left(3a+\frac{1}{2}b\right)^2=\left(3a\right)^2+2.3a.\frac{1}{2}b+\left(\frac{1}{2}b\right)^2=9a^2+3ab+\frac{1}{4}b^2\)
2,
\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)
\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)
\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)
\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
a)
\(a^2+b^2+c^2+d^2+m^2-a(b+c+d+m)\)
\(=\frac{4a^2+4b^2+4c^2+4d^2+4m^2-4a(b+c+d+m)}{4}\)
\(=\frac{(a^2+4b^2-4ab)+(a^2+4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4m^2-4am)}{4}\)
\(=\frac{(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2m)^2}{4}\geq 0\) (đpcm)
Dấu "=" xảy ra khi \(a=2b=2c=2d=2m\)
b)
Xét hiệu
\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}=\frac{x+y}{xy}-\frac{4}{x+y}=\frac{(x+y)^2-4xy}{xy(x+y)}\)
\(=\frac{x^2+y^2-2xy}{xy(x+y)}=\frac{(x-y)^2}{xy(x+y)}\geq 0, \forall x,y>0\)
\(\Rightarrow \frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\) (đpcm)
Dấu "=" xảy ra khi $x=y$
c)
Xét hiệu:
\((a^2+c^2)(b^2+d^2)-(ab+cd)^2\)
\(=(a^2b^2+a^2d^2+c^2b^2+c^2d^2)-(a^2b^2+2abcd+c^2d^2)\)
\(=a^2d^2-2abcd+b^2c^2=(ad-bc)^2\geq 0\)
\(\Rightarrow (a^2+c^2)(b^2+d^2)\geq (ab+cd)^2\) (đpcm)
Dấu "=" xảy ra khi \(ad=bc\)
d)
Xét hiệu:
\(a^2+b^2-(a+b-\frac{1}{2})=a^2+b^2-a-b+\frac{1}{2}\)
\(=(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})\)
\(=(a-\frac{1}{2})^2+(b-\frac{1}{2})^2\geq 0\)
\(\Rightarrow a^2+b^2\geq a+b-\frac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(=>a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)=\left(ax\right)^2+2axby+\left(by\right)^2\)
\(=>a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2axby-b^2y^2=0\)
\(=>a^2y^2+b^2x^2-2axby=0=>\left(ay-bx\right)^2=0\)
=>ax-by=0=>ax=by
Vậy .....................
2) b)
Xét hiệu :
\(100^2+103^2+105^2+94^2-\left(101^2+98^2+96^2+107^2\right)\)
\(=100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)
\(=\left(100^2-98^2\right)+\left(103^2-101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)
\(=\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+1\right)-\left(107-105\right)\left(107+105\right)\)\(-\left(96-94\right)\left(96+94\right)\)
\(=2.198+2.204-2.212-2.190=2\left(198+204-212-190\right)=2.0=0\)
Vậy 1002+1032+1052+942=1012+982+962+1072
a)\(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b)\(\left(a-2b^2\right)^2=a^2-4ab^2+4b^4\)
c)\(\left(m+\dfrac{1}{2}\right)^2=m^2+m+\dfrac{1}{4}\)
d) \(m^2-4n^4=\left(m-2n^2\right)\left(m+2n^2\right)\)
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