\(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2\)

\(=\left(x+y+2\right)^2\)

a) ta có

ˆHBA+ˆHAB=900;ˆHAB+ˆHAF=900⇒ˆHBA=ˆHAF(1)HBA^+HAB^=900;HAB^+HAF^=900⇒HBA^=HAF^(1)

ˆBHE+ˆEHA=900;ˆEHA+ˆFHA=900⇒ˆBHE=ˆFHA(2)BHE^+EHA^=900;EHA^+FHA^=900⇒BHE^=FHA^(2)

xét △BEH và △AFH có

(1) và (2)

⇒ △BEH ~ △AFH(g - g)

b) xét △AHB và △CAB có

ˆH=ˆA=900;ˆBH^=A^=900;B^ chung

⇒ △AHB ~ △CAB (g - g)

⇒BHBA=AHAC⇒BHAH=ABAC⇒BHBA=AHAC⇒BHAH=ABAC

từ câu a ⇒ EHFH=BHAHEHFH=BHAH

⇒ ABAC=EHFH⇒ABEH=ACFH(3)ABAC=EHFH⇒ABEH=ACFH(3)

xét △CAB và △FHE có

(3); ˆA=ˆH=900A^=H^=900

⇒ △CAB ~ △FHE (g - g)

⇒ ABHE=BCEF⇒AB.EF=HE.BCABHE=BCEF⇒AB.EF=HE.BC ⇒ đpcm

\(a^6-a^4+2a^3+2a^2=a^4\left(a^2-1\right)+2a^2\left(a+1\right)\)

\(=a^4\left(a-1\right)\left(a+1\right)+2a^2\left(a+1\right)=\left(a+1\right)\left(a^5-a^4+2a^2\right)\)

F thuộc đoạn nào ??

cm: HEF = BCA

a) \(x^3+8y^3=x^3+\left(2y\right)^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

b) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

c) \(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)

a) x^3 + 8y^3 = (x + 8y)(x^2 - 8xy + (8y)^2) = (x + 8y)(x^2 - 8xy + 64y^2)

b) a^6 + b^3 = (a^2)^3 + b^3 = (a^2 + b)((a^2)^2 - a^2b + b^2) = (a^2 + b)(a^4 - a^2b + b^2)

c) 8y^3 - 125 = 8y^3 - 5^3 = (8y - 5)((8y)^2 + 8y x 5 + 5^2) = (8y - 5)(64y^2 + 40y + 25)

cố lên nha

\(=bc\left(b+c\right)\)\(+ac^2\)\(-a^2\)\(c\)\(-a^2\)\(b\)\(-ab^2\)

\(=bc\left(b+c\right)\)\(-a^2\)\(\left(b+c\right)\)\(-a\left(b+c\right)\)\(\left(b-c\right)\)

\(=\left(b+c\right)\)\(\left(bc-a^2-ab+ac\right)\)

\(=\left(b+c\right)\)\(\orbr{c\left(b+a\right)-a\left(a+b\right)}\)

\(=\left(b+c\right)\)\(\left(b+a\right)\)\(\left(c-a\right)\)

Tỉ lệ sai 1%

a) \(xy\left(y-7\right)+7y\left(1+x\right)\)

\(=xy^2-7xy+7y+7xy=xy^2+7y\)

Thay vào ta được:

\(=\left(-6\right).1^2+7.1=\left(-6\right)+7=1\)

b) \(xy-7x+y-7\)

\(=xy+y-7x-7=y\left(x+1\right)-7\left(x+1\right)=\left(y-7\right)\left(x+1\right)\)

Thay vào ta được:

\(=\left(10-7\right)\left(9+1\right)=3.10=30\)

c) \(xy\left(y-2\right)+2x\left(1+x\right)\)

Thay vào ta được:

\(\left(-1\right).2\left(2-2\right)+2\left(-1\right)[1+\left(-1\right)]=0+0=0\)

\(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\left(\frac{x}{2}\right)^2-y^2=\frac{x^2}{4}-y^2\)

Đề bài là gì vậy ??

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