Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
***Hình bạn tự vẽ nha***
a, Xét tam giác ABC và tam giác BHA có :
Góc ABC chung
Góc BAC = góc BHA ( =90°)
==> Tam giác ABC đồng dạng tam giác HBA ( g.g )
==> AB/HB = BC/AB ==> AB^2 = HB. BC
Câu 1:
a: \(=a^2+2ab+b^2-a^2-2ab-b^2=0\)
b: \(=x^3+27-54-x^3=-27\)
Câu 4:
\(\Leftrightarrow3x^3+x^2+9x^2+3x-3x-1-4⋮3x+1\)
\(\Leftrightarrow3x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{0;1\right\}\)
#)Giải :
a) \(ab-ac-b+c=a\left(b-c\right)-\left(b-c\right)=\left(a-1\right)\left(b-c\right)\)
b) \(5a^2-5=5\left(a^2-1\right)=5\left(a-1\right)\left(a+1\right)\)
c) \(x^2-2x+1-a^2-2ab-b^2=\left(x-1\right)^2-\left(a+b\right)^2\)
\(=\left(x-1-a-b\right)\left(x-1+a+b\right)\)
d) \(7x^2-14x+7=7\left(x^2-2x+1\right)=7\left(x-1\right)^2\)
e) \(81x^4+4=81x^4+36x^2+4-36x^2=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)
f) \(x^7+x^2+1=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+...+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+...+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x+1\right)\left(x^2+x+1\right)\)
g) \(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(a^2-b^2+c^2-a^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(a+b\right)-\left(a-b\right)\left(a-c\right)\left(a+c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
1. \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{3}{2}\end{cases}}\)
Vậy \(S=\left\{5;\frac{3}{2}\right\}\)
A B C H 9cm 12cm K I
a. Xét \(\Delta ABC\)và \(\Delta HAC\)có:
Góc C: chung (gt)
Góc HAC = Góc ABC ( cùng phụ với góc ACB)
\(\Rightarrow\Delta ABC\infty\Delta HAC\)
b.Ta có: \(\Delta ABC\infty\Delta HAC\)(cmt)
\(\Rightarrow\frac{BC}{AC}=\frac{AC}{HC}\Rightarrow AC^2=BC.HC=\left(BH+HC\right).HC=\left(9+12\right).12=252cm.\Rightarrow AC=\sqrt{252}=6\sqrt{7}\)
giúp mình với nha m.n