Bài 2: 

a: \(f\left(x\right)=-9x^3-2x^2+6x-3\)

\(G\left(x\right)=9x^3-6x+53\)

b: \(H\left(x\right)=9x^3-6x+53-9x^3-2x^2+6x-3=-2x^2+50\)

c: Đặt H(x)=0

=>2x²-50=0

=>x=5 hoặc x=-5

a) Xét ∆ ABM(<A=90°(gt)) và ∆NDM(<N=90°(gt)), ta có:
<BMA=<DMN( đối đỉnh)
BM=DM(gt)
⟹∆ABM=∆NDM(c.h=g.n)
b) Ta có: 
<ABM=<MDN(Vì ∆ABM=∆NDM(CM ở a))
mà <ABM=<CBM(gt)
⟹<MDN=<CBM
⟹∆EBD cân tại E
⟹ BE=DE
c)Áp dụng định lý Py-ta-go vào ∆ABC(<A=90°(gt)), ta có:
   BC²=AB²+AC²
⟹AB²=BC²-AC²=15²-12²=225-144=81
⟹AB=√81=9cm
mà AB=DN(Vì ∆ABM=∆NDM(CM ở a))
⟹AB=DN=9cm

Câu 2:

\(R1=R_{nt}-R2=9-6=3\Omega\)

\(=>R_{ss}=\dfrac{R1\cdot R2}{R1+R2}=\dfrac{3\cdot6}{3+6}=2\Omega\)

Chọn A

\(\dfrac{2A}{2A+16.5}=\dfrac{43,66}{100}\)

=> \(200A=43,66.\left(2A+16.5\right)\)

=> \(200A-87,32A=3492,8\)

=> \(112,68A=3492,8\)

=> A= 31 

Cái đó là tìm ra A là bn ạ

a: Δ=(m-2)^2-4(m-4)

=m^2-4m+4-4m+16

=m^2-8m+20

=m^2-8m+16+4

=(m-2)^2+4>=4>0

=>Phương trình luôn có 2 nghiệm pb

b: x1^2+x2^2

=(x1+x2)^2-2x1x2

=(m-2)^2-2(m-4)

=m^2-4m+4-2m+8

=m^2-6m+12

=(m-3)^2+3>=3

Dấu = xảy ra khi m=3

b: \(BC=\sqrt{89}\left(cm\right)\)

\(\sin\widehat{B}=\dfrac{5\sqrt{89}}{89}\)

\(\Leftrightarrow\widehat{B}\simeq32^0\)

\(\widehat{C}=58^0\)

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Gọi vận tốc của ô tô là x

=>Vận tốc xe máy là x-10

Theo đề, ta có: 120/(x-10)-120/x=1

=>(120x-120x+1200)/x(x-10)=1

=>x^2-10x=1200

=>x^2-10x-1200=0

=>x=40

Câu 10:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\notin\left\{2;-1\right\}\\y\ne-5\end{matrix}\right.\)

\(A=\dfrac{y+5}{x^2-4x+4}\cdot\dfrac{x^2-4}{x+1}\cdot\dfrac{x-2}{y+5}\)

\(=\dfrac{y+5}{y+5}\cdot\dfrac{\left(x^2-4\right)}{x^2-4x+4}\cdot\dfrac{x-2}{x+1}\)

\(=\dfrac{\left(x^2-4\right)\cdot\left(x-2\right)}{\left(x+1\right)\left(x^2-4x+4\right)}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)\cdot\left(x-2\right)}{\left(x+1\right)\left(x-2\right)^2}=\dfrac{x+2}{x+1}\)

b: \(A=\dfrac{x+2}{x+1}\)

=>A không phụ thuộc vào biến y

Khi x=1/2 thì \(A=\left(\dfrac{1}{2}+2\right):\left(\dfrac{1}{2}+1\right)=\dfrac{5}{2}:\dfrac{3}{2}=\dfrac{5}{2}\cdot\dfrac{2}{3}=\dfrac{5}{3}\)

Câu 12:

a: \(A=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{x^2-9}\)

\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)+9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{x^2-3x+2x^2+6x+9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x-3}\)

b: Khi x=1 thì \(A=\dfrac{3}{1-3}=\dfrac{3}{-2}=-\dfrac{3}{2}\)

\(x+\dfrac{1}{3}=\dfrac{10}{3}\)

=>\(x=\dfrac{10}{3}-\dfrac{1}{3}\)

=>\(x=\dfrac{9}{3}=3\left(loại\right)\)

Vậy: Khi x=3 thì A không có giá trị

c: \(B=A\cdot\dfrac{x-3}{x^2-4x+5}\)

\(=\dfrac{3}{x-3}\cdot\dfrac{x-3}{x^2-4x+5}\)

\(=\dfrac{3}{x^2-4x+5}\)

\(x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1>=1\forall x\) thỏa mãn ĐKXĐ

=>\(B=\dfrac{3}{x^2-4x+5}< =\dfrac{3}{1}=3\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-2=0

=>x=2