Câu 2 :

Ta có : \(AH=HB.HC\) (Hệ thức lượng trong \(\Delta ABC\))

\(\Rightarrow AH=3.9=27\left(cm\right)\)

Xét \(\Delta ABH\perp\) tại H có :

\(AB^2=HB^2+AH^2\) \(\left(d/lPytago\right)\)

\(\Rightarrow AB^2=9^2+27^2=810\)

\(\Rightarrow AB=\sqrt{810}=9\sqrt{10}\left(cm\right)\)

Vậy \(AB=9\sqrt{10}cm\)

Fruits and vegetables are ..better....than sweets and cakes

\(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ =\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}\\ =\dfrac{x+2\sqrt{x}}{x-4}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để \(A>1\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}>1\Leftrightarrow\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}>0\Leftrightarrow2>0\left(LD\right)\)

\(\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow x>4\left(tm\right)\)

Vậy \(x>4\) thì \(A>1\).

\(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\\ \Rightarrow\dfrac{2\left(2x-3\right)-3.3-\left(5-3x\right)+2}{6}=0\\ \Rightarrow4x-6-9-5+3x+2=0\)

\(\Rightarrow7x=18\\ \Rightarrow x=\dfrac{18}{7}\)

Vậy ...

\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\left(dk:x\ge0,x\ne25\right)\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)+20-2\sqrt{x}}{x-25}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{x-25}\\ =\dfrac{\sqrt{x}+35}{x-25}\)

\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\left(dk:x\ge0,x\ne9\right)\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x+5\sqrt{x}-2}{x-9}\)

\(c,\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{x-4}\\ =\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}\\ =\dfrac{x+2\sqrt{x}}{x-4}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(d,\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(dk:x\ge0,x\ne1\right)\\ =\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

1.Would you like to join the summer camp with us ?

2.Owen is plays basketball and sporty well.

3.At present my sister is helping my physics homework with me.

\(A=\left\{x\in Z|\left(x^2-9\right)\left(x^2-7\right)\left(3x+5\right)=0\right\}\)

Giải pt \(\left(x^2-9\right)\left(x^2-7\right)\left(3x+5\right)=0\) \(\left(dk:x\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-7=0\\3x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm3\left(tm\right)\\x=\pm\sqrt{7}\left(ktm\right)\\x=-\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

Vậy \(A=\left\{-3;3\right\}\)

\(\sqrt{x^2+6x+9}=2x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=2x-1\\ \Leftrightarrow\left|x+3\right|=2x-1\\ TH_1:x\ge-3\\ x+3=2x-1\Leftrightarrow-x=-4\Leftrightarrow x=4\left(tm\right)\\ TH_2:x< -3\\ -x-3=2x-1\Leftrightarrow-3x=2\Leftrightarrow x=-\dfrac{2}{3}\left(tm\right)\)

Vậy \(S=\left\{-\dfrac{2}{3};4\right\}\)

Ta có : \(A=3cm\)

\(x=3sin2\left(\pi t-\dfrac{\pi}{4}\right)\left(cm;s\right)\)

Quỹ đạo \(L=2A=2.3=6cm\)

Chu kì \(T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\pi}=2s\)

Tần số \(f=\dfrac{1}{T}=\dfrac{1}{2}Hz\)

Pha ban đầu \(\phi=-\dfrac{\pi}{4}rad\)

Sắp xếp

showed/her/year/girl/has/results/of/study/the/mother/The/183/that/a/rows/with/teenage/each/typical.