Tìm số tự nhiên bé nhất có 2 chữ số ? số đó là 10

GTLN=\(y=\frac{8}{9^3}\)

a) \(\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)

\(=\left(\sqrt{2-\sqrt{3}}\right)^2-2\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}+\left(\sqrt{2+\sqrt{3}}\right)^2\)

\(=2-\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2+\sqrt{3}\)

\(=4-2\sqrt{1}\)

\(=2\)

để A ko phải là p/s tối giản, ý bn là thế phải ko

là 15 viên

a. \(M=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{1-\sqrt{a}}{\sqrt{a}+1}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}+\dfrac{\sqrt{a}}{1-a}\right)\)

\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}-\dfrac{\left(\sqrt{a}-1\right)^2}{a-1}\right):\left(\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}-\dfrac{\sqrt{a}}{a-1}\right)\)\(=\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}:\dfrac{\left(\sqrt{a}+1\right)^2+\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}}{a-1}\)

\(=\dfrac{\left(a+2\sqrt{a}+1\right)-\left(a-2\sqrt{a}+1\right)}{a-1}:\dfrac{\left(a+2\sqrt{a}+1\right)+a-\sqrt{a}-\sqrt{a}}{a-1}\)

\(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1}{a-1}:\dfrac{a+2\sqrt{a}+1+a-\sqrt{a}-\sqrt{a}}{a-1}\)

\(=\dfrac{4\sqrt{a}}{a-1}:\dfrac{2a+1}{a-1}\)

\(=\dfrac{4\sqrt{a}}{a-1}.\dfrac{a-1}{2a+1}\)

\(=\dfrac{4\sqrt{a}\left(a-1\right)}{\left(a-1\right)\left(2a+1\right)}\)

\(=\dfrac{4\sqrt{a}}{2a+1}\)

á mình nhầm rùi

A. đúng , đúng 100%

ủng hộ nhé

a. Xét \(\Delta ABC\) vuông tại A ta có:

\(+AB^2=BH^2+AH^2=9^2+12^2=225\)

=> AB = 15cm

\(+AH^2=BH.HC\)

=> \(HC=\dfrac{AH^2}{BH}=\dfrac{12^2}{9}=16cm\)

Ta có: BC = BH + HC = 9 + 16 = 25 (cm)

Xét \(\Delta ABC\) vuông tại A ta có:

\(AB^2+AC^2=BC^2\)

=> \(AC^2=BC^2-AB^2=25^2-15^2=400\)

=> AC = 20(cm)

b. Diện tích \(\Delta ABC\) là:

\(S_{ABC}=\dfrac{1}{2}BC.AH=\dfrac{1}{2}.25.12=150cm^2\)