10.D

11.C

 a) Xét ∆AHB(<H=90°(gt))  và ∆AHC(<H=90°(gt)), ta có:
AB=AC(gt)
<B=<C(gt)
⟹∆AHB=∆AHC(c.h-g.n)
b)  Xét ∆AHM(<M=90°(gt))  và ∆AHN(<N=90°(gt)), ta có:
AH cạnh chung
<MAH=NAH( vì ∆AHB=∆AHC(CM ở a))
⟹∆AHM=∆AHN(c.h-g.n)
⟹AM=AN ( 2 cạnh tương ứng)
⟹∆AMN cân tại A
c)Ta có: <M=<N=(180°-<A)/2
               <B=<C=(180°-<A)/2
⟹ <M=<N=<B=<C
⟹<M=<B mà 2 góc này lại ở vị trí đồng vị
⟹MN//BC
 

a, \(2^3.2^5=2^8=256\)   

    \(\left(-3\right)^9:\left(-3\right)^5=\left(-3\right)^4=81\)     

    \(\left(-6\right)^9.6^5=\left(-1\right)^9.6^9.6^5=\left(-1\right).6^{14}\\ \left(\dfrac{1}{2}\right)^5=\dfrac{1}{32}\)

b, \(\left(\dfrac{3}{5}\right)^6.\left(\dfrac{5}{3}\right)^6=\left(\dfrac{3}{5}. \dfrac{5}{3}\right)^6=1^6=1\\ \left(-\dfrac{7}{8}\right)^9:\left(\dfrac{7}{4}\right)^9=\left(-\dfrac{7}{8}:\dfrac{7}{4}\right)^9=\left(-\dfrac{1}{2}\right)^9=-\dfrac{1}{512}\\ \left(\left(-\dfrac{1}{2}\right)^2\right)^3=\left(\dfrac{1}{2}\right)^{...}\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)...\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)^6\)

c, \(\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\left(\dfrac{2}{3}\right)^4\right)^2=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^2\\ \left(\dfrac{1}{3}\right)^{12}:\left(-\dfrac{3}{9}\right)^{12}=\left(\dfrac{1}{3}.\left(-3\right)\right)^{12}=\left(-1\right)^{12}=1\\ \left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{3}\right)^{10}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)

???  Sao tui chẳng thấy j nhể

Cho mình hỏi từ câu C trở xuống đc ko ạ tại mắt mình yếu nên nhìn không rõ ấy

a: Xét ΔABD và ΔACD có

AB=AC

BD=CD

AD chung

Do đó: ΔABD=ΔACD

b: ΔABD=ΔACD

=>\(\widehat{ADB}=\widehat{ADC}\)

mà \(\widehat{ADB}+\widehat{ADC}=180^0\)(hai góc kề bù)

nên \(\widehat{ADB}=\widehat{ADC}=\dfrac{180^0}{2}=90^0\)

c: Ta có: \(\widehat{ADB}=90^0\)

=>AD\(\perp\)BC tại D

D là trung điểm của BC

=>\(DB=DC=\dfrac{BC}{2}=\dfrac{24}{2}=12\left(cm\right)\)

ΔADB vuông tại D

=>\(AD^2+DB^2=AB^2\)

=>\(AD^2=20^2-12^2=256\)

=>\(AD=\sqrt{256}=16\left(cm\right)\)

Xét ΔABC có

AD là đường trung tuyến

G là trọng tâm

Do đó: \(AG=\dfrac{2}{3}AD=\dfrac{2}{3}\cdot16=\dfrac{32}{3}\left(cm\right)\)

1: \(\sqrt{11}\) là số vô tỉ

2:

a: 4,9(18)=4,91818...

mà 4,91818<4,928

nên 4,9(18)<4,928

b: 4,315<4,318

=>-4,315>-4,318

=>-4,315...>-4,318...

c: \(\sqrt{3}=\sqrt{\dfrac{6}{2}}< \sqrt{\dfrac{7}{2}}\)

3: 

a: \(6=\sqrt{3};-1,7=-\sqrt{2,89}\)

0<2,89<3

=>\(0< \sqrt{2,89}< \sqrt{3}\)

=>\(-\sqrt{3}< -\sqrt{2,89}< 0\)

0<35<36<47

=>\(0< \sqrt{35}< \sqrt{36}< \sqrt{47}\)

=>\(-\sqrt{3}< -\sqrt{2,89}< 0< \sqrt{35}< \sqrt{36}< \sqrt{47}\)

=>\(-\sqrt{3}< -\sqrt{2,89}< 0< \sqrt{35}< 6< \sqrt{47}\)

b: \(-\sqrt{2\dfrac{1}{3}}=-\sqrt{2,\left(3\right)}\)

\(-1,5=-\sqrt{2,25}\)

2,25<2,3<2,(3)

=>\(\sqrt{2.25}< \sqrt{2.3}< \sqrt{2.\left(3\right)}\)

=>\(0>-1.5>-\sqrt{2.3}>-\sqrt{2\dfrac{1}{3}}\)

\(0< \sqrt{5\dfrac{1}{6}}=\sqrt{5,1\left(6\right)}< \sqrt{5,3}\)

=>\(\sqrt{5.3}>\sqrt{5\dfrac{1}{6}}>0>-1.5>-\sqrt{2.3}>-\sqrt{2\dfrac{1}{3}}\)

a) Ta có:

\(\widehat{yOu}+\widehat{xOy}=180^o\) (kề bù) 

\(\Rightarrow\widehat{yOu}=180^o-\widehat{xOy}\)

\(\Rightarrow\widehat{yOu}=180^o-60^o=120^o\) 

Mà: \(\widehat{xOt}+\widehat{tOu}=180^o\) (kề bù) 

\(\Rightarrow\widehat{xOt}=180^o-\widehat{tOu}\)

\(\Rightarrow\widehat{xOt}=180^o-30^o=150^o\)

b) Ta có:

\(\widehat{xOy}+\widehat{yOt}+\widehat{tOu}=\widehat{xOu}\)

\(\Rightarrow\widehat{yOt}=\widehat{xOu}-\widehat{xOy}-\widehat{tOu}\)

\(\Rightarrow\widehat{yOt}=180^o-60^o-30^o\)

\(\Rightarrow\widehat{yOt}=90^o\)

a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\) 

               = \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)

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

               = \(\dfrac{1}{2}\)

tách re đc hơm, chỗ này nhìn mún lười

giúp với ạ huhu