Hình hơi mờ á

1A

2B

3C

4A

5C

6A

a) Xét \(\Delta ABD\) và \(\Delta AND\) có AD - cạnh chung, \(\widehat{ABD}=\widehat{AND}=90^o\), \(\widehat{BAD}=\widehat{NAD}\) (gt)

Do đó \(\Delta ABD=\Delta AND\left(ch-gn\right)\).

b) Ta có \(DI>DB=ND\)

200 + 50 = 250     5000 + 25000 = 52500    5 + 1 = 6 + 10 = 16 + 1 = 17 hỏi mẹ đi chứ tui hong béc tui mới có lớp 2 thui ....... 

A\(\frac{7}{15}+\frac{4}{5}=\frac{7}{15}+\frac{12}{15}=\frac{19}{15}\)

B\(\frac{2}{3}-\frac{3}{8}=\frac{16}{24}-\frac{9}{24}=\frac{7}{24}\)

C\(\frac{4}{5}\times\frac{5}{8}=\frac{20}{40}=\frac{1}{2}\)

D\(4:\frac{3}{2}=\frac{4}{1}\times\frac{2}{3}=\frac{8}{3}\)

a)\(\text{xét ΔABH và ΔACH có:}\)

\(\text{AB=AC(gt)}\)

\(\text{AH cạnh chung}\)

\(\text{BH=CH(gt)}\)

\(\Delta ABH=\Delta ACH\)\(\left(c.c.c\right)\)

b)

\(\Delta ABH=\Delta ACH\\ \Rightarrow\widehat{H_1}=\widehat{H_2}\)

=>AH là phân giác\(\widehat{BAC}\)

c) do \(\Delta ABH=\Delta ACH\\ \Rightarrow\widehat{H_1}=\widehat{H_2}\)

⇒\(\widehat{H_1}\)+\(\widehat{H_2}\)\(=180^o\)

mà \(\widehat{H_1}\)= \(\widehat{H_2}\) 

\(\Rightarrow\widehat{H_1}=\widehat{H_2}=\dfrac{180^o}{2}=90^o\)

\(AH\)⊥\(BC\)

\(a,\left\{{}\begin{matrix}AB=AC\\BH=HC\\AH\text{ chung}\end{matrix}\right.\Rightarrow\Delta AHB=\Delta AHC\left(c.c.c\right)\\ \Rightarrow\widehat{ABH}=\widehat{ACH}\\ b,\Delta AHB=\Delta AHC\\ \Rightarrow\widehat{BAH}=\widehat{CAH}\\ \Rightarrow AH\text{ là p/g }\widehat{BAC}\\ c,\Delta AHB=\Delta AHC\\ \Rightarrow\widehat{AHB}=\widehat{AHC}\\ \text{Mà }\widehat{AHB}+\widehat{AHC}=180^0\\ \Rightarrow\widehat{AHB}=\widehat{AHC}=90^0\\ \text{Vậy }AH\bot BC\)

ét o ét

⛇

giải chi tiết nhoa, mới tick

Câu 1: B

Câu 2: A

Câu 3: A

Câu 4: D