Question

[nguyễn thị nga]

-1/3+1/3(-1/2+1/6)^2-1/12-(-1/3)^0

1/4x:3=5/6:(0,125)

Cho tam giac ABC, co A=90°.Ke AH vuong goc voi BC (H€BC), tren tia doi cua tia HA lay diem D sao cho AH=HD.

C/m ∆AHB=∆DHB

C/m goc BAD=goc BCD

C/m BD vuong goc DC

Answer 1

[✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°...]

Bài 1

  \(-\frac{1}{3}+\frac{1}{3}\left(-\frac{1}{2}+\frac{1}{6}\right)^2-\frac{1}{12}-\left(-\frac{1}{3}\right)^0\)

= \(-\frac{1}{3}+\frac{1}{3}\left(-\frac{1}{2}\right)^2+\left(\frac{1}{6}\right)^2-\frac{1}{2}+1\)

= \(-\frac{1}{3}+\frac{1}{3}.\left(-\frac{1}{4}\right)+\left(\frac{1}{36}\right)-\frac{1}{2}+1\)

= \(-\frac{1}{3}+\left(-\frac{1}{12}\right)+\frac{1}{36}+\left(-\frac{1}{2}\right)+1\)

= \(-\frac{12}{36}+\left(-\frac{3}{36}\right)+\frac{1}{36}+\left(-\frac{18}{36}\right)+\frac{36}{36}\)

= \(\frac{\left(-12\right)+\left(-3\right)+1+\left(-18\right)+36}{36}\)

= \(\frac{13}{36}\)

Bài 2

 \(\frac{1}{4}xX:3=\frac{5}{6}:\left(0,125\right)\)

\(\frac{1}{4}xX:3=\frac{5}{6}:\left(\frac{5}{4}\right)\)

\(\frac{1}{4}xX:3=\frac{2}{3}\)

\(\frac{1}{4}xX=\frac{2}{3}x3\)

\(\frac{1}{4}xX=2\)

       \(X=2:\frac{1}{4}\)

      \(X=2x4\)

     \(X=8\)

Vậy \(X=8\)

Answer 2

[✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°...]

Bài làm 

a) Xét tam giác AHB và tam giác DHB

Ta có: AH = HD ( giả thiết )

 góc BHA = góc BHD ( = 90 độ )

        HB là cạnh chung

 => Tam giác AHB = tam giác DHB ( c.g.c )