Xét ΔABC vuông tại A có \(\widehat{ABC}+\widehat{ACB}=90^0\)

=>\(2\cdot\left(\widehat{OBC}+\widehat{OCB}\right)=90^0\)

=>\(\widehat{OBC}+\widehat{OCB}=45^0\)

Xét ΔOBC có \(\widehat{BOC}+\widehat{OBC}+\widehat{OCB}=180^0\)

=>\(\widehat{BOC}+45^0=180^0\)

=>\(\widehat{BOC}=135^0\)

\(\widehat{A}=180^o-\widehat{B}-\widehat{C}=180^o-40^o-40^o=100^o\)

=> \(\widehat{A_{ngoai}}=180^o-100^o=80^o\) 

=> \(\widehat{DAB}=\dfrac{1}{2}\widehat{A_{ngoai}}=\dfrac{1}{2}\cdot80^o=40^o\)

Ta có: \(\widehat{DAB}=\widehat{ABC}\left(=40^o\right)\)

Mà 2 góc này ở vị trí so le trong

=> AD//BC 

cậu giúp mik nhiều ghê, cám ơn nha

\(\left(2,5x-\dfrac{4}{7}\right):\dfrac{8}{21}=-1,5\)

=>\(\left(\dfrac{5}{2}x-\dfrac{4}{7}\right):\dfrac{8}{21}=-\dfrac{3}{2}\)

=>\(\dfrac{5}{2}x-\dfrac{4}{7}=\dfrac{-3}{2}\cdot\dfrac{8}{21}=\dfrac{-24}{42}=\dfrac{-4}{7}\)

=>\(\dfrac{5}{2}x=-\dfrac{4}{7}+\dfrac{4}{7}=0\)

=>x=0

\(2^2+3^2+...+2021^2\)

\(=\left(1^2+2^2+...+2021^2\right)-1\)

\(=\dfrac{2021\cdot\left(2021+1\right)\left(2\cdot2021+1\right)}{6}=1\)

\(=2753594310\)

1: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)

=>\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)

=>\(2\cdot5^{x+3}=2\cdot5^{11}\)

=>x+3=11

=>x=8

2: \(\dfrac{1}{2}\cdot2^x+4\cdot2^x=9\cdot2^5\)

=>\(2^x\cdot\left(\dfrac{1}{2}+4\right)=9\cdot2^5\)

=>\(2^x\cdot\dfrac{9}{2}=9\cdot2^5\)

=>\(2^x=2^6\)

=>x=6

3: \(9^{2x+1}=27^3\)

=>\(3^{4x+2}=3^9\)

=>4x+2=9

=>4x=7

=>\(x=\dfrac{7}{4}\)

4: \(2^{-1}\cdot2^x+4\cdot2^x=9\cdot2^5\)

=>\(2^x\left(4+\dfrac{1}{2}\right)=9\cdot2^5\)

=>\(2^x\cdot\dfrac{9}{2}=9\cdot2^5\)

=>\(2^x=9\cdot2^5:\dfrac{9}{2}=2^6\)

=>x=6

5: \(\left(2x-1\right)^3=\dfrac{8}{27}\)

=>\(\left(2x-1\right)^3=\left(\dfrac{2}{3}\right)^3\)

=>\(2x-1=\dfrac{2}{3}\)

=>\(2x=\dfrac{2}{3}+1=\dfrac{5}{3}\)

=>\(x=\dfrac{5}{3}:2=\dfrac{5}{6}\)

Bài 6: Oz là phân giác của góc xOy

=>\(\widehat{xOz}=\dfrac{\widehat{xOy}}{2}=\dfrac{142^0}{2}=71^0\)

Ta có: \(\widehat{xOz}+\widehat{x'Oz}=180^0\)(hai góc kề bù)

=>\(\widehat{x'Oz}+71^0=180^0\)

=>\(\widehat{x'Oz}=109^0\)

Bài 7:

Ta có: Oz là phân giác của góc xOy

=>\(\widehat{xOz}=\widehat{yOz}=\dfrac{\widehat{xOy}}{2}=\dfrac{180^0}{2}=90^0\)

Ot là phân giác của góc xOz

=>\(\widehat{zOt}=\dfrac{\widehat{xOz}}{2}=\dfrac{90^0}{2}=45^0\)

Ov là phân giác của góc yOz

=>\(\widehat{vOz}=\dfrac{90^0}{2}=45^0\)

\(\widehat{vOt}=\widehat{zOv}+\widehat{zOt}=45^0+45^0=90^0\)

sửa đề chia hết 31 nhé 

\(S=5+5^2+5^3+...+5^{2019}=5\left(1+5+5^2+5^3\right)+...+5^{2016}\left(1+5+5^2+5^3\right)\)

\(=31\left(5+...+5^{2016}\right)⋮31\)

Vậy ta có đpcm 

Bài 2:

a) \(\dfrac{-7}{-13}=\dfrac{7}{13}\) là số hưu tỉ dương

b) \(\dfrac{2}{-17}=-\dfrac{2}{17}\) là số hưu tỉ âm

c) \(-\dfrac{-6}{5}=\dfrac{6}{5}\) là số hưu tỉ dương

Bài 3:

a) \(-2\dfrac{1}{4}=-\left(2+\dfrac{1}{4}\right)=-\dfrac{9}{4}\)

b) \(6\dfrac{2}{3}=6+\dfrac{2}{3}=\dfrac{20}{3}\)

c) \(-3\dfrac{1}{4}=-\left(3+\dfrac{1}{4}\right)=-\dfrac{13}{4}\)

        Bài 6:

\(\dfrac{9^5.9^7}{3^{22}}\) = \(\dfrac{3^{15}.3^{21}}{3^{22}}\) = \(\dfrac{3^{36}}{3^{22}}\)  = 3¹⁴

  Bài 7:

\(\dfrac{9^{16}.8^{11}}{6^{33}}\) =  \(\dfrac{3^{32}.2^{33}}{3^{33}.2^{33}}\) = \(\dfrac{1}{3}\)