Ta có : A=2²+2⁴+2⁶+...+2²⁰

=(2²+2⁴)+(2⁶+2⁸)+...+(2¹⁸+2²⁰)

=2²(1+2²)+2⁶(1+2²)+...+2¹⁸(1+2²)

=2².5+2⁶.5+...+2¹⁸.5 chia hết cho 5

Vậy A chia hết cho 5.

\(A=2^2+2^4+2^6..+2^{18}+2^{20}\)

\(\Leftrightarrow A=\left(2^2+2^4\right)+\left(2^6+2^8\right)+...+\left(2^{18}+2^{20}\right)\)

\(\Leftrightarrow A=20+2^4.\left(2^2+2^4\right)+...+2^{16}.\left(2^2+2^4\right)\)

\(\Leftrightarrow A=20+2^4.20+..+2^{16}.20\)

\(\Leftrightarrow A=20\left(1+2^4+..+2^{16}\right)\)

Vì \(20⋮5\)

\(\Rightarrow A=20\left(1+2^4+..+2^{16}\right)⋮5\)

Vậy \(A⋮5\)

hok tốt!!

\(A=2^2+2^4+2^6+2^8+....+2^{18}+2^{20}\)

\(A=\left(2^2+2^4\right)+\left(2^6+2^8\right)+....+\left(2^{18}+2^{20}\right)\)

\(A=2^2\left(1+2^2\right)+2^6\left(1+2^2\right)+....+2^{18}\left(1+2^2\right)\)

\(A=2^2\cdot5+2^6\cdot5+...+2^{18}\cdot5\)

\(A=5\left(2^2+2^6+...+2^{18}\right)\)

=> A chia hết cho 5

đề bài là j bạn?? 

so sánh à bạn 

là tính tổng á

(21/6+5/26)+(7/9+4/180 còn bằng bao nhiêu thì tự tính 

hơ hơ tốt ghê ha HO TRONG HIEU :<

Bài 1:

Vì AD là p/g góc A nên \(\widehat{A_1}=\widehat{A_2}=\dfrac{1}{2}\widehat{BAC}=30^0\)

Mà \(\widehat{A_2}+\widehat{C}+\widehat{D_1}=180^0\Rightarrow\widehat{D_1}=180^0-30^0-40^0=110^0\)

Mà AE//BC nên \(\widehat{EAD}=\widehat{D_1}=110^0\left(so.le.trong\right)\)

Vì DE//AC nên \(\widehat{A_2}=\widehat{D_2}=30^0\left(so.le.trong\right);\widehat{D_3}=\widehat{C}=40^0\left(đồng.vị\right)\)

Vì AE//BC nên \(\widehat{D_3}=\widehat{E}=40^0\)

Vậy các góc tg ADE là \(\widehat{A}=110^0;\widehat{D}=30^0;\widehat{E}=40^0\)

Bài 2:

a, Xét tg ABH và tg DBH có 

\(\left\{{}\begin{matrix}\widehat{BAH}=\widehat{BDH}=90^0\\\widehat{ABH}=\widehat{DBH}\left(BH.là.p/g\right)\\BH.chung\end{matrix}\right.\\ \Rightarrow\Delta ABH=\Delta BDH\left(ch-gn\right)\\ \Rightarrow HA=HD\)

b, Xét tg AHE và tg DHC có

\(\left\{{}\begin{matrix}\widehat{AHE}=\widehat{DHC}\left(đối.đỉnh\right)\\\widehat{HAE}=\widehat{HDC}=90^0\\HA=HD\left(cmt\right)\end{matrix}\right.\\ \Rightarrow\Delta AHE=\Delta DHC\left(g.c.g\right)\\ \Rightarrow CD=AE\)

Mà \(AB=BD\left(\Delta ABH=\Delta DBH\right)\)

\(\Rightarrow AB+AE=BD+CD\\ \Rightarrow BE=BC\)

Vì \(AD=AB\) nên tg ABD cân tại B

Do đó \(\widehat{BAD}=\dfrac{180^0-\widehat{ABC}}{2}\left(1\right)\)

Vì \(BE=BC\) nên tg BEC cân tại B

Do đó \(\widehat{BEC}=\dfrac{180^0-\widehat{ABC}}{2}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\widehat{BAD}=\widehat{BEC}\) mà 2 góc này ở vị trí đồng vị nên AD//EC

a) \(\dfrac{3}{8}+\dfrac{15}{-25}+\dfrac{3}{5}\)

\(=\dfrac{-9}{40}+\dfrac{3}{5}\)

\(=\dfrac{3}{8}\)

b) \(\dfrac{-5}{18}+\dfrac{23}{45}-\dfrac{9}{10}\)

\(=\dfrac{7}{30}-\dfrac{9}{10}\)

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

c) \(\dfrac{-5}{12}+\dfrac{15}{18}-2,25\)

\(=\dfrac{5}{12}-2,25\)

\(=\dfrac{-11}{6}\)

d) \(\dfrac{5}{6}+\dfrac{2}{3}-0,5\)

\(=\dfrac{3}{2}-0,5\)

\(=1\)

tick cho mink nhé bn ✔

a)8/5-18/27=216/135-90/135=126/135

b)3/5-1/3=9/15-5/15=4/15(rút gọn p/s)

c)1/3-1/5=5/15-3/15=2/15

d)6/7-12/27=162/189-84/189=78/189

a) 8/5 - 18/27

= 8/5 - 2/3

= 24/15 - 10/15

= 14/15

b) 12/20 - 6/18

= 3/5 - 1/3

= 9/15 - 5/15

= 4/15

c) 8/24 - 5/25

= 1/3 - 1/5

= 5/15 - 3/15

= 2/15

d) 18/21 - 12/27

= 6/7 - 4/9

= 54/63 - 28/63

= 26/63

-1/3;-1/2;-5/9;1/3;7/18;5/12

Xếp:

-1/3; -1/2; -5/9; 1/3; 7/18; 5/12

1) Thay x = -8 , ta có ;

(-8) - 2 x (-8) - 3 x (-8) - 4 x (-8 ) - 5 x (-8) = 104

2) Thay x = -8 , ta có ;

(-8 + 1)-2(-8 + 1)-3( -8 + 1 )- 4( -8 + 1 ) - 5( -8 + 1 ) =91

Đúng thì tick ko đúng thì thôi 

\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2-18=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)-18=0\)

Đặt \(x^2+2x+1=a\ge0\)

\(\Rightarrow a\left(4a-1\right)-18=0\)

\(\Leftrightarrow4a^2-a-18=0\)

\(\Leftrightarrow\left(4a^2+8a\right)+\left(-9a-18\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(4a-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=\frac{9}{4}\end{cases}}\)

\(\Rightarrow x^2+2x+1=\frac{9}{4}\)

\(\Leftrightarrow4x^2+8x-5=0\)

\(\Leftrightarrow\left(4x^2-2x\right)+\left(10x-5\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\)