0,(6).\(x\) = 1

Ta có: vì 0,(6) = \(\dfrac{2}{3}\) 

Vậy 0,(6).\(x\) = 1 ⇔ \(\dfrac{2}{3}\)\(x\) = 1

     ⇒ \(\dfrac{2}{3}\)\(x\) = 1

        \(x\) = 1 : \(\dfrac{2}{3}\)

        \(x\) = \(\dfrac{3}{2}\)

Vậy \(x=\dfrac{3}{2}\)

co cai corn card , money dau tao giai cho

1: Ta có: \(\widehat{ABC}+\widehat{ABM}=180^0\)(hai góc kề bù)

\(\widehat{ACB}+\widehat{ACN}=180^0\)(hai góc kề bù)

mà \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)

nên \(\widehat{ABM}=\widehat{ACN}\)

Xét ΔABM và ΔACN có

AB=AC

\(\widehat{ABM}=\widehat{ACN}\)

BM=CN

Do đó: ΔABM=ΔACN

2: ΔABM=ΔACN

=>\(\widehat{BAM}=\widehat{CAN};\widehat{AMB}=\widehat{ANC}\); AM=AN

Xét ΔAHB vuông tại H và ΔAKC vuông tại K có

AB=AC

\(\widehat{BAH}=\widehat{CAK}\)

Do đó: ΔAHB=ΔAKC

=>AH=AK

3: ΔAHB=ΔAKC

=>\(\widehat{ABH}=\widehat{ACK}\)

Ta có: \(\widehat{ABH}+\widehat{ABC}+\widehat{OBC}=180^0\)

\(\widehat{ACK}+\widehat{ACB}+\widehat{OCB}=180^0\)

mà \(\widehat{ABH}=\widehat{ACK};\widehat{ABC}=\widehat{ACB}\)

nên \(\widehat{OBC}=\widehat{OCB}\)

=>ΔOBC cân tại O

\(\dfrac{1}{99}-\dfrac{1}{97.99}-\dfrac{1}{95.97}-\dfrac{1}{93.95}-...-\dfrac{1}{3.5}-\dfrac{1}{1.3}\\ =\dfrac{1}{99}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}+\dfrac{1}{97.99}\right)\\ \)

\(=\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ \)

\(=\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}+\dfrac{1}{198}=-\dfrac{16}{33}\)

Đăng Tùng em làn đúng rồi đó

Do you go to school?

Dữ liệu cuối cùng nhìn khó hiểu thế em? là phân số, số thập phân em ơi?

vì 5/2 = 2,5 nên những số đo chiều cao của tầng hầm phù hợp với dự định của cô Hạnh là: 2,56m;2,59m;2,6m.

a) Do M là trung điểm của BC (gt)

\(\Rightarrow MB=MC\)

Xét \(\Delta AMB\) và \(\Delta DMC\) có:

\(MB=MC\left(cmt\right)\)

\(\widehat{AMB}=\widehat{DMC}\) (đối đỉnh)

\(MA=MD\left(gt\right)\)

\(\Rightarrow\Delta AMB=\Delta DMC\left(c-g-c\right)\)

b) Do N là trung điểm của AC (gt)

\(\Rightarrow NA=NC\)

Xét \(\Delta ANB\) và \(\Delta CNE\) có:

\(NA=NC\left(cmt\right)\)

\(\widehat{ANB}=\widehat{CNE}\) (đối đỉnh)

\(NB=NE\left(gt\right)\)

\(\Rightarrow\Delta ANB=\Delta CNE\left(c-g-c\right)\)

\(\Rightarrow\widehat{ABN}=\widehat{CEN}\) (hai góc tương ứng)

Mà \(\widehat{ABN}\) và \(\widehat{CEN}\) là hai góc so le trong

\(\Rightarrow AB\) // \(CE\)

c) Do \(\Delta AMB=\Delta DMC\left(cmt\right)\)

\(\Rightarrow\widehat{BAM}=\widehat{CDM}\) (hai góc tương ứng)

Mà \(\widehat{BAM}\) và \(\widehat{CDM}\) là hai góc so le trong

\(\Rightarrow AB\) // \(CD\)

Mà \(AB\) // \(CE\left(cmt\right)\)

Theo tiên đề Ơclit \(\Rightarrow E,C,D\) thẳng hàng

a: \(A=\left(x^2y\right)\cdot\left(xy^2\right)\cdot\left(-x^3y^2\right)\)

\(=-x^2\cdot x\cdot x^3\cdot y\cdot y^2\cdot y^2\)

\(=-x^6y^5\)

Bậc là 6+5=11

ai cứu em với

Hệ số tỉ lệ k là \(k=\dfrac{y}{x}=\dfrac{8}{17}\)

Để \(\dfrac{3-x}{5}>0\) thì 3-x>0

=>x<3

=>\(x\in\left\{...;1;2;3\right\}\)

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

2) \(\dfrac{25^5}{125^3}=\dfrac{\left(5^2\right)^5}{\left(5^3\right)^3}=\dfrac{5^{10}}{5^9}=5\)

3) \(\dfrac{27^6}{9^9}=\dfrac{\left(3^3\right)^6}{\left(3^2\right)^9}=\dfrac{3^{18}}{3^{18}}=1\) 

4) \(\dfrac{16^{13}}{32^{10}}=\dfrac{\left(2^4\right)^{13}}{\left(2^5\right)^{10}}=\dfrac{2^{52}}{2^{50}}=2^2-4\)

5) \(\dfrac{16^5}{64^4}=\dfrac{\left(4^2\right)^5}{\left(4^3\right)^4}=\dfrac{4^{10}}{4^{12}}=\dfrac{1}{4^2}=\dfrac{1}{16}\)

6) \(\dfrac{81^8}{27^{11}}=\dfrac{\left(3^4\right)^8}{\left(3^3\right)^{11}}=\dfrac{3^{32}}{3^{33}}=\dfrac{1}{3}\)

7) \(\dfrac{6^3}{2^3}=\dfrac{2^3\cdot3^3}{2^3}=3^3=27\)

8) \(\dfrac{5^4}{15^3}=\dfrac{5^4}{3^3\cdot5^3}=\dfrac{5}{3^3}=\dfrac{5}{27}\)

9) \(\dfrac{7^{15}}{14^{13}}=\dfrac{7^{15}}{7^{13}\cdot2^{13}}=\dfrac{7^2}{2^{13}}=\dfrac{49}{2^{13}}\)

10) \(\dfrac{\left(-2\right)^6}{24^2}=\dfrac{2^6}{8^2\cdot3^2}=\dfrac{2^6}{\left(2^3\right)^2\cdot3^2}=\dfrac{2^6}{2^6\cdot3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

11: \(\dfrac{27^2}{\left(-18\right)^3}=\dfrac{-3^6}{\left(3^2\cdot2\right)^3}=\dfrac{-3^6}{3^6\cdot2^3}=\dfrac{-1}{8}\)

12: \(\dfrac{\left(-10\right)^8}{8^3\cdot25^4}=\dfrac{2^8\cdot5^8}{2^6\cdot5^8}=2^2=4\)

13: \(\dfrac{4^4\cdot8^3}{16^4}=\dfrac{2^8\cdot2^9}{2^{16}}=2\)

14: \(\dfrac{5^7\cdot9^2}{15^5}=\dfrac{5^7\cdot3^4}{5^5\cdot3^5}=\dfrac{5^2}{3}=\dfrac{25}{3}\)

15: \(\dfrac{21^{13}}{49^6\cdot\left(-27\right)^4}=\dfrac{-7^{13}\cdot3^{13}}{7^{12}\cdot3^{12}}=-7\cdot3=-21\)

16: \(\dfrac{\left(-18\right)^{21}\cdot27^4}{81^{13}\cdot16^5}=\dfrac{-3^{42}\cdot2^{21}\cdot3^{12}}{3^{52}\cdot2^{20}}=\dfrac{-3^{54}}{3^{52}}\cdot2=-3^2\cdot2=-18\)

17: \(\dfrac{45^{14}\cdot8^2}{6^5\cdot125^4\cdot81^6}=\dfrac{3^{28}\cdot5^{14}\cdot2^6}{2^5\cdot3^5\cdot3^{24}\cdot5^{12}}=\dfrac{3^{28}}{3^{29}}\cdot\dfrac{5^{14}}{5^{12}}\cdot\dfrac{2^6}{2^5}=\dfrac{5^2\cdot2}{3}=\dfrac{50}{3}\)

18: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot2^2}=3\cdot\dfrac{8}{4}=3\cdot2=6\)

19: \(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{16^4\cdot5^7+20^8}\)

\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\cdot\left(2-5\right)}{2^{16}\cdot5^7\cdot\left(1+5\right)}=\dfrac{1}{4}\cdot5\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)