Sửa đề: \(\left(-1\dfrac{1}{5}\right)\left(-1\dfrac{1}{6}\right)\left(-1\dfrac{1}{7}\right)\left(-1\dfrac{1}{8}\right)\left(-1\dfrac{1}{9}\right)\left(-1\dfrac{1}{10}\right)\)

\(=\dfrac{-6}{5}\cdot\dfrac{-7}{6}\cdot...\cdot\dfrac{-11}{10}\)

\(=\dfrac{6}{5}\cdot\dfrac{7}{6}\cdot...\cdot\dfrac{11}{10}=\dfrac{11}{5}\)

Bài 4:

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=k\)

=>\(\left\{{}\begin{matrix}a=bk\\b=ck\\c=ak\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=ak\\b=ak\cdot k=ak^2\\a=ak^2\cdot k=ak^3\end{matrix}\right.\)

=>ak³=a

mà a<>0

nên k³=1

=>k=1

=>\(\left\{{}\begin{matrix}a=b\cdot1=b\\b=c\cdot k=c\\c=a\cdot k=a\cdot1=a\end{matrix}\right.\Leftrightarrow a=b=c\)

\(A=\left(a+2b-3c\right)^{2024}\cdot\left(a+b\right)^{100}\)

\(=\left(a+2a-3a\right)^{2024}\cdot\left(a+a\right)^{100}\)

=0

Đảo ngữ câu điều kiện loại 3:

-Had + S + PII, S + would/could/... + have + PII + ....

-Had it not been for/If it hadn't been for/But for + N, S + would/could/... + have + PII +....: Nếu không nhờ cái gì thì đã...

1 Which planet is the biggest in our solar system?
2 He is the worst doctor I've ever known.
3 She is the happiest woman in our office.
4 This is the most boring film I've ever seen this week.
5 We have many books, but this is the most interesting one.
6 We have many subjects, but I think English is the most important.
7 That is the cheapest hat in our shop, Madam.
8 She is the best singer in America.
9 Maths is the most difficult for me. I like English and Art.
10 He is the most careful student in our school.

1 the biggest

2 the worst

3 the happiest

4 the most boring

5 the most interesting

a: \(\dfrac{-7}{8}+\dfrac{5}{12}=\dfrac{-21}{24}+\dfrac{10}{24}=-\dfrac{11}{24}\)

b: \(-\dfrac{5}{7}-\dfrac{8}{21}=-\dfrac{15}{21}-\dfrac{8}{21}=-\dfrac{23}{21}\)

c: \(0,25+1\dfrac{5}{12}=\dfrac{3}{12}+\dfrac{17}{12}=\dfrac{20}{12}=\dfrac{5}{3}\)

d: \(-1,4-\dfrac{3}{5}=-\dfrac{7}{5}-\dfrac{3}{5}=-\dfrac{10}{5}=-2\)

e: \(\left(-7\right)-\dfrac{-5}{8}=-7+\dfrac{5}{8}=\dfrac{-56+5}{8}=\dfrac{-51}{8}\)

g: -21,25+13,3=-(21,25-13,3)=-7,95

Văn bản này viết về đề tài cuộc sống của những đứa trẻ nơi phố chợ nghèo trong con gió lạnh đầu mùa.

Cho hai góc \(\widehat{xOy};\widehat{yOz}\) là hai góc kề bù có hai tia phân giác lần lượt là \(Om;On\).

Ta có: 

+) \(\widehat{xOy}+\widehat{yOz}=180^o\) (theo giả thiết) (1)

+) \(\left\{{}\begin{matrix}\widehat{mOy}=\dfrac{\widehat{xOy}}{2}\\\widehat{nOy}=\dfrac{\widehat{yOz}}{2}\end{matrix}\right.\) (theo giả thiết) (2) 

Từ (1) và (2) suy ra:

\(\widehat{mOy}+\widehat{nOy}=\dfrac{\widehat{xOy}+\widehat{yOz}}{2}=\dfrac{180^o}{2}=90^o\)

\(\Rightarrow\widehat{mOn}=90^o\) hay \(Om\text{⊥}On\) (đpcm)

Vậy...

OM là phân giác của góc AOC

=>\(\widehat{AOM}=\widehat{COM}=\dfrac{1}{2}\cdot\widehat{AOC}\)

\(\widehat{MOB}=110^0\)

=>\(\widehat{MOC}+\widehat{COB}=110^0\)

=>\(\dfrac{1}{2}\cdot\widehat{AOC}+\widehat{COB}=110^0\)

mà \(\widehat{AOC}+\widehat{COB}=180^0\)(hai góc kề bù)

nên \(\widehat{AOC}+\widehat{COB}-\dfrac{1}{2}\cdot\widehat{AOC}-\widehat{COB}=70^0\)

=>\(\dfrac{1}{2}\cdot\widehat{AOC}=70^0\)

=>\(\widehat{AOC}=140^0\)

mà \(\widehat{AOC}=\widehat{BOD}\)(hai góc đối đỉnh)

nên \(\widehat{BOD}=140^0\)

Tìm \(x;\) y nguyên hay như nào em nhỉ?

Dạ x,y thuộc Z ạ