Bn cần bài nào trong 2 bài nhỉ?

e tách câu hỏi ra nhe tạm thời cj giúp mụt câu nhe

1 poverty

2 decision

3 worried

4 awake

5 singers

7 northen

8 disappeared

9 difference

10 unimportant

11 windy

12 portable

13 sensibly

14 painting

15 mainly

SOS😥👉🏳👈

ngu the bn oi bai day lop 4 cung lamf dc day

ngu the ban ( t chui ko dau do xuc pham)

1824 : 32 = 57

15 622 : 73 = 214

1824 : 32 = 57

15622 : 73 = 214

\(3n-2\inƯ\left(15\right)\) \(=\left\{1;-1;3;-3;5;-5;15;-15\right\}.\)

\(\Leftrightarrow n\in\left\{1;\dfrac{1}{3};\dfrac{5}{3};\dfrac{-1}{3};\dfrac{7}{3};-1;\dfrac{17}{3};\dfrac{-13}{3}\right\}.\)

Mà \(n\ne\dfrac{2}{3};n\in Z.\)

\(\Rightarrow n\in\left\{1;-1\right\}.\)

Ta có: \(\frac{1}{5^2}<\frac{1}{4\cdot6}\)

\(\frac{1}{7^2}<\frac{1}{6\cdot8}\)

...

\(\frac{1}{103^2}<\frac{1}{102\cdot104}\)

Do đó: \(\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{103^2}<\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\cdots+\frac{1}{102\cdot104}\)

=>\(S<\frac12\left(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\cdots+\frac{2}{102\cdot104}\right)\)

=>\(S<\frac12\left(\frac14-\frac16+\frac16-\frac18+\cdots+\frac{1}{102}-\frac{1}{104}\right)\)

=>\(S<\frac12\left(\frac14-\frac{1}{104}\right)=\frac12\cdot\frac{26-1}{104}=\frac12\cdot\frac{25}{104}\)

=>\(S<\frac{25}{208}<\frac{25}{160}\)

=>S<5/32

\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

Ta có: \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{20}{41}\)

Lại có: \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{31}\)

\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)

\(B=\dfrac{10}{31}\)

Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên...

5:

a: Chiều rộng là 18*1/6=3m

S=18*3=54m²

Số viên gạch cần mua là:

54*10^4:30^2=600 viên

b: Số viên gạch phải giảm đi là:

600*5/6=500 viên