\({2x{}\over3} = {2y{}\over4} = {4z{}\over5}\) và x+y+z=49
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\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)
\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\)( đpcm )
Bài làm:
Sửa đề:
Ta có: \(B=2x\left(y-z\right)+\left(z-y\right)\left(x+y\right)\)
\(B=2x\left(y-z\right)-\left(y-z\right)\left(x+y\right)\)
\(B=\left(y-z\right)\left(2x-x-y\right)\)
\(B=\left(x-y\right)\left(y-z\right)\)
Với x=18 ; y=24 ; z=10 ta được:
\(B=\left(18-24\right)\left(24-10\right)\)
\(B=\left(-6\right).14=-84\)
I
1)Mai usually listen to K-pop music in her freetime.
2) When I was a child I enjoyed playing computer games.
3) My father spends most spare time to look after the garden.
4) Is watching TV the most popular leisure activity in Britain?
Chịu thôi em mới lớp 6 😥😥😥
I>
- Mai usually listens to K - pop music in her free time .
- When I was a child, I enjoyed playing computer games.
- My father spends most spare time looking after the garden.
- Is watching T.V the most popular leisure activity in Britain
- Many teenagers addicted the Internet and the computer games.
- ......
- Most my friends prefer playing sports to sufting the net.
- Today's world, teenagers are rely the technology more the past.
Xin lỗi nha, em mới học lớp 7 ^^"
Sai đâu bỏ qua nhé, hơi to mới lại mk tính máy tính ra : \(\frac{77}{30}\)nên ko chắc nhé
\(2+\frac{1}{1+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}}=2+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{13}{4}}}}\)
\(=2+\frac{1}{1+\frac{1}{1+\frac{4}{13}}}=2+\frac{1}{1+\frac{1}{\frac{17}{3}}}\)
\(=2+\frac{1}{1+\frac{3}{17}}=2+\frac{1}{\frac{20}{17}}=2+\frac{17}{20}=\frac{57}{20}\)
\(\frac{2}{3}.\frac{4}{7}=\frac{8}{21}\)
\(\frac{3}{11}.2=\frac{6}{11}\)
\(4.\frac{2}{7}=\frac{8}{7}\)
\(\frac{8}{21}:\frac{2}{3}=\frac{8}{21}.\frac{3}{2}=\frac{21}{2.21}=\frac{1}{2}\)
\(\frac{3}{7}.\frac{7}{3}=\frac{21}{21}=1\)
\(\frac{3}{7}:\frac{3}{7}=\frac{3}{7}.\frac{7}{3}=\frac{21}{21}=1\)
lỡ tay bấm gửi trả lời luôn
\(\frac{2}{3}.\frac{1}{6}.\frac{9}{11}=\frac{2.9}{18.11}=\frac{2.9}{2.9.11}=\frac{1}{11}\)
\(\frac{2.3.4}{2.3.4.5}=\frac{6.4}{6.4.5}=\frac{24}{24.5}=\frac{1}{5}\)
có : Tam giác ABC đều
Góc ABC = (180 - BAC) / 2 (1)
Xét tam giác AEH và tam giác AFH có :
EAH = FAH ( vì AH là tia phân giác của BAC )
AEH = AFH (cùng bằng 90 đọ )
AH ( cạnh chung )
Tam giác AEH = Tam Giác AFH (g-c-g)
AE=AF suy ra Tam giác AEF cân tại A suy ra AEF = ( 180 - EAF) / 2 (2) Từ (1) và (2) suy ra AEF=ABC suy ra EF song song BC