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\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
Bài 1:
a) \(\left(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}\right)-\left(-\frac{1}{511}-\frac{1}{2}+\frac{5}{19}\right)\)
= \(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}+\frac{1}{511}+\frac{1}{2}-\frac{5}{19}\)
= \(\left(\frac{5}{19}-\frac{5}{19}\right)+\left(\frac{1}{511}-\frac{1}{511}\right)+\left(\frac{7}{12}+\frac{1}{2}\right)\)
= 0 + 0 + \(\frac{13}{12}\)
= \(\frac{13}{12}\).
b) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
= \(-\frac{13}{25}+\frac{4}{191}-\frac{2}{51}+\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\)
= \(\left(-\frac{13}{25}+\frac{3}{5}\right)+\left(\frac{4}{191}+\frac{4}{191}\right)+\left(\frac{2}{51}-\frac{2}{51}\right)\)
= \(\frac{2}{25}+\frac{8}{191}+0\)
= \(\frac{582}{4775}\).
Mình chỉ làm câu a) và câu b) thôi nhé.
Chúc bạn học tốt!
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
câu b sai đề bạn ơi
a)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n}{n+1}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot n}{2\cdot3\cdot4\cdot...\cdot\left(n+1\right)}\)
\(=\frac{1}{n+1}\)
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)
\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{n+1}{n+1}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n}{n+1}=\frac{1}{n+1}\)
b) sửa đề :
\(\frac{1}{2000\cdot1999}-\frac{1}{1999\cdot1998}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=\frac{1}{1999\cdot2000}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{1998\cdot1999}\right)\)
Xét VP :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{1998\cdot1999}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\)
\(=1-\frac{1}{1999}\)
\(\Rightarrow\frac{1}{1999\cdot2000}-1+\frac{1}{1999}\)
\(=\frac{1}{1999\cdot2000}-\frac{1998}{1999}\)
Bài 1 : Tính :
a) ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ... . ( 1 - 1/n+1 )
= 1/2 . 2/3 . 3/4 . ... . n/n+1
= 1 . 2 . 3 . .... . n / 2 . 3 . 4 . ... . n+1
= 1/ n+1
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n+1-1}{n+1}=\frac{1.2.3...n}{2.3.4....\left(n+1\right)}=\frac{n}{n+1}\)
b) Mình sửa đề câu b tí nha
\(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}\right)\)
= 1/2000.1999 - (1 - 1/1999)
= 1/2000 - 1/1999 - 1 + 1/1999
= 1/2000 - 1
giải tiếp
c) \(-3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{3}}}}\)
Ta có: \(\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{3}}}}=\frac{1}{1+\frac{1}{3+\frac{1}{\frac{4}{3}}}}=\frac{1}{1+\frac{1}{3+\frac{3}{4}}}=\frac{1}{1+\frac{1}{\frac{15}{4}}}=\frac{1}{1+\frac{4}{15}}\)
\(=\frac{1}{\frac{19}{15}}=\frac{15}{19}\)
\(\Rightarrow-3+\frac{15}{19}=\frac{-42}{19}\)
Vậy ...
câu b cô giáo tôi ra đề sao sai được với lại câu này cô lấy từ trong sách tham khảo của cô mà