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\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
Vậy \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\left(đpcm\right)\)
\(-5,6x+2,9x-3,86=-9,8\)
\(-2,7x=-9,8+3,86\)
\(-2,7x=-5,94\)
\(x=-5,94:\left(-2,7\right)\)
\(x=2,2\)
\(A=-5,13:\left(5\frac{5}{28}-1\frac{8}{9}.1,25+1\frac{16}{63}\right)\)
\(A=-5,13:\left(\frac{145}{28}-\frac{17}{9}.\frac{5}{4}+\frac{79}{63}\right)\)
\(A=-5,13:\left(\frac{145}{28}-\frac{85}{36}+\frac{79}{63}\right)\)
\(A=-5,13:\left(\frac{355}{126}+\frac{79}{63}\right)\)
\(A=-5,13:\frac{57}{14}\)
\(A=-1,26=\frac{-63}{50}\)
\(B=\left(3\frac{1}{3}.1,9+19,5:4\frac{1}{3}\right).\left(\frac{62}{75}-\frac{4}{25}\right)\)
\(B=\left(\frac{10}{3}.\frac{19}{10}+\frac{39}{2}.\frac{3}{13}\right).\frac{2}{3}\)
\(B=\left(\frac{19}{3}+\frac{9}{2}\right).\frac{2}{3}\)
\(B=\frac{65}{6}.\frac{2}{3}\)
\(B=\frac{65}{9}\)
học tốt
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
Mấy cái này là bài tìm x mày mò một tẹo là ra mà. Câu a thì tính ra được căn bậc 2 của 16/9 là 4/3. Sẽ tính ra được giá trị tuyệt đối của x + 1/2. Từ đó suy ra 2 trường hợp. Làm tương tự với câu b.
Câu c tính ra được x bằng 3 mũ 7 (3^12 / 3^5 = 3^7)
Câu d đổi hỗn số ra phân số rồi làm như bình thường.
\(1,\frac{7x-3}{x-1}=\frac{2}{3}\) ĐKXĐ : \(x\ne1\)
\(\Leftrightarrow\frac{3\left(7x-3\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Rightarrow21x-2x=9-2\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\frac{7}{19}\)(TM)
kl :....
\(3,\frac{1}{x-2}+3=\frac{x-3}{2-x}\) ĐKXĐ : \(x\ne2\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)
\(\Leftrightarrow1+3x-6=3-x\)
\(\Leftrightarrow3x+x=-1+6-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=2\)(TM)
KL : ....
\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)
\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{9.11}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(S=\frac{5}{11}\)
\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)
\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(Q=\frac{1007}{2015}\)
~ Đấng Ed :) ~
Ta có:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Leftrightarrow x=305\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\left(\text{đpcm}\right)\)
\(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99.100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}=VP\) (đpcm)