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\(\frac{-3}{14}\) - \(\frac{5}{-14}\)
= \(\frac{-3}{14}+\frac{5}{14}\)
= \(\frac{2}{14}\)
= \(\frac17\)
- \(\frac54\) - \(\frac34\)
= -(\(\frac54+\frac34\))
= - \(\frac84\)
= - 2
\(\frac{15}{6}-\frac{-10}{20}\)
= \(\frac52\) + \(\frac12\)
= \(\frac62\)
= 3
\(\frac{26}{-35}\) - \(\frac{6}{35}\)
= - (\(\frac{26}{35}\) + \(\frac{6}{35}\))
=- \(\frac{32}{35}\)
\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\frac{36}{5\cdot7\cdot9}+...+\frac{36}{25\cdot27\cdot29}\)
\(=9\left[\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right]\)
\(=9\left[\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right]\)
\(=9\left[\frac{1}{3}-\frac{1}{783}\right]=9\cdot\frac{260}{783}=\frac{260}{87}\)
Đặt \(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{261}{783}-\frac{1}{783}\)
\(\Rightarrow\frac{1}{9}A=\frac{260}{783}\)
\(\Rightarrow A=\frac{260}{783}\div\frac{1}{9}\)
\(\Rightarrow A=\frac{2340}{783}=\frac{260}{87}\)
Bài làm:
Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)
=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)
Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)
Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)
Tìm x
\(a,2x-25\%=\frac{1}{2}\)
\(b,\left(\frac{3x}{7}+1\right).\left(-0,25\right)=\frac{-1}{28}\)
\(\)
a,1/5+2/5+3/5+4/5+...+9/5
=(1+2+3+4+...+9)/5
=45/5
=9
b,17,8(3,7+5,7)-7,8(4,6+4,8)
=17,8.9,4-7,8.9,4
=9,4(17,8-7,8)
=9,4.10
=94
a=bao nhiêu bạn
xin lỗi vì thếu đề
a bài 1 là bằng 3/11 nha
1
a)\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\frac{4}{55}\right]=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
xl các bạn mk chép bị thiếu
trên ghi nhầm là thếu
trả lời giùm mk nha
ai nhanh, đầy đủ, đúng mk k nha
có bạn nào chs discord k?
kb vs mk nha
bạn ơi câu b 1/36 hay 1/35
Lưu ý "*"là dấu nhân nha
Bài 1
a) x-(20/11*13+20/13*15+.....+20/53*55)=3/11
x-(20/11-20/13+20/13-20/15+.......+20/53-20/55)=3/11
x-(20/11-0-20/55)=3/11
x-16/11=3/11
x=3/11+16/11
x=19/11
x=?
PHẦN CÒN LẠI HÔNG BÍT NHA BẠN
****************************Học Tốt**********************************
36 nha Kaito Kid
hoặc có thể mk chép sai đề của thầy giáo
Giang à mk nghĩ phải đặt 10 ra ngoài để tử = 2 và hiệu 2 thừa số ở mẫu cx = 2 chứ nhỉ
bạn xem lại ik
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
=> \(2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
=> \(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
=> \(\frac{1}{3}-\frac{2}{x+1}=\frac{2}{9}\)
=> \(\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}=\frac{2}{18}\)
=> x + 1 = 18 => x = 17
Bài 2 : a) \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{18\cdot19\cdot20}< \frac{1}{4}\)chứ ta??
\(A=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{18\cdot19\cdot20}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{19\cdot20}\right)=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}< \frac{189}{756}=\frac{1}{4}\)
b) \(B=9\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right)\)
\(B=9\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right)\)
\(B=9\left(\frac{1}{3}-\frac{1}{27\cdot29}\right)=9\left(\frac{1}{3}-\frac{1}{783}\right)=9\cdot\frac{260}{783}=\frac{260}{87}< \frac{261}{87}=3\)