Giúp tớ làm bài này:
Tìm thương A chia B biết :
A =\(\frac{9}{1}\)+\(\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)
B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
Tìm thương A:B biết:
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9};B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
A = \(10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
A : B = 10
Tìm thương A:B, biết:
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
Tìm A:B, biết:
A=\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)
B=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\)
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
bài 1:
\(a,\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)
\(b,\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)
\(c,13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)
Các bạn làm giúp mk nha
Làm đc phần nào thì làm
Ko cần làm hết 3 phần cũng đc
a) \(\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)
= \(\frac{1}{3}-\left(\frac{4}{8}+\frac{1}{8}\right)\)
= \(\frac{1}{3}-\frac{5}{8}\)
= \(\frac{8}{24}-\frac{15}{24}\)
= \(\frac{-7}{24}\)
b) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)
= \(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)\)+ \(\frac{1}{13}\)
= \(\left(\frac{4}{8}-\frac{2}{8}+\frac{1}{8}\right)+\frac{1}{13}\)
= \(\frac{1}{8}+\frac{1}{13}\)
= \(\frac{13}{104}+\frac{8}{104}\)
= \(\frac{23}{104}\)
c) \(13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)
= \(\left(13\frac{2}{7}+2\frac{5}{7}\right):\left(\frac{-8}{9}\right)\)
= \(16:\left(\frac{-8}{9}\right)\)
= -18
a)\(\frac{7}{x}<\frac{x}{4}<\frac{10}{x}\)
b) Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\). Chứng tỏ: \(\frac{8}{9}>A>\frac{2}{5}\)
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
Bài 1:
a,\(A=\frac{5}{27}-\frac{8}{9}+\frac{3}{5}+\frac{22}{27}-\frac{1}{9}\)
b, \(B=0,375.1\frac{3}{5}+60\%.\frac{2}{7}+\frac{3}{5}.\frac{5}{7}\)
Bài 2:
a,\(\frac{1}{5}+\frac{4}{5}:x=-1\)
b, \(1-\left(x-\frac{1}{2}\right)^2=\frac{21}{25}\)
c, \(|x+\frac{1}{3}|.3\frac{2}{5}=\frac{2}{5}\)
Bài 3: Một lớp học có 45 HS cùng làm một bài kiểm tra, số bài lớp đạt điểm giỏi bằng \(\frac{1}{3}\) tổng số bài. Số bài lớp đạt điểm khá bằng \(\frac{9}{10}\) số bài còn lại. Tính số bài đạt điểm TB, biết rằng lớp ko có bài điểm yếu, kém.
Bài 4:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2018^2}\)
\(B=75\%\)
So sánh A và B
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Bài 1:Tìm x biết Bài 2:So sánh
a, \(x+\frac{1}{2}=\frac{3}{8}.\frac{4}{5}\) a, \(A=\frac{10^{10}-1}{10^{11}-1}vaB=\frac{10^9-1}{10^{10}-1}\)
b, \(\frac{5}{16}:x-\frac{1}{4}=\frac{5}{8}\) b, B =\(\frac{10^{10}}{10^{10}+1}vaB=\frac{10^{10}+1}{10^{10}+2}\)
c, \(\frac{-1}{4}.x+\frac{3}{7}.x=2\)
d, \(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
e, \(\left|\frac{1}{4}-x\right|+5\frac{1}{8}=6\frac{1}{8}\)
Bai 1: a) \(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)
b) \(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
bài 2:
a) \(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)
b) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
c)
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)