A = ( 6 : \(\dfrac{3}{5}\) - 1\(\dfrac{1}{6}\) x \(\dfrac{6}{7}\) ) : ( 4\(\dfrac{1}{5}\) x \(\dfrac{10}{11}\) + 5\(\dfrac{2}{11}\) )
Gấp nè mí máaa
Tính:
a) (6 : \(\dfrac{3}{5}\) \(1\dfrac{1}{6}\) x \(\dfrac{6}{7}\) ) : ( \(4\dfrac{1}{5}\) x \(\dfrac{10}{11}\) + \(5\dfrac{2}{11}\) )
b) (\(1-\dfrac{1}{2}\)) x (\(1-\dfrac{1}{3}\)) x (\(1-\dfrac{1}{4}\)) x ..... x (\(1-\dfrac{1}{2003}\)) x (\(1-\dfrac{1}{2007}\))
\(\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{6}x\dfrac{6}{7}}{\dfrac{4}{5}x\dfrac{10}{11}+5\dfrac{2}{12}}\)
\(\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{6}x\dfrac{6}{7}}{\dfrac{4}{5}x\dfrac{10}{11}+5\dfrac{2}{12}}=\left(10-x\right):\left(\dfrac{8}{11}x+\dfrac{31}{6}\right)\)
\(\dfrac{10-\dfrac{7}{6}\cdot x\cdot\dfrac{6}{7}}{\dfrac{8}{11}\cdot x+\dfrac{31}{6}}=\dfrac{10-1\cdot x}{\dfrac{8}{11}\cdot x+\dfrac{31}{6}}=\dfrac{10-\dfrac{11}{11}\cdot x}{\dfrac{8}{11}\cdot x+\dfrac{31}{6}}=\dfrac{10-\dfrac{3}{11}\cdot x}{\dfrac{31}{6}}=>10-\dfrac{3}{11}\cdot x=\dfrac{31}{6} =>\dfrac{3}{11}\cdot x=10-\dfrac{31}{6}=\dfrac{29}{6}=>x=\dfrac{29}{6}:\dfrac{3}{11}=\dfrac{319}{18}\)
a,\(\dfrac{1}{7}\text{x}\dfrac{2}{7}+\dfrac{1}{7}\text{x}\dfrac{5}{7}+\dfrac{6}{7}\) b,\(\dfrac{6}{11}\text{x}\dfrac{4}{9}+\dfrac{6}{11}\text{x}\dfrac{7}{9}-\dfrac{6}{11}\text{x}\dfrac{2}{9}\)
c, \(\dfrac{4}{25}\text{x}\dfrac{5}{8}\text{x}\dfrac{25}{4}\text{x}24\)
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)
\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)
c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)
\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)
\(=1.15=15\)
Tính nhanh:
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\)\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\)
giúp mình với
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)
\(\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)
\(\Rightarrow\) \(x=22\)
b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)
\(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)
\(1< x\le5\)
\(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}
c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)
Ko biết làm
d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)
\(4\le x\le9\)
\(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}
3. Tính :
a/ \(\dfrac{-1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{1}{3}\) b/ \(\dfrac{-3}{8}\) + \(\dfrac{7}{4}\) - \(\dfrac{1}{12}\) c/ \(\dfrac{3}{5}\) : (\(\dfrac{1}{4}\) . \(\dfrac{7}{5}\)) d/ \(\dfrac{10}{11}\) + \(\dfrac{4}{11}\) : 4 - \(\dfrac{1}{8}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
tìm x
\([\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{16}.\dfrac{16}{7}}{4\dfrac{1}{5}-\dfrac{10}{11}+5\dfrac{2}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{5}\right).\dfrac{12}{49}}{3\dfrac{1}{3}+\dfrac{2}{9}}].x=2\dfrac{23}{96}\)
giúp mình với
gấp gấp lắm
1/ tìm x
a) \(x.\left(x+\dfrac{2}{3}\right)=0\)
2/tìm giá trị nhỏ nhất của các biểu thức
a) \(\left|x+\dfrac{4}{6}\right|\)
b) \(\left|x-\dfrac{1}{3}\right|\)
3/ tính
a) \(\left(1+2+3+...90\right).\left(12.34-668\right):\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}\right)\)
b) \(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
Nghiệm X của pt sau đc viết dưới dạng phân số A/B.Khi đó B=?\(\dfrac{1}{x+\dfrac{2}{3+\dfrac{4}{5+\dfrac{6}{7+\dfrac{8}{9+\dfrac{10}{11}}}}}}=\dfrac{1}{1-\dfrac{x}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{5+\dfrac{1}{6}}}}}}\)
Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)