\(1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
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JV
18 tháng 10 2017
Giải:
\(A=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2.\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(\Leftrightarrow A=1-\dfrac{5}{14}-\dfrac{5}{2^2.\left(\sqrt{21}\right)^2}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(\Leftrightarrow A=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(\Leftrightarrow A=\dfrac{5}{5}-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(\Leftrightarrow A=5\left(\dfrac{1}{5}-\dfrac{1}{14}-\dfrac{1}{84}-\dfrac{1}{204}-\dfrac{1}{374}\right)\)
\(\Leftrightarrow A=5.\dfrac{6}{55}\)
\(\Leftrightarrow A=\dfrac{6}{11}\)
Vậy giá trị của biểu thức A là \(\dfrac{6}{11}\).
Chúc bạn học tốt!
8 tháng 1 2018
Lần sau nhớ đăng đúng chỗ nhá
M=\(1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(M=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
M=\(\dfrac{15708-5610-935-385-210}{15708}\)
\(\Rightarrow M=\dfrac{6}{11}\)
DP
19 tháng 12 2017
\(\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}\)+\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{3}\right)}\)
=\(\frac{3}{5}\)+\(\frac{1}{\frac{5}{2}}\)
=\(\frac{3}{5}\)+\(\frac{2}{5}\)
=1 !!!
25 tháng 3 2022
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
AH
Akai Haruma
Giáo viên
5 tháng 1 2019
Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)
\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)
\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)
AH
Akai Haruma
Giáo viên
5 tháng 1 2019
Bài 2:
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)
\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)
11 tháng 2 2022
a: \(=\dfrac{-12}{56}+\dfrac{35}{56}-\dfrac{28}{56}=-\dfrac{5}{56}\)
b: \(=\dfrac{5}{12}-\dfrac{4}{5}=\dfrac{25-48}{60}=\dfrac{-23}{60}\)
d: SỐ cần tìm là:
-24:3/8=-24x8:3=-64
11 tháng 2 2022
a \(\dfrac{-5}{56}\)
b \(\dfrac{-23}{60}\)
c \(\dfrac{-23}{60}\)
d \(\dfrac{-1}{64}\)
K
7 tháng 4 2022
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
\(1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\) =\(1-\left(\dfrac{1}{1.14}+\dfrac{1}{14.6}+\dfrac{1}{6.34}+\dfrac{1}{34.11}\right)\)
= \(1-\left(\dfrac{1}{1}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{11}\right)\)
= 1 - \(\left(1-\dfrac{1}{11}\right)\) = \(1-\dfrac{10}{11}=\dfrac{1}{11}\)
\(=1-\dfrac{5}{2\cdot7}-\dfrac{5}{7\cdot12}-\dfrac{5}{12\cdot17}-\dfrac{5}{17\cdot22}\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=5-\dfrac{11-1}{22}=5-\dfrac{10}{22}=5-\dfrac{5}{11}=\dfrac{50}{11}\)