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23 tháng 3 2018
\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)
\(C=\frac{2}{3}\)
Bạn Cô nàng Thiên Bình làm đúng hết òi =.=
TX
23 tháng 3 2018
a=7.[1/8+1/27-1/49]
------------------------
11.[1/8+1/27-1/49]
=7/11
cau b,c tuong tu nha h mk
ML
2
KK
14 tháng 3 2019
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
KN
14 tháng 3 2019
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B
TA
22 tháng 2 2018
\(\frac{\left(\frac{5}{8}+\frac{5}{27}-\frac{5}{49}\right)\cdot8\cdot27\cdot49}{\left(\frac{11}{8}+\frac{11}{27}-\frac{11}{49}\right)\cdot8\cdot27\cdot49}+\frac{6}{11}\)
\(=\frac{8+27-49}{8+27-49}+\frac{6}{11}\)
\(=1+\frac{6}{11}\)
\(=\frac{11}{11}+\frac{6}{11}=\frac{17}{11}\)
22 tháng 2 2018
\(\frac{\frac{5}{8}+\frac{5}{27}-\frac{5}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}+\frac{6}{11}\)
\(=\frac{5\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}{11\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}+\frac{6}{11}\)
\(=\frac{5}{11}+\frac{6}{11}=\frac{11}{11}=1\)
LV
1
YN
14 tháng 2 2023
a) \(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{-11}{2}\right)\)
\(=\dfrac{17}{8}.\dfrac{-8}{17}\)
\(=-1\)
b) \(7,63.21,15+7,63.\left(-121,15\right)\)
\(=7,63.\left(21,15-121,15\right)\)
\(=7,63.\left(-100\right)\)
\(=-763\)
c) \(\left(-\dfrac{5}{24}-0,75\right):\left(-2\dfrac{7}{8}\right)\)
\(=\left(-\dfrac{5}{24}-\dfrac{3}{4}\right):\left(-\dfrac{7}{4}\right)\)
\(=\dfrac{-23}{24}.\dfrac{-4}{7}\)
\(=\dfrac{23}{42}\).
NT
Nguyễn thành Đạt
CTVHS
14 tháng 2 2023
\(a.\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{-11}{2}\right)\text{=}\dfrac{17}{8}:\dfrac{-17}{8}\text{=}-1\)
\(b.7,63.21,15+7,63.-121,15\text{=}7,63.\left(21,15-121,15\right)\text{=}7,63.-100\text{=}-763\)
\(c.\left(\dfrac{-5}{24}-0,75\right):\dfrac{-23}{8}\text{=}\dfrac{-23}{24}.\dfrac{-8}{23}\text{=}\dfrac{1}{3}\)
LT
2
30 tháng 1 2022
b: \(\Leftrightarrow3n+6+5⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
c: \(\Leftrightarrow n+3+5⋮n+3\)
\(\Leftrightarrow n+3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-2;-4;2;-8\right\}\)
d: \(\Leftrightarrow2n+2+1⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1\right\}\)
hay \(n\in\left\{0;-2\right\}\)
e: \(\Leftrightarrow n-8-4⋮n-8\)
\(\Leftrightarrow n-8\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{9;7;10;6;12;4\right\}\)
31 tháng 1 2022
b: ⇔3n+6+5⋮n+2⇔3n+6+5⋮n+2
⇔n+2∈{1;−1;5;−5}⇔n+2∈{1;−1;5;−5}
hay n∈{−1;−3;3;−7}n∈{−1;−3;3;−7}
c: ⇔n+3+5⋮n+3⇔n+3+5⋮n+3
⇔n+3∈{1;−1;5;−5}⇔n+3∈{1;−1;5;−5}
hay n∈{−2;−4;2;−8}n∈{−2;−4;2;−8}
d: ⇔2n+2+1⋮n+1