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a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)
=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)
=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)
=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)
b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)
=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))
=99+1=100
c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Bài 1:
a) Ta có: \(13A=\dfrac{13^{16}+13}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=\dfrac{13^{17}+13}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Vì \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(1999C=\dfrac{1999^{2000}+1999}{1999^{2000}+1}=1+\dfrac{1998}{1999^{2000}+1}\)
\(1999D=\dfrac{1999^{1999}+1999}{1999^{1999}+1}=1+\dfrac{1998}{1999^{1999}+1}\)
\(\dfrac{1998}{1999^{2000}+1}< \dfrac{1998}{1999^{1999}+1}\Rightarrow1+\dfrac{1998}{1999^{2000}+1}< 1+\dfrac{1999}{1999^{1999}+1}\)
\(\Rightarrow1999C< 1999D\)
\(\Rightarrow C< D\)
Vậy C < D