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1 )Ta có
\(M=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).....\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{4}{3}\cdot\dfrac{-3}{4}\cdot\dfrac{5}{4}\cdot\cdot\cdot\cdot\dfrac{-99}{100}\cdot\dfrac{101}{100}\)
\(=\dfrac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot3\cdot\left(-4\right)\cdot4\cdot\left(-5\right)\cdot5....\cdot\left(-100\right)\cdot100\cdot101}{2^2\cdot3^2\cdot4^2....\cdot100^2}\)
\(=-\dfrac{101}{200}< \dfrac{1}{2}\)
2 ) Số phân số của biểu thức B là 180 phân số
Ta có
\(\dfrac{1}{20}>\dfrac{1}{200};\dfrac{1}{21}>\dfrac{1}{200};\dfrac{1}{22}>\dfrac{1}{200};....;\dfrac{1}{199}>\dfrac{1}{200}\)
\(\Rightarrow B=\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}\cdot180=\dfrac{9}{10}\)
a) Ta có:\(\dfrac{31}{67}>\dfrac{31}{73}\) (1)
\(\dfrac{31}{73}>\dfrac{29}{73}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{31}{67}>\dfrac{31}{73}>\dfrac{29}{73}\)
\(\Rightarrow\dfrac{31}{67}>\dfrac{29}{73}\)
Vậy:...............
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530
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: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
2)
\(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3}{3}+\dfrac{1}{3}+\dfrac{3^2}{3^2}+\dfrac{1}{3^2}+\dfrac{3^3}{3^3}+\dfrac{1}{3^3}+...+\dfrac{3^{98}}{3^{98}}+\dfrac{1}{3^{98}}\\ D=1+\dfrac{1}{3}+1+\dfrac{1}{3^2}+1+\dfrac{1}{3^3}+...+1+\dfrac{1}{3^{98}}\\ D=\left(1+1+1+...+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ D=98+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\)
Gọi \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\) là \(C\)
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ 3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ 2C=1-\dfrac{1}{3^{98}}\\ C=\left(1-\dfrac{1}{3^{98}}\right):2\\ C=1:2-\dfrac{1}{3^{98}}:2\\ C=\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}\)
\(D=98+C=98+\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}=98\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}< 100\)
Vậy \(D< 100\)
Câu1:
a: \(=2008^2-\left(2008-2\right)\left(2008+2\right)\)
\(=2008^2-\left(2008^2-4\right)\)
=4
b: \(=\dfrac{23\cdot29\cdot10101}{23\cdot29\cdot10101}=1\)
c: \(=\dfrac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\cdot\left(16-16\right)}{15^2+5^3+67^7}\)
=0