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a: -1/200<0<1/2000
b: \(\dfrac{-11}{56}=\dfrac{-275}{56\cdot25}=\dfrac{-275}{1400}\)
\(\dfrac{-25}{124}=\dfrac{-275}{124\cdot11}=\dfrac{-275}{1364}\)
mà 1400>1364
nên \(\dfrac{-11}{56}>-\dfrac{25}{124}\)
a)
Khi a, b cùng dấu:
\(\Rightarrow\dfrac{a}{b}\ge0\) (Luôn luôn nhận giá trị không âm)
b)
Khi a, b khác dấu:
\(\Rightarrow\dfrac{a}{b}< 0\) (Luôn luôn nhận giá trị âm)
P/s: Đề phải là thế này nhé:
Cho số hữu tỉ abab ( a;b∈Z∈Z;b≠0≠0).
So sánh ababvới 0 khi
a) a, b cùng dấu.
b) a, b khác dấu.
Chúc bạn học tốt!
a ) khi a , b cùng dấu thì :
\(\dfrac{a}{b}\) \(\ge\) 0 ( vì luôn nhận giá trị dương hoặc = 0 )
b ) khi a , b khác dấu thì :
\(\dfrac{a}{b}\) \(\le\) 0 ( vì luôn nhận giá trị âm hoặc = 0 )
1/ a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...........
b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\left(\dfrac{1}{3}\right)^5\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{24}{30}\)
Vậy ....
ta có: 1+\(\dfrac{-99}{100}=1-\dfrac{99}{100}=\dfrac{1}{100}\)
\(1+\dfrac{-100}{101}=1-\dfrac{100}{101}=\dfrac{1}{101}\)
Nhận thấy \(\dfrac{1}{100}>\dfrac{1}{101}\) \(\Rightarrow x>y\)
Phân tích ra số thập phân nhé bạn, hoặc là lấy x - y:
+ Nếu ra kết quả là số dương thì x > y.
+ Nếu ra kết quả là số âm thì x < y.
Giải:
Ta có:
\(x=-\dfrac{99}{100}\)
\(y=-\dfrac{100}{101}\)
Vì \(-\dfrac{99}{100}-\left(-\dfrac{100}{101}\right)=-\dfrac{1}{10100}\)
=> \(x< y\)
a,
\(3-\left|\dfrac{-1}{2}\right|\\ =3-\dfrac{1}{2}\\ =\dfrac{6}{2}-\dfrac{1}{2}\\ =\dfrac{5}{2}\)
b,
\(\left|\dfrac{-1}{4}\right|+\dfrac{3}{4}-\left|-1\right|\\ =\dfrac{1}{4}+\dfrac{3}{4}-1\\ =1-1\\ =0\)
c,
\(\left|0,25\right|=-\left(-0,25\right)\\ 0,25=0,25\)
1/a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{4}{5}\)
Vậy .........
2/ a/
Ta có :
\(5^{222}=\left(5^2\right)^{111}=25^{111}\)
\(2^{555}=\left(2^5\right)^{111}=32^{111}\)
Vì \(25^{111}< 32^{111}\Leftrightarrow5^{222}< 2^{555}\)
b/ Ta có :
\(3^{48}=\left(3^4\right)^{12}=81^{12}\)
\(4^{36}=\left(4^3\right)^{12}=64^{12}\)
Vì \(81^{12}>64^{12}\Leftrightarrow3^{48}>4^{36}\)
\(a,\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
\(\Rightarrow2x=-\dfrac{30}{11}\)
\(\Rightarrow x=-\dfrac{15}{11}\)
\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow2x-1=-\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)
\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}+\dfrac{1}{3^{2013}}\)
\(\Rightarrow A-\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{2012}}-\dfrac{1}{3^{2013}}\)\(\Rightarrow\dfrac{2}{3}A=\dfrac{1}{3}-\dfrac{1}{3^{2013}}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}A< \dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{3}.\dfrac{3}{2}=\dfrac{1}{2}\)
Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}< \dfrac{1}{2}\)
Đặt: \(A=\dfrac{222}{222^2+1}>0,B=\dfrac{223}{223^2+1}>0\)
Xét:
\(\dfrac{1}{A}=\dfrac{222^2+1}{222}=222+\dfrac{1}{222}\\ \dfrac{1}{B}=\dfrac{223^2+1}{223}=223+\dfrac{1}{223}\)
Dễ dàng nhận thấy: \(\dfrac{1}{A}=222+\dfrac{1}{222}< 222+1< 222+1+\dfrac{1}{223}=\dfrac{1}{B}\)
hay \(\dfrac{1}{A}< \dfrac{1}{B}\Rightarrow A>B\)
Vậy: \(\dfrac{222}{222^2+1}>\dfrac{223}{223^2+1}\)