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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 ....
\(a,\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
\(\Rightarrow2x=-\dfrac{30}{11}\)
\(\Rightarrow x=-\dfrac{15}{11}\)
\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow2x-1=-\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)
\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)
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 \(81^{12}>64^{12}\Leftrightarrow3^{48}>4^{36}\)
4/ \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{24}\end{matrix}\right.\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\) (đặt k)
Suy ra \(x=15k;y=20k;z=24k\)
Thay vào,ta có:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
1. Câu hỏi của Cuber Việt ( Câu b í -.- )
2. Quy đồng mẫu số:
\(\dfrac{a}{b}=\dfrac{a.\left(b+2018\right)}{b.\left(b+2018\right)}=\dfrac{ab+2018a}{b.\left(b+2018\right)}\)
\(\dfrac{a+2018}{b+2018}=\dfrac{\left(a+2018\right).b}{\left(b+2018\right).b}=\dfrac{ab+2018b}{b.\left(b+2018\right)}\)
Vì \(b>0\) \(\Rightarrow\) Mẫu 2 phân số ở trên dương.
So sánh \(ab+2018a\) và \(ab+2018b\):
. Nếu \(a< b\Rightarrow\) Tử số phân số thứ 1 < Tử số phân số thứ 2.
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)
. Nếu \(a=b\) \(\Rightarrow\) Hai phân số bằng 1.
. Nếu \(a>b\Rightarrow\) Tử số phân số thứ 1 > Tử số phân số thứ 2.
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)
3. \(\dfrac{x}{6}-\dfrac{1}{y}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{x}{6}-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{x-3}{6}\)
\(\Rightarrow y.\left(x-3\right)=6\)
Ta có: \(6=1.6=2.3=(-1).(-6)=(-2).(-3)\)
Tự lập bảng ...
Vậy ta có những cặp x,y thỏa mãn là:
\(\left(1,7\right);\left(6,2\right);\left(2,4\right);\left(3,3\right);\left(-1,-5\right);\left(-6,0\right);\left(-2,-2\right);\left(-3,-1\right)\)
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a\left(b+2018\right)}{b\left(b+2018\right)}\\\dfrac{a+2018}{b+2018}=\dfrac{b\left(a+2018\right)}{b\left(b+2018\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{ab+2018a}{b^2+2018b}\\\dfrac{a+2018}{b+2018}=\dfrac{ab+2018b}{b^2+2018b}\end{matrix}\right.\)
Cần so sánh:
\(ab+2018a\) với \(ab+2018b\)
Cần so sánh \(2018a\) với \(2018b\)
Cần so sánh \(a\) với \(b\)
\(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2018}{b+2018}\)
\(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)
\(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2018}{b+2018}\)
\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(A=1-1+1=1\)
\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)
\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)
\(A=-1+1=0\)
\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=-36-\dfrac{1}{4}\)
B = \(-\dfrac{145}{4}\)
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: \(\dfrac{222}{222^2+1}>\dfrac{223}{223^2+1}\)