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Ta có: |x| ≥ 0 ;\(\forall\)x
=> |x| + 1996≥ 1996 ;
=> \(\dfrac{|x|+1996}{-1997}\) ≥ \(\dfrac{-1996}{1997}\) ;\(\forall\)x
=> A \(\ge\) \(\dfrac{-1996}{1997}\)
Dấu = xảy ra <=> |x| =0
<=> x=0
Vậy GTLN của A là \(\dfrac{-1996}{1997}\) tại x = 0
a: \(\left|x\right|+1996>=1996\)
\(\Leftrightarrow\dfrac{\left|x\right|+1996}{1997}\ge\dfrac{1996}{1997}\)
\(\Leftrightarrow A\le-\dfrac{1996}{1997}\)
Dấu '=' xảy ra khi x=0
b: \(\left|x\right|+1>=1\)
\(\Leftrightarrow\dfrac{1}{\left|x\right|+1}\le1\)
\(\Leftrightarrow B\ge-1\)
Dấu '=' xảy ra khi x=0
1.\(\frac{1996}{\left|x\right|+1997}\)có GTLN \(\Leftrightarrow\left|x\right|+1997\)có GTNN.
Mà \(\left|x\right|+1997\ne0\)
Ta thấy: \(\left|x\right|\ge0\forall x\in R\Rightarrow\left|x\right|+1997\ge1997\)
\(\Rightarrow\left|x\right|=0\)thì \(\left|x\right|+1997\)có GTNN là \(1997\)
\(\Rightarrow\)GTLN của \(\frac{1996}{\left|x\right|+1997}\)là \(\frac{1996}{1997}\)khi x=0
2.\(\frac{\left|x\right|+1996}{-1997}=\frac{-\left(\left|x\right|+1996\right)}{1997}\)
\(\Rightarrow\left|x\right|+1996\)phải có GTNN thì \(\frac{\left|x\right|+1996}{-1997}\)đạt GTLN
Mà \(\left|x\right|\ge0\forall x\in R\Rightarrow x=0\)thì \(\left|x\right|+1996\)có GTNN là \(1996\)
Vậy GTLN của \(\frac{\left|x\right|+1996}{-1997}\)là \(\frac{1996}{-1997}\)khi x=0
\(\dfrac{x}{y}=\dfrac{z}{t}\\ \Rightarrow\dfrac{x}{z}=\dfrac{y}{t}\\ \Rightarrow\dfrac{x}{z}=\dfrac{y}{t}=\dfrac{x-y}{z-t}\\ \Rightarrow\dfrac{x^{1996}}{z^{1996}}=\dfrac{y^{1996}}{t^{1996}}=\left(\dfrac{x-y}{z-t}\right)^{1996}\\ \dfrac{x^{1996}}{z^{1996}}=\dfrac{y^{1996}}{t^{1996}}=\dfrac{x^{1996}+y^{1996}}{z^{1996}+t^{1996}}\\ \Rightarrow\left(\dfrac{x-y}{z-t}\right)^{1996}=\dfrac{x^{1996}+y^{1996}}{z^{1996}+t^{1996}}\)
a) Ta có : \(\dfrac{-1}{5}< 0< \dfrac{1}{1000}\)
\(\Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)
b) Ta có : \(\dfrac{267}{268}< 1< \dfrac{1347}{1343}\)
=> \(\dfrac{267}{-268}< -\dfrac{1347}{1343}\)
c) \(\dfrac{13}{38}>\dfrac{13}{39}=\dfrac{1}{3}=\dfrac{19}{87}>\dfrac{29}{88}\)
=> \(-\dfrac{13}{38}< \dfrac{29}{-88}\)
d) \(\dfrac{181818}{313131}=\dfrac{18}{31}\)
=> \(-\dfrac{18}{31}=-\dfrac{181818}{313131}\)
Áp dụng 1.5 ta có:
a) \(\dfrac{4}{9}< 1\Rightarrow\dfrac{4}{9}< \dfrac{4+9}{9+9}=\dfrac{13}{18}\).
b) \(\dfrac{-15}{7}< 1\Rightarrow\dfrac{-15}{7}< \dfrac{-15+3}{7+3}=\dfrac{-12}{10}=\dfrac{-6}{5}\).
c) \(\dfrac{278}{37}>1\Rightarrow\dfrac{278}{37}>\dfrac{278+9}{37+9}=\dfrac{278}{46}\).
d) \(\dfrac{-157}{623}< 1\Rightarrow\dfrac{-157}{623}< \dfrac{-157+16}{623+16}=\dfrac{-141}{639}=\dfrac{-47}{213}\);
a) \(\dfrac{4}{9}\)và \(\dfrac{13}{18}\)
Ta có : \(\dfrac{4}{9}\)=\(\dfrac{4.2}{9.2}\)=\(\dfrac{8}{18}\)
\(\Rightarrow\)\(\dfrac{8}{18}\)>\(\dfrac{13}{18}\)
giữ nguyên \(\dfrac{13}{18}\) (vì \(8>13\))
- Vậy \(\dfrac{4}{9}>\dfrac{13}{18}\)
b)\(-\dfrac{15}{7}\)và \(-\dfrac{6}{5}\)
Ta có :\(-\dfrac{15}{7}=\dfrac{-15.5}{7.5}=\dfrac{-75}{35}\)
\(\Rightarrow\)\(-\dfrac{75}{35}< \dfrac{-42}{35}\)
\(-\dfrac{6}{5}=\dfrac{-6.7}{5.7}=\dfrac{-42}{35}\) (vì - 75>-42)
- vậy \(\dfrac{-15}{7}< \dfrac{-6}{5}\)
c)\(\dfrac{278}{37}\)và \(\dfrac{287}{46}\)
Ta có :\(\dfrac{278}{37}=\dfrac{278.46}{37.46}=\dfrac{12788}{1702}\)
\(\Rightarrow\)\(\dfrac{12788}{1702}>\dfrac{10286}{1702}\)
\(\dfrac{287}{46}=\dfrac{287.37}{46.37}=\dfrac{10286}{1702}\) (vì 12788>10286)
- vậy \(\dfrac{278}{37}>\dfrac{10286}{46}\)
d)\(-\dfrac{157}{623}\)và \(-\dfrac{47}{213}\)
\(-\dfrac{157}{623}=\dfrac{-157.213}{623.213}=\dfrac{-33441}{132699}\)
\(\Rightarrow\dfrac{-33441}{132699}< \dfrac{-29281}{132699}\)
\(\dfrac{-47}{213}=\dfrac{-47.623}{213.623}=\dfrac{-29281}{132699}\) (vì -33441<-29281)
-vậy \(-\dfrac{157}{623}< -\dfrac{47}{213}\)
Cô làm rồi em nhé:
https://olm.vn/cau-hoi/giup-em-voiii.8161766187032
\(\frac{1997}{1996}>1;\frac{1998}{1999}< 1\Rightarrow\frac{1997}{1996}>\frac{1998}{1999}\)
Ta thấy : \(\frac{1997}{1996}>1;\frac{1998}{1999}< 1\)
\(\rightarrow\frac{1997}{1996}>\frac{1998}{1999}\)
~ ai tk mk mk tk lại cho nha ~
\(a,\dfrac{1997}{1996}>1>\dfrac{1996}{1997}\\ b,\dfrac{3}{5}< 1< \dfrac{15}{13}\)