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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:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
\(LINH_1=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+....+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=LINH_2\left(đpcm\right)\)
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 \(-\dfrac{157}{623}< -\dfrac{47}{213}\)
Ta có:
\(\overline{abcabc}=1001\overline{abc}\)
\(=143.7.\overline{abc}\)
\(\Rightarrow1001\overline{abc}⋮7\Rightarrow\overline{abcabc}⋮7\)
\(\rightarrowđpcm\)
\(\overline{aaa}=111a\)
\(=37.3.a\)
\(\Rightarrow111a⋮37\Rightarrow\overline{aaa}⋮37\)
\(\rightarrowđpcm\)
\(\overline{1ab1}-\overline{1ba1}\)
\(=1000+\overline{ab}+1-1000-\overline{ba}-1\)
\(=\overline{ab}-\overline{ba}\)
\(=10a+b-10b-a\)
\(=9a-9b\)
\(=9\left(a-b\right)⋮9\)
Mà \(\overline{1ab1}-\overline{1ba1}=\overline{...0}⋮10\)
\(\Rightarrow\overline{1ab1}-\overline{1ba1}⋮9;10\Rightarrow⋮90\)
\(\rightarrowđpcm\)
bn ơi câu b mk ghi nhầm đề là 4 chữ a mới đúng bn giải lại giùm mk nhoa
a) 45<1<1,1⇒45<1,145<1<1,1⇒45<1,1
b) -500 < 0 < 0,001 => -500 < 0,001
c) −12−37=1237<1236=13=1339<1338⇒−12−37<1338
a)Ta có :
\(\dfrac{4}{5}< 1< 1,1\Rightarrow\dfrac{4}{5}< 1,1\)
b)Ta có :
\(-500< 0< 0,001\Rightarrow-500< 0,001\)
c)Ta có :
\(\dfrac{-12}{-37}=\dfrac{12}{37}< \dfrac{12}{36}=\dfrac{1}{3}=\dfrac{13}{39}< \dfrac{13}{38}\Rightarrow\dfrac{-12}{-37}< \dfrac{13}{38}\)
Lời giải:
Ta có \(\frac{2016c-2017b}{2015}=\frac{2017a-2015c}{2016}=\frac{2015b-2016a}{2017}\)
\(\Rightarrow \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\( \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
\(=\frac{2015.2016c-2015.2017b+2016.2017a-2016.2015c+2017.2015b-2017.2016a}{2015^2+2016^2+2017^2}=0\)
\(\Rightarrow \left\{\begin{matrix} 2015.2016c-2015.2017b=0\\ 2016.2017a-2016.2015c=0\\ 2017.2015b-2016.2016a=0\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2016c=2017b\\ 2017a=2015c\\ 2015b=2016a\end{matrix}\right.\Rightarrow \frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)
Ta có đpcm.
Giải:
\(\dfrac{bz-xy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)
\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)
\(\Rightarrow\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}\)
\(=\dfrac{0}{a^2+b^2+c^2}\)
\(=0\)
Ta có: \(\dfrac{bz-cy}{a}\)
\(\Rightarrow bz-cy=0\)
\(\Rightarrow\dfrac{z}{c}=\dfrac{b}{y}\)\(\left(1\right)\)
Ta có:\(\dfrac{cx-az}{b}=0\)
\(\Rightarrow cx-az=0\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{z}{c}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)(đpcm)
Ta có :
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}:\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}\cdot\dfrac{2}{1}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{b}{ab}+\dfrac{a}{ab}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{2}{c}\)
\(\Rightarrow2ab=\left(a+b\right)c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Ta có: \(\overline{aaa}\) = 111.a (a < 10)
mà 111 ⋮ 37 nên 111.a ⋮ 37
Vậy \(\overline{aaa}\) ⋮ 37
Ta có: \(\overline{aaaaaa}\) = 111111.a (a < 10)
mà 111111 ⋮ 37 nên 111111.a ⋮ 37
Vậy \(\overline{aaaaaa}\) ⋮ 37
Ta có: \(\dfrac{ }{aaa}=100a+10a+a=111a=37.3.a,⋮37\)
\(\dfrac{ }{aaaaaa}=\dfrac{ }{aaa}.1000+\dfrac{ }{aaa},\) cả hai số hạng của tổng đều \(⋮37\) nên tổng của chúng cũng \(⋮37\)
Ví dụ: Số \(555⋮37\) ( thương là 15 )
Số \(777777⋮37\) ( thương là 21021 )