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Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\)
\(\Rightarrow ad+cd< bc+dc\)
\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\) (1)
\(ad< bc\)
\(\Rightarrow ad+ab< bc+ab\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\left(đpcm\right)\)
Ta có :
\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\)
\(\Rightarrow ad+ab< bc+ab\Rightarrow a\left(d+b\right)< b\left(c+a\right)\)
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)
Lại có :
\(ad< bc\Rightarrow ad+cd< bc+cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\rightarrowđpcm\)
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt
a) \(\dfrac{x}{48}=-\dfrac{4}{7}\Rightarrow x=-\dfrac{192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\Rightarrow x+\dfrac{4}{5}=1\)
\(\Rightarrow x=\dfrac{1}{5}\)
c) \(2\left|x-1\right|^2=72\Rightarrow\left|x-1\right|^2=36\)
\(\Rightarrow\left|x-1\right|=6\)
TH1: x - 1 = -6 => x = -5
TH2: x - 1 = 6 => x = 7
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\Rightarrow x=2\)
f) | x - 2 | = 1 + 4 = 5
TH1: x - 2 = -5 => x = -3
TH2: x - 2 = 5 => x = 7
a) \(\dfrac{x}{48}=\dfrac{-4}{7}\)
⇒ x.7=48.(-4)
7x = -192
x=\(\dfrac{-192}{7}\) Vậy x=\(\dfrac{-192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\)
\(\left(x+\dfrac{4}{5}\right)=\dfrac{3}{5}+\dfrac{2}{5}\)
\(x+\dfrac{4}{5}=1\)
\(x=1-\dfrac{4}{5}\)
\(x=\dfrac{1}{5}\)
c) chưa từng gặp dạng với giá trị tuyệt đối sory
d) \(\dfrac{1}{6}x-\dfrac{2}{3}=2\)
\(\dfrac{1}{6}x=2+\dfrac{2}{3}\)
\(\dfrac{1}{6}x=\dfrac{8}{3}\)
\(x=\dfrac{8}{3}:\dfrac{1}{6}\)
\(x=16\)
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\)
=> x.5 = 4.2,5
5x=10
x=10:5
x=2
f) |x-2|-4=1
|x-2|=1+4
|x-2|=5
=>\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=5+2\\x=-5+2\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
đôi khi cũng có sai sót , hãy xem lại thật kĩ
Bài 1:
a: \(=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{4}{3}-1+\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)
b: \(=\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{1}{9}-1-\dfrac{2}{5}+\dfrac{5}{4}=2-1+\dfrac{1}{9}=\dfrac{10}{9}\)
c: \(=\left(\dfrac{-3}{2}\cdot\dfrac{4}{3}\right)\cdot\dfrac{-9}{2}-\dfrac{1}{2}=9-\dfrac{1}{2}=8.5\)
a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15
=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)
=1 - 3/4
=1/4
b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2
=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)
=-8 +3/2 +1/3 -3/10
=-97/15
a: \(\Leftrightarrow\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)
=>10<=x<=-13/7
hay \(x\in\varnothing\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=22/36
hay \(x\in\left\{-1;0\right\}\)
\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\\ \Rightarrow\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\\\Rightarrow\dfrac{7}{4}:x=\dfrac{1}{6} \\ \Rightarrow x=\dfrac{7}{4}:\dfrac{1}{6}\\ \Rightarrow x=\dfrac{21}{2}\)
Vậy \(x=\dfrac{21}{2}\)
\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\)
=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\)
=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{4}{6}\)
=> \(\dfrac{7}{4}:x=\dfrac{1}{6}\)
=> \(x=\dfrac{7}{4}:\dfrac{1}{6}\)
=> x = \(\dfrac{7}{4}.6\)
=> \(x=\dfrac{21}{2}\)
Vậy ...