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Bài 3:
a) \(\frac{x}{1,2}=\frac{5}{6}\)
⇒ \(x.6=5.1,2\)
⇒ \(x.6=6\)
⇒ \(x=6:6\)
⇒ \(x=1\)
Vậy \(x=1.\)
b) \(\frac{5}{9}:x=\frac{7}{4}:\frac{3}{10}\)
⇒ \(\frac{5}{9}:x=\frac{35}{6}\)
⇒ \(x=\frac{5}{9}:\frac{35}{6}\)
⇒ \(x=\frac{2}{21}\)
Vậy \(x=\frac{2}{21}.\)
Bài 5:
Ta có: \(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)
\(\Rightarrow\left(a+b\right).\left(d+a\right)=\left(b+c\right).\left(c+d\right)\)
\(\Rightarrow ad+a^2+bd+ba=bc+bd+c^2+cd\)
\(\Rightarrow a^2+a.\left(b+d\right)=c^2+c.\left(b+d\right)\)
\(\Rightarrow a.\left(b+d\right)=c.\left(b+d\right)\)
\(\Rightarrow a=c\left(đpcm\right).\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)
\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)
Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)
b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)
Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)
- Ta có:
\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)
- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)
a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)
=>Bằng nhau
b: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) a) Ta có: \(\frac{x}{-15}=\frac{-60}{x}\) \(\Rightarrow x^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow x=30\)
b) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\) \(\Rightarrow x.\left(-x\right)=\left(-2\right).\frac{8}{25}\)
\(\Rightarrow x.\left(-x\right)=\frac{-16}{25}\)
\(\Rightarrow x.\left(-x\right)=\left(\frac{-4}{5}\right).\frac{4}{5}\)
Vậy \(x=\frac{4}{5}\)
2) a) \(3,8: \left(2x\right)=\frac{1}{4}:2\frac{2}{3}\)
\(\Rightarrow3,8: \left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=\frac{3}{32}:3,8=\frac{15}{608}\)
\(x=\frac{15}{608}:2=\frac{15}{1216}\)
Vậy \(x=\frac{15}{1216}\)
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(\Rightarrow\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}.3=20\)
\(\Rightarrow x=20:0,25=80\)
Vậy x = 80
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(\Rightarrow\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Leftrightarrow0,75x=\frac{1}{250}.0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
Vậy \(x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{5}{3}.\frac{2}{3}=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}:0,1=\frac{100}{9}\)
Vậy \(x=\frac{100}{9}\)
a) \(\frac{x}{-15}=\frac{-60}{x}\Leftrightarrow x.x=-15.\left(-60\right)\Leftrightarrow x^2=900\Leftrightarrow x^2=\orbr{\begin{cases}30^2\\\left(-30\right)^2\end{cases}}\Leftrightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{a}{b}=\frac{c}{d}\)
\(\frac{a}{b}+1=\frac{c}{d}+1\)
\(\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)
\(\frac{a+b}{b}=\frac{c+d}{d}\)
Vậy \(\frac{a+b}{b}=\frac{c+d}{d}\).
b) \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}\)
\(\frac{b}{a}+1=\frac{d}{c}+1\)
\(\frac{b}{a}+\frac{a}{a}=\frac{d}{c}+\frac{c}{c}\)
\(\frac{a+b}{a}=\frac{c+d}{c}\)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
Vậy \(\frac{a}{a+b}=\frac{c}{c+d}\)
a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\Leftrightarrow\frac{a+b}{a}=\frac{c+d}{c}\Rightarrow\frac{a}{a+b}=\)\(\frac{c}{c+d}\)