Tìm hai phân số : \(\frac{a}{b}\)và \(\frac{c}{d}\)biết rằng \(\frac{a}{b}\)-\(\frac{c}{d}\) = \(\frac{1}{15}\) và \(\frac{a}{b}\):\(\frac{c}{d}\)=\(\frac{6}{5}\)
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Bài 1:suy ra 5*(44-x)=3*(x-12)
220-5x=3x-36
-5x-3x=-36-220
-8x =-256
x=32
Bài 2 :Đặt a/3=b/4=k
suy ra a=3k ; b=4k
Ta có a*b=48
suy ra 3k*4k=48
12k =48
k=4
suy ra a=3*4=12
b=4*4 =16
Bài 3: áp dụng tính chất dãy số bằng nhau ta được
a+b+c+d/3+5+7+9 = 12/24=0,5
suy ra a=1,5; b=2,5; c=3,5; d=4,
1)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ac-ad=ac-bc\Leftrightarrow a\left(c-d\right)=c\left(a-b\right)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
2) Gọi độ dài các cạnh của tam giác đó là a,b,c thì a : b : c = 3 : 4 : 5 ; a + b + c = 36
\(\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{36}{12}=3\Rightarrow\hept{\begin{cases}a=3.3=9\\b=3.4=12\\c=3.5=15\end{cases}}\).Vậy tam giác đó có 3 cạnh là 9 cm ; 12 cm ; 15 cm
3)\(\hept{\begin{cases}a:b:c:d=3:4:5:6\\a+b+c+d=3,6\end{cases}\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{d}{6}=\frac{a+b+c+d}{3+4+5+6}=\frac{3,6}{18}=0,2}\)
=> a = 0,2.3 = 0,6 ; b = 0,2.4 = 0,8 ; c = 0,2.5 = 1 ; d = 0,2.6 = 1,2
4)\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}:5=\frac{y}{2}:5\Leftrightarrow\frac{x}{15}=\frac{y}{10}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}:2=\frac{z}{7}:2\Leftrightarrow\frac{y}{10}=\frac{z}{14}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{14}=\frac{x+y+z}{15+10+14}=\frac{184}{39}=4\frac{28}{39}\Rightarrow\hept{\begin{cases}x=4\frac{28}{39}.15=70\frac{10}{13}\\y=4\frac{28}{39}.10=47\frac{7}{39}\\z=4\frac{28}{39}.14=66\frac{2}{39}\end{cases}}\)
Bài 1 :
Từ \(\frac{1}{4}< \frac{1}{3}\) suy ra \(\frac{1}{4}< \frac{1+1}{4+3}< \frac{1}{3}\) hay \(\frac{1}{4}< \frac{2}{7}< \frac{1}{3}\)
Từ \(\frac{1}{4}< \frac{2}{7}\)suy ra \(\frac{1}{4}< \frac{1+2}{4+7}< \frac{1}{3}\)hay \(\frac{1}{4}< \frac{3}{11}< \frac{1}{3}\)
Từ \(\frac{2}{7}< \frac{1}{3}\)suy ra \(\frac{2}{7}< \frac{2+1}{7+3}< \frac{1}{3}\)hay \(\frac{2}{7}< \frac{3}{10}< \frac{1}{3}\)
Vậy ta có : \(\frac{1}{4}< \frac{3}{11}< \frac{2}{7}< \frac{3}{10}< \frac{1}{3}\)
Chúc bạn học tốt ( -_- )
Bài 2 :
\(\frac{a}{a+b+c+d}< \frac{a}{a+b+c}< \frac{a}{a+c}\left(1\right)\)
\(\frac{b}{a+b+c+d}< \frac{b}{b+c+d}< \frac{b}{b+d}\left(2\right)\)
\(\frac{c}{a+b+c+d}< \frac{c}{c+d+a}< \frac{c}{c+a}\left(3\right)\)
\(\frac{d}{a+b+c+d}< \frac{d}{d+a+b}< \frac{d}{d+b}\left(4\right)\)
Cộng ( 1 ), ( 2 ) , (3 ) và ( 4 ) theo từng vế ta được :
\(1=\frac{a+b+c+d}{a+b+c+d}< \frac{a}{a+b+c}+\frac{b}{b+c+d}\)\(+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{a+c}{a+c}+\frac{b+d}{b+d}\)
Chúc bạn học tốt ( -_- )
1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)
Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)
Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
\(\Rightarrow A=4\)
2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)
Bài 2 :
a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy ...
b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy ...
Đặt \(\frac{a}{2}=\frac{c}{4}=\frac{e}{5}\) = k => a = 2k; c = 4k ; e = 5k
\(\frac{b}{3}=\frac{d}{5}=\frac{g}{6}\)= h => b = 3h; d = 5h; g = 6h
Khi đó: \(\frac{a}{b}+\frac{c}{d}+\frac{e}{g}=\frac{2k}{3h}+\frac{4k}{5h}+\frac{5k}{6h}=\left(\frac{2}{3}+\frac{4}{5}+\frac{5}{6}\right).\frac{k}{h}=2\frac{3}{10}\)
=> \(\frac{23}{10}.\frac{k}{h}=\frac{23}{10}\)=> \(\frac{k}{h}=1\)=> k = h
Vậy \(\frac{a}{b}=\frac{2k}{3h}=\frac{2}{3};\frac{c}{d}=\frac{4}{5};\frac{e}{g}=\frac{5}{6}\)
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
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)