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bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)
áp dụng dảy tỉ số bằng nhau
ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)
\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)
\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)
vậy \(a=-3;b=-11;c=-7\)
b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)
áp dụng dảy tỉ số bằng nhau
ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)
\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)
vậy \(a=40;b=60;c=48\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-5c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3+}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)
\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)
Mà \(a+b+c=-50\)
\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)
\(\Leftrightarrow2c=-50\)
\(\Leftrightarrow c=-25\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-15\end{matrix}\right.\)
Vậy ...
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}=\dfrac{5\left(3a-2b\right)+3\left(2c-5a\right)+2\left(5b-3c\right)}{25+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3a-2b=0\\2c-5a=0\\5b-3c=0\end{matrix}\right.\)
⇔ 15a= 10b = 6c ⇔ \(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a+b+c}{\dfrac{1}{15}+\dfrac{1}{10}+\dfrac{1}{6}}=-\dfrac{50}{\dfrac{1}{3}}=-150\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)
Bài 1 :
a, \(-1\dfrac{2}{3}\)= \(\dfrac{-5}{3}\)
Dựa vào tính chất của Tỉ lệ thức :
Ta có : \(\dfrac{x}{y}=\dfrac{-5}{3}\rightarrow\dfrac{x}{-5}=\dfrac{y}{3}\)
Dựa vào tính chất của dãy tỉ số = nhau
Ta có : \(\dfrac{x}{-5}=\dfrac{y}{3}=\dfrac{x+y}{\left(-5\right)+3}=\dfrac{18}{-2}=-9\)
\(\rightarrow\dfrac{x}{-5}=-9\rightarrow x=\left(-5\right).\left(-9\right)\Rightarrow x=45\\ \rightarrow\dfrac{y}{3}=-9\rightarrow y=3.\left(-9\right)\Rightarrow y=-27\)b,
Ta có :
( x + 4 ) . 7 = ( y + 7 ) . 4
\(\rightarrow\) 7x + 28 = 4y + 28
\(\rightarrow\) 7x = 4y
Vì 7x = 4y
\(\Rightarrow\) x = 22 / ( 4 + 7 ) . 7 = 14
\(\Rightarrow\) y = 22 - 14 = 8
Đợi mk lm câu 2 nha
ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}\)
Áp dụng tính chất dãy tỉ sốbằng nhau ta có:
\(\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}=\dfrac{a^6+c^6}{b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
=\(\dfrac{c^6+3a^6}{d^6+3b^6}\)
\(\Rightarrow\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (ĐPCM)
\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
=> \(\dfrac{3a+4b}{3c+4d}=\dfrac{5a-6b}{5c-6d}\)
ta có
\(\dfrac{3a+4b}{3c+4d}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a}{b}=\dfrac{c}{d}\)(đpcm)
Ta có:
\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
\(\Leftrightarrow\left(3a+4b\right)\left(5c-6d\right)=\left(3c+4d\right)\left(5a-6b\right)\)
\(\Rightarrow15ac-18ad+20bc-24bd=15ac-18bc+20ad-24bd\)
\(\Rightarrow15ac-15ac-18ad-20ad=-24bd+24bd-18bc-20bc\)
\(\Rightarrow-38ad=-38bc\)
\(\Rightarrow ad=bc\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
<=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{3a}{3c}=\dfrac{2b}{2d}\)
<=>\(\dfrac{5a-3b}{5c-3d}=\dfrac{3a-2b}{3c-2d}\)(đpcm)
Các câu sau tương tự
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
@Tuấn Anh Phan Nguyễn
-_-