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\(\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ - 3}}{4}\\x = \dfrac{{( - 3).6}}{4}\\x = \dfrac{{ - 9}}{2}\end{array}\)
Vậy \(x = \dfrac{{ - 9}}{2}\)
\(\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ - 20}}\\x = \dfrac{{5.( - 20)}}{{15}}\\x = \dfrac{{ - 20}}{3}\end{array}\)
Vậy \(x = \dfrac{{ - 20}}{3}\)
Bài 1:
Ta có: \(3x=2y\)
nên \(\dfrac{x}{2}=\dfrac{y}{3}\)
mà x+y=-15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)
Vậy: (x,y)=(-6;-9)
Bài 2:
a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)
mà x+y-z=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)
Vậy: (x,y,z)=(40;30;50)
`x/(-4) = (-11)/2`
`=> 2x=-4.(-11)`
`=> 2x=44`
`=>x=44:2`
`=>x=22`
`---`
`(15-x)/(x+9) =3/5`
`=> (15-x).5=(x+9).3`
`=> 75-5x =3x+27`
`=> -5x -3x=27 -75`
`=> -8x=-48`
`=>x=-48:(-8)`
`=>x=6`
a) x−4=−112
x=(−11).(−4)2
x=22.
b) 15−xx+9 =35
(15−x).5 =(x+9).3
75−5x =3x+27
8x=48
x=6.
a, \(\dfrac{x}{-3}\)= \(\dfrac{7}{4}\) ⇒ x = \(\dfrac{7}{4}\)x (-3) ⇒ x = - \(\dfrac{21}{4}\)
b, \(\dfrac{x+9}{15-x}\) = \(\dfrac{2}{3}\) ⇒ 3(x+9) = 2( 15-x) ⇒ 3x + 27 = 30 - 2x
⇒ 3x + 2x = 30 - 27 ⇒
5x = 3 ⇒ x = 3 : 5 ⇒ x = \(\dfrac{3}{5}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a)\(\dfrac{x}{7}=\dfrac{18}{14}\)
\(\Rightarrow x=\dfrac{7.18}{14}=9\)
b)\(6:x=1\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{6}{x}=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{6.4}{7}=\dfrac{24}{7}\)
c)5,7:0,35=(-x):0,45
\(\Leftrightarrow\dfrac{114}{7}=\dfrac{-x}{0,45}\)
\(\Rightarrow\left(-x\right)=\dfrac{114.0,45}{7}=\dfrac{-513}{70}\)
a) Ta có:\(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=21+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=46:2=23\)
\(a,\dfrac{x-3}{x+5}=\dfrac{5}{7}\\ \Leftrightarrow7x-21=5x+25\\ \Leftrightarrow2x=46\\ \Leftrightarrow x=23\)
Vậy......
\(b,\dfrac{x+4}{20}=\dfrac{5}{x+4}\\ \Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow x+4=\pm10\\ \Leftrightarrow x\in\left\{-14;6\right\}\)
Vậy.........
a; \(\dfrac{x}{6}\) = \(\dfrac{-3}{4}\)
\(x=\dfrac{-3}{4}.6\)
\(x\) = - \(\dfrac{9}{2}\)
Vậy \(x=-\dfrac{9}{2}\)
b; \(\dfrac{5}{x}\) = \(\dfrac{15}{-20}\) (đk \(x\ne0\))
\(x\) = 5 : \(\dfrac{15}{-20}\)
\(x=-\dfrac{20}{3}\)
Vậy \(x=-\dfrac{20}{3}\)
c; \(\dfrac{x+11}{14-x}\) = \(\dfrac{2}{3}\) (đk \(x\ne14\))
3.(\(x+11\)) = 2.(14 - \(x\))
3\(x\) + 33 = 28 - 2\(x\)
3\(x\) + 2\(x\) = 28 - 33
5\(x\) = -5
\(x\) = -1
Vậy \(x\) = -1