C

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

\(\dfrac{y}{3}=2\Rightarrow y=6\)

\(\dfrac{z}{7}=2\Rightarrow z=14\)

a) \(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)

⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

⇒\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c: Ta có: 5x=8y=20z

nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)

Do đó: x=24; y=15; z=6

1,a/ Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-2\\\dfrac{y}{5}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\)

Vậy ...

b, Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=4\\\dfrac{y}{5}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=20\end{matrix}\right.\)

Vậy ...

2/a, Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{5}=4\\\dfrac{z}{7}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\\z=28\end{matrix}\right.\)

Vậy ...

b/ \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Leftrightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}\)

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{6}=4\\\dfrac{y}{5}=4\\\dfrac{z}{8}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=24\\y=20\\z=32\end{matrix}\right.\)

Vậy ..

Bài Giải:

Bài 1:

a) Theo đề bài, ta có:

\(\dfrac{x}{2}=\dfrac{y}{5}\)và x+y=-4

Áp dụng tính chất của dãy tỉ số bằng nhau

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)

Suy ra: x = 2 . (-2) =-4

y = 5 . (-2) =-10

Vậy: x = -4 và y = -10

Mấy câu sau cậu cứ dựa vào bài trên để giải nhé!

Tick cho Phong nhé:>

Yêu nhiều>3

#Phong_419

Bài 1
a) \(\dfrac{x}{2}=\dfrac{y}{5}\)
\(\text{Mà }x+y=-14\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)
\(\Rightarrow\dfrac{x}{2}=-2\Rightarrow x=2.\left(-2\right)=-4\)
\(\Rightarrow\dfrac{y}{5}=-2\Rightarrow y=5.\left(-2\right)=-10\)
\(\text{Vậy }\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\)
b) \(\dfrac{x}{7}=\dfrac{y}{5}\)
\(\text{Mà }x-y=8\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)
\(\Rightarrow\dfrac{x}{7}=4\Rightarrow x=4.7=28\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\text{Vậy }\left\{{}\begin{matrix}x=28\\y=20\end{matrix}\right.\)
Bài 2:
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)
\(\text{Mà }x+y+z=56\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\dfrac{x}{2}=4\Rightarrow x=4.2=8\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{7}=4\Rightarrow z=4.7=28\)
\(\text{Vậy }\left\{{}\begin{matrix}x=8\\y=20\\z=28\end{matrix}\right.\)
b) \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)
\(\Rightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}\)
\(\text{Mà }2x+y-z=12\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)
\(\Rightarrow\dfrac{2x}{6}=4\Rightarrow2x=4.6=24\Rightarrow x=24:2=12\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{8}=4\Rightarrow z=4.8=32\)
\(\text{Vậy }\left\{{}\begin{matrix}x=12\\y=20\\z=32\end{matrix}\right.\)

=>x/35=y/28=z/32

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{32}=\dfrac{x-y+z}{35-28+32}=\dfrac{7}{39}\)

=>x=245/39; y=196/39; z=224/39

x/5 = y/4 => 4x = 5y => x = 5y/4   (1)

y/7 = z/8 => 7z = 8y => z = 8y/7    (2)

có: x-y+z=7

từ (1) và (2) => 5y/4 - y + 8y/7 = 7

<=> 39y/28 = 7

=> y = 196/39

=> x = 245/39

=> z = 224/39

\(\dfrac{x}{y}=\dfrac{3}{7}.\\ \Rightarrow x=\dfrac{3}{7}y.\\ x-y=16.\\\Rightarrow\dfrac{3}{7}y-y=16.\\ \Rightarrow y=-28.\\ \Rightarrow x=-12.\)

\(\dfrac{x}{1,8}=\dfrac{y}{3,2}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{1,8}{3,2}=\dfrac{9}{16}.\\ \Rightarrow x=\dfrac{9}{16}y.\\ y-x=7.\\ \Rightarrow y-\dfrac{9}{16}y=7.\\ \Leftrightarrow y=16.\\ \Leftrightarrow x=9.\)

\(\dfrac{x}{5}=\dfrac{y}{8}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{8}.\\ \Rightarrow x=\dfrac{5}{8}y.\\ x+2y=42.\\ \Rightarrow\dfrac{5}{8}y+2y=42.\\ \Leftrightarrow y=16.\\ \Rightarrow x=10.\)

\(\dfrac{x}{5}=\dfrac{y}{7}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{7}.\\ \Rightarrow x=\dfrac{5}{7}y.\\ x.y=35.\\ \Rightarrow\dfrac{5}{7}y.y=35.\\ \Leftrightarrow y^2=49.\\ \Leftrightarrow u=\pm7.\\ \Rightarrow x=\pm5.\)

a) \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{z}{3}=\dfrac{y}{7}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y-z}{9-7-3}=\dfrac{-15}{-1}=15\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.9\\y=15.7\\z=15.3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=135\\y=105\\z=45\end{matrix}\right.\)

Vậy, x = 135, y = 105, z = 45

b, \(\dfrac{x}{-3}=\dfrac{y}{-8}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{64}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{64}=\dfrac{x^2-y^2}{9-64}=-\dfrac{44}{\dfrac{5}{-55}}=-\dfrac{44}{5}:\left(-55\right)=-\dfrac{44}{5}.-\dfrac{1}{55}=\dfrac{44}{275}=0,16\)

+) \(\dfrac{x^2}{9}=0,16\Rightarrow x^2=1,44\Rightarrow x=\pm1,2\)

+) \(\dfrac{y^2}{64}=0,16\Rightarrow y^2=10,24\Rightarrow y=\pm3,2\)

Vậy ...

Ta có: \(\dfrac{x}{6}=\dfrac{y}{6}:\dfrac{y}{8}=\dfrac{z}{7}\)

\(\Leftrightarrow\dfrac{x}{6}=\dfrac{4}{3}=\dfrac{z}{7}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\z=\dfrac{28}{3}\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot6}{2}=12\\y=4\cdot5=20\\z=8\cdot4=32\end{matrix}\right.\)

x/3=y/5=z/8

áp dung dãy tỉ số bằng nhau

=

>x/3=y/5=z/8=2.x+y-z/2.3+5-8=12/3=4

x=12,y=20,z=32