a) Vì \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) (1)

Vì \(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)

Mà \(3x-7y+5z=30\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{21}=2\\\dfrac{y}{14}=2\\\dfrac{z}{10}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=20\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=42\\y=28\\z=20\end{matrix}\right.\).

b) Cách làm giống y hệt câu a nhé! Không khác đâu vì \(3x-7y+5z=3x+5z-7y\), nó chỉ đổi đổi vị trí các số hạng thoy.

2x=3y => \(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\)

                                                                              \(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)

5y=7z => \(\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\) 

áp dụng tính chất của dãy tỉ số bằng nhau(còn lại tự tính)

kích nha Phạm Quang Huy

Theo bài ra ta có :

\(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{3}=\dfrac{7y}{14}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) \(\left(1\right)\)

\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{2y}{14}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) \(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\Rightarrow\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)

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

\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x+5z-7y}{63+50-98}=\dfrac{30}{15}=2\\ \)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x}{63}=2\Rightarrow\dfrac{x}{21}=2\Rightarrow x=42\\\dfrac{7y}{98}=2\Rightarrow\dfrac{y}{14}=2\Rightarrow y=28\\\dfrac{5z}{50}=2\Rightarrow\dfrac{z}{10}=2\Rightarrow z=20\end{matrix}\right.\\ \)

\(\text{Vậy }x=42\\ y=28\\ z=20\)

Ta có:

\(2x=3y\Rightarrow10x=15y\)

\(5y=7z\Rightarrow15y=21z\)

\(\Rightarrow10x=15y=21z\Rightarrow\dfrac{10x}{210}=\dfrac{15y}{210}=\dfrac{21z}{210}\)

\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)

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

\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x+5z-7y}{3.21+5.14-7.10}\)

\(=\dfrac{30}{63+70-70}=\dfrac{30}{63}=\dfrac{10}{21}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{21}.21=10\\y=\dfrac{10}{21}.14=\dfrac{20}{3}\\z=\dfrac{10}{21}.10=\dfrac{100}{21}\end{matrix}\right.\)

Chúc bạn học tốt!!!

2x = 3y => 10x=15y
5y = 7z => 15y=21z
=> 10x=15y=21z =>x=2,1z
y=1,4z
Mà : 3x - 7y + 5z = 30 => 6,3z - 9,8z + 5z=30 =>1,5z=30
=>z=20
y=28
x=42

Từ \(2x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{2}=\frac{x}{3}.\frac{1}{7}=\frac{y}{2}.\frac{1}{7}=\frac{x}{21}=\frac{y}{14}\)( 1 )

Từ \(5y=7z\)\(\Rightarrow\)\(\frac{y}{7}=\frac{z}{5}=\frac{y}{7}.\frac{1}{2}=\frac{z}{5}.\frac{1}{2}=\frac{y}{14}=\frac{z}{10}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Đặt \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=k\)

\(\Rightarrow\hept{\begin{cases}x=21k\\y=14k\\z=10k\end{cases}}\)

Thay vào \(3x+5z-7y=30\)ta có ;

\(3.21k+5.10k-7.14k=30\)

\(63k+50k-98k=30\)

\(15k=30\)

\(k=2\)

Thay vào ta được :

\(\Rightarrow\hept{\begin{cases}x=21.2\\y=14.2\\z=10.2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)

2x=3y;5y=7z

=>x/3=y/2;y/7=z/5

=>x/21=x/14;y/14=z/10

=>x/21=y/14=z/10

=>3x/63=7y/98=5z/50

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

3x/63=7y/98=5z/50=3x-7y+5z/63-98+50=30/15=2

suy ra : 3x/63=2 =>3x=126 =>x=126:3=42

7y/98=2 =>7y =196 =>y=196:7=28

5z/50=2 =>5z = 100 => z=100:5=20

\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\left(1\right)\)

\(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\left(2\right)\)

Từ 1 và 2 

=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Áp dụng tính chất dãy tỉ số = nhau ta có

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\frac{x}{21}=2\Rightarrow x=42\)

\(\frac{y}{14}=2\Rightarrow y=28\)

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

\(Ta\ có:\)

\(2x=3y\)

⇒\(\frac{x}{21}\)=\(\frac{y}{14}\)(1)

\(5y=7z\)

⇒\(\frac{y}{14}\)=\(\frac{z}{10}\)(2)

\(Từ\ (1)\ và\ (2)\ suy\ ra: \)\(\frac{x}{21}\)=\(\frac{y}{14}\)=\(\frac{z}{10}\)

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

\(\frac{x}{21}\)=\(\frac{y}{14}\)=\(\frac{z}{10}\)=\(\frac{3x}{63}\)=\(\frac{7x}{98}\)=\(\frac{5z}{50}\)=\(\frac{3x-7y+5z}{63-98+50}\)=\(\frac{30}{15}\)=\(2\)

⇒\(\hept{\begin{cases}x=2.21\\y=2.14\\z=2.10\end{cases}}\text{⇒}\)\(\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)

lời giải mấy một hai câu em có tự làm dựa theo đề bài

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Vậy $x=42$ ; $y=28$ ; $z=20$ 

y=7/5 z

x=3/2 y=21/10 z

3x - 7y + 5z =30

<=> 63/10 z - 49/5 z + 5z = 30

<=>                 5z - 35/10 z = 30

<=>                        15/10 z = 30

<=>                                   z = 30 : 15/10

<=>                                    z = 20

=> y = 28

=> x = 42

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       Câu hỏi của Phạm Quang Huy     

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