a) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và \(x-y+z=-49\)

Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)

\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)

Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)

b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2-y^2+2z^2=10\)

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\)

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

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{10}{27}\)

Vậy ... (tự tính x, y, z nhé!)

vãi ***** làm bài

Bài 1:

a) ta có: \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=\frac{2y-4}{6}\)

ADTCDTSBN

có: \(\frac{x-1}{5}=\frac{2y-4}{6}=\frac{z-2}{2}=\frac{x-1+2y-4-z+2}{5+6-2}\)\(=\frac{\left(x+2y-z\right)-\left(1+4-2\right)}{9}=\frac{6-3}{9}=\frac{3}{9}=\frac{1}{3}\)

=>...

bn tự tính típ nhé!

b) ta có: \(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}\)

ADTCDTSBN

có: \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{52}{13}=4\)

=>...

Bài 2:

a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)

\(\Rightarrow\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a+b}{b}=\frac{c+d}{b}\left(đpcm\right)\)

b) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\) (*)

mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

Từ (*) \(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)

\(A=2x+2y+3xy\left(x+y\right)+5\left(x^3y^2+x^2y^3\right)\)

\(\Rightarrow A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)\)

\(\Rightarrow A=0\) ( do x+y = 0 )

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\Rightarrow\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{16}\)\(=\dfrac{x^2-2y^2+z^2}{4-18+16}=\dfrac{8}{2}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{4}=4\\\dfrac{y^2}{9}=4\\\dfrac{z^2}{16}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\\z^2=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-4\\y=-6\\z=-8\end{matrix}\right.\)

b: \(3x^2y^3=\dfrac{1}{9}\)

\(\Leftrightarrow3x^2=\dfrac{1}{9}:\dfrac{1}{27}=3\)

=>x=1 hoặc x=-1

a: \(A=\dfrac{-2}{3}\cdot\left(-27\right)\cdot4\cdot\dfrac{1}{2}=18\cdot2=36\)

2. Tham khảo thêm tại đây nha bạn

https://hoc24.vn/hoi-dap/question/417550.html

Bài 1.

a) Nhân 2 vào tỉ số thứ 2 rồi áp dụng tính chất của dãy tỉ số bằng nhau.

Kết quả:

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

b) \(\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)

Theo tính chất dãy tỉ số bằng nhau:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2+y^2}{4+9}=\dfrac{52}{13}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm4\\y=\pm6\end{matrix}\right.\)

Vậy ...

Bài 2.

a) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}\)

\(\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)

Vậy ...

2:

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=i\Rightarrow\left\{{}\begin{matrix}a=bi\\c=di\end{matrix}\right.\)

Ta có:

\(\dfrac{ac}{bd}=\dfrac{c^2i}{d^2i}=\dfrac{c^2}{d^2}=\left(\dfrac{c}{d}\right)^2=i^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2i^2+d^2i^2}{b^2+d^2}=\dfrac{i^2\left(b^2+d^2\right)}{b^2+d^2}=i^2\)

Từ đó suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\) (đpcm)

Bài 1:

Giải:

Ta có: \(\left\{{}\begin{matrix}3x=4y\\5y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Mà \(xyz=30\)

\(\Rightarrow240k^3=30\)

\(\Rightarrow k^3=\dfrac{1}{8}\)

\(\Rightarrow k=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=2,5\end{matrix}\right.\)

Vậy...

Bài 2: sai đề

Bài 3:

Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Ta có: \(x+2y+3z=38\)

\(\Rightarrow2k+1+8k-6+18k+15=38\)

\(\Rightarrow28k=28\)

\(\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\\z=11\end{matrix}\right.\)

Vậy...

1) Ta có :

\(3x=4y\Rightarrow\dfrac{3x}{12}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\) <=> \(\dfrac{x}{8}=\dfrac{y}{6}\)

\(5y=6z\Rightarrow\dfrac{5y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\)

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

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Thay vào đẳng thức xyz = 30

=> 8k.6k.5k = 30

<=> 240k³ = 30

<=> k³ = 8

<=> k = 2

\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)

b) Câu này cũng tương tự câu 1 nha ! Đặt k luôn , còn không bình phương lên rồi dùng tính chất dãy tỉ số bằng nhau .

c) Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)

=> \(\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Thay vào đẳng thức , ta có :

x + 2y + 3z = 2k + 1 + 2(4k - 3) + 3(6k + 5) = 38

=> 28k = 38

=> k = \(\dfrac{19}{14}\)

Vậy .....