Question

Giúp ạ !!!!

Answer 1

a.

ĐKXĐ: \(x\ne\left\{-7;1\right\}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x-7}=\dfrac{x-2-\left(x+4\right)}{x-1-\left(x-7\right)}=\dfrac{-6}{6}=-1\)

\(\Rightarrow x-2=-1\left(x-1\right)\)

\(\Rightarrow x-2=1-x\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\dfrac{3}{2}\)

b.

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}=\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{-2z}{-42}=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.2=20\\y=6.2=12\\z=21.2=42\end{matrix}\right.\)

c.

\(4x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\)

\(7y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{-3y}{-60}=\dfrac{z}{28}=\dfrac{2x-3y+z}{30-60+28}=\dfrac{6}{-2}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.15=-45\\y=-3.20=-60\\z=-3.48=-144\end{matrix}\right.\)

Answer 2

d.

\(x:y:z=12:9:5\Rightarrow\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=12k\\y=9k\\z=5k\end{matrix}\right.\)

\(xyz=20\Rightarrow12k.9k.5k=20\)

\(\Rightarrow k^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=12k=4\\y=9k=3\\z=5k=\dfrac{5}{3}\end{matrix}\right.\)

e.

Câu này đề sai, ko thể giải được trong chương trình phổ thông.

f.

\(2x^3-1=15\Rightarrow2x^3=16\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

\(\Rightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y-25}{16}=2\\\dfrac{z+9}{25}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=16.2+25=57\\z=25.2-9=41\end{matrix}\right.\)