Question

[Hà]

Câu1.timf x

a)4x=3y;5y=3z và 2x-3y+z=6

B) |x-1/2| - 1/9 bình phương = 1/4 nình phương

Answer 1

[Vũ Minh Tuấn]

Câu 1:

b) \(\left|x-\frac{1}{2}\right|-\left(\frac{1}{9}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Rightarrow\left|x-\frac{1}{2}\right|-\frac{1}{81}=\frac{1}{16}\)

\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{16}+\frac{1}{81}\)

\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{97}{1296}.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{2}=\frac{97}{1296}\\x-\frac{1}{2}=-\frac{97}{1296}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{97}{1296}+\frac{1}{2}\\x=\left(-\frac{97}{1296}\right)+\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{745}{1296}\\x=\frac{551}{1296}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{745}{1296};\frac{551}{1296}\right\}.\)

Chúc bạn học tốt!

Answer 2

[Bùi Hùng Cừơng]

a)4x=3y;5y=3z và 2x-3y+z=6

Có : \(4x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{4}\Leftrightarrow\frac{1}{3}\times\frac{x}{3}=\frac{1}{3}\times\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)

\(5y=3z\Leftrightarrow\frac{y}{3}=\frac{z}{5}\Leftrightarrow\frac{1}{4}\times\frac{y}{3}=\frac{1}{4}\times\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Leftrightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)

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

\(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

-Do đó :

\(\frac{2x}{18}=2\Leftrightarrow\frac{x}{9}=2\Rightarrow x=2\times9=18\)

\(\frac{3y}{36}=2\Leftrightarrow\frac{y}{12}=2\Rightarrow y=2\times12=24\)

\(\frac{z}{20}=2\Rightarrow z=2\times20=40\)

Vậy x = 18 , y = 24 , z = 40