Question

[Nguyễn Bùi Minh Thư]

Tìm x, biết :

a. \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

b. \(|x.\left(x^2-\frac{5}{4}\right)|=x\)

Answer 1

[Huỳnh Quang Sang]

\(a,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Leftrightarrow\left[\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right]x=-1\)

\(\Leftrightarrow0x=-1\Leftrightarrow x\in\varnothing\)

\(b,\left|x\cdot\left[x^2-\frac{5}{4}\right]\right|=x\)

Vì vế trái \(\left|x\left[x^2-\frac{5}{4}\right]\right|\ge0\)với mọi x nên vế phải \(x\ge0\)

Ta có : \(x\left|x^2-\frac{5}{4}\right|=x\)vì \(x\ge0\)

Nếu x = 0 thì \(0\left|0^2-\frac{5}{4}\right|=0\)đúng

Nếu \(x\ne0\)thì ta có \(\left|x^2-\frac{5}{4}\right|=1\Leftrightarrow x^2-\frac{5}{4}=\pm1\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

Answer 2

[Edogawa Conan]

a) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

=> \(\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x+1=0\)

=> \(0x+1=0\)

=> \(1=0\)(vô lí)

b) |x . (x² - 5/4)| = x

TH1: \(x.\left(x^2-\frac{5}{4}\right)=x\)

=> \(x^3-\frac{5}{4}x-x=0\)

=> \(x^3-\frac{9}{4}x=0\)

=> \(x\left(x^2-\frac{9}{4}\right)=0\)

=> \(\orbr{\begin{cases}x=0\\x^2-\frac{9}{4}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^2=\left(\frac{3}{2}\right)^2\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

TH2: \(x\left(x^2-\frac{5}{4}\right)=-x\)

=> \(x^3-\frac{5}{4}x+x=0\)

=> \(x^3-\frac{1}{4}x=0\)

=> \(x\left(x^2-\frac{1}{4}\right)=0\)

=> \(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^2=\left(\frac{1}{2}\right)^2\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}\)

Do |x.(x² - 5/4)| \(\ge\)0 => x\(\ge\)0 => x thuộc {0; 1/2; 3/2}