Question

Tìm x biết \(\frac{3}{7}< \left|x-\frac{6}{7}\right|< \frac{8}{7}\)

Answer 1

\(\frac{3}{7}< \left|x-\frac{6}{7}\right|< \frac{8}{7}\)

\(\Rightarrow\hept{\begin{cases}\left|x-\frac{6}{7}\right|< \frac{8}{7}\\\left|x-\frac{6}{7}\right|>\frac{3}{7}\end{cases}}\)  \(\Rightarrow\orbr{\begin{cases}x-\frac{6}{7}< \frac{8}{7}\\x-\frac{6}{7}< \frac{-8}{7}\end{cases}}\)   và \(\orbr{\begin{cases}x-\frac{6}{7}>\frac{3}{7}\\x-\frac{6}{7}>-\frac{3}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x< \frac{14}{7}\\x< \frac{-2}{7}\end{cases}}\)  và \(\orbr{\begin{cases}x>\frac{9}{7}\\x>\frac{3}{7}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{9}{7}< x< \frac{14}{7}\\x< \frac{-2}{7}\end{cases}}\)

Answer 2

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

Answer 3

vì \(\frac{3}{7}< \left|x-\frac{6}{7}\right|< \frac{8}{7}\)

nên \(\left|x-\frac{6}{7}\right|\in\left\{\frac{4}{7};\frac{5}{7};\frac{6}{7};\frac{7}{7}\right\}\)

nên \(\left(x-\frac{6}{7}\right)\in\left\{\pm\frac{4}{7};\pm\frac{5}{7};\pm\frac{6}{7};\pm\frac{7}{7}\right\}\)

Ta có bảng sau:

\(x-\frac{6}{7}\)	\(x\)
\(\frac{4}{7}\)	\(\frac{10}{7}\)
\(-\frac{4}{7}\)	\(\frac{2}{7}\)
\(\frac{5}{7}\)	\(\frac{11}{7}\)
\(-\frac{5}{7}\)	\(\frac{1}{7}\)
\(\frac{6}{7}\)	\(\frac{12}{7}\)
\(-\frac{6}{7}\)	\(0\)
\(\frac{7}{7}\)	\(\frac{13}{7}\)
\(-\frac{7}{7}\)	\(\frac{-1}{7}\)