\(\dfrac{14}{17}:\dfrac{17}{32}-\dfrac{14}{17}:\dfrac{17}{15}\\ =\dfrac{14}{17}\times\dfrac{32}{17}-\dfrac{14}{17}\times\dfrac{15}{17}\\ =\dfrac{14}{17}\times\left(\dfrac{32}{17}-\dfrac{15}{17}\right)\\ =\dfrac{14}{17}\times\dfrac{17}{17}\\ =\dfrac{14}{17}\times1\\ =\dfrac{14}{17}\)

\(\Rightarrow\dfrac{14}{17}\times\dfrac{32}{17}-\dfrac{14}{17}\times\dfrac{15}{17}=\dfrac{14}{17}\times\left(\dfrac{32}{17}-\dfrac{15}{17}\right)=\dfrac{14}{17}\times\dfrac{17}{17}=\dfrac{14}{17}\)

1 14/17 là phân số hay hỗn số zậy bn

14/17 + -5/13+(-8/13) - 14/17=(14/17-14/17) +{(-5/13) + (-8/13)]

=0 + -13/13

=(-1)

\(1\dfrac{14}{15}-\dfrac{5}{17}-\dfrac{12}{17}-\dfrac{14}{15}\\ =\left(1\dfrac{14}{15}-\dfrac{14}{15}\right)-\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\\ =\left[1+\left(\dfrac{14}{15}-\dfrac{14}{15}\right)\right]-1=1-1=0\)

22

24

26

28

30

32

34

36

11+11= 22       15+15=30

12+12= 24       16+16=32

13+13=     26   17+17=34

14+14=    28    18+18=36

22             30

24              32

26               34

28                36

$\frac{14}{17}<\frac{...}{17}<\frac{15}{17}$

Ta gọi tử số cần tìm là $x$ .

Theo bài ra , ta có :

$\frac{14}{17}$ = $\frac{14\times 2}{17\times 2}$ = $\frac{28}{34}$

$\frac{15}{17}$ = $\frac{15\times 2}{17\times 2}$ = $\frac{30}{34}$

$\frac{...}{17}=\frac{x}{34}$

 Vậy : $\frac{28}{34}<\frac{x}{34}<\frac{30}{34}$ nên $x=29$

$\frac{14}{17}<\frac{...}{17}<\frac{15}{17}$

Ta gọi tử số cần tìm là $x$ .

Theo bài ra , ta có :

$\frac{14}{17}$ = $\frac{14\times 2}{17\times 2}$ = $\frac{28}{34}$

$\frac{15}{17}$ = $\frac{15\times 2}{17\times 2}$ = $\frac{30}{34}$

$\frac{...}{17}=\frac{x}{34}$

 Vậy : $\frac{28}{34}<\frac{x}{34}<\frac{30}{34}$ nên $x=29$

a) Ta có: \(\left|x-9\right|=8\)

         \(\Leftrightarrow x-9=\pm8\)

         \(\Leftrightarrow\orbr{\begin{cases}x-9=8\\x-9=-8\end{cases}}\)

         \(\Leftrightarrow\orbr{\begin{cases}x=17\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)

Vậy \(x=17\)hoặc \(x=1\)

b);c) Ta có: \(17-\left|x+5\right|=14\)

            \(\Leftrightarrow\left|x+5\right|=3\)

            \(\Leftrightarrow x+5=\pm3\)

            \(\Leftrightarrow\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)

            \(\Leftrightarrow\orbr{\begin{cases}x=-2\left(TM\right)\\x=-8\left(TM\right)\end{cases}}\)

Vậy \(x=-2\)hoặc \(x=-8\)