Tìm số nguyên x, biết: a) -5/7+1+30/-7<=x<=-1/6+1/3+5/6 b) -8/13+7/17+21/13<=x<=-9/14+3+5/-14

Giải SBT Toán 6 Cánh diều Bài 3: Phép cộng, phép trừ phân số  

Bài 31 trang 38 SBT Toán 6 Tập 2:

Tìm số nguyên x, biết:

a) $\frac{-5}{7}+1+\frac{30}{-7}\le x\le \frac{-1}{6}+\frac{1}{3}+\frac{5}{6}$;

b) $\frac{-8}{13}+\frac{7}{17}+\frac{21}{13}\le x\le \frac{-9}{14}+3+\frac{5}{-14}$.

Lời giải

a) Ta có:

$\frac{-5}{7}+1+\frac{30}{-7}\le x\le \frac{-1}{6}+\frac{1}{3}+\frac{5}{6}$

$\frac{-5}{7}+\frac{7}{7}+\frac{-30}{7}\le x\le \frac{-1}{6}+\frac{2}{6}+\frac{5}{6}$

$\frac{-28}{7}\le x\le \frac{6}{6}$

–4 ≤ x ≤ 1

Mà x ∈ ℤ, suy ra x ∈ {–4; –3; –2; –1; 0; 1}.

Vậy x ∈ {–4; –3; –2; –1; 0; 1}.

b) Ta có:

$\frac{-8}{13}+\frac{7}{17}+\frac{21}{13}\le x\le \frac{-9}{14}+3+\frac{5}{-14}$

$\left(\frac{-8}{13}+\frac{21}{13}\right)+\frac{7}{17}\le x\le \left(\frac{-9}{14}+\frac{-5}{14}\right)+3$

$\frac{13}{13}+\frac{7}{17}\le x\le \frac{-14}{14}+3$

$1+\frac{7}{17}\le x\le -1+3$

$1+\frac{7}{17}\le x\le 2$

Do đó 1 < x ≤ 2.

Mà x ∈ ℤ, suy ra $x=2$.

Vậy x = 2.