Câu hỏi của Jenny_2690

Question

Tính:

B=[-$\sqrt{2,25}$+4$\sqrt{\left(-2,15\right)^2}$-(3$\sqrt{\dfrac{7}{6}}$)^2].

Fa Châu De · Accepted Answer

Tính:

C = $3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right).2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)$

= $\dfrac{7}{2}.\dfrac{4}{49}-\left[\left(2+0,\left(1\right).4\right).\dfrac{27}{5}\right].\dfrac{-5}{42}$

= $\dfrac{1.2}{1.7}-\left[\left(2+\dfrac{1.4}{9}\right).\dfrac{27}{5}\right].\dfrac{-5}{42}$

= $\dfrac{2}{7}-\left[\dfrac{18+4}{9}.\dfrac{27}{5}\right].\dfrac{-5}{42}$

= $\dfrac{2}{7}-\left[\dfrac{22.9}{3.5}\right].\dfrac{-5}{42}$

= $\dfrac{2}{7}-\dfrac{198}{15}.\dfrac{-5}{42}=\dfrac{2}{7}-\dfrac{11}{3}.\dfrac{-1}{7}$

= $\dfrac{2}{7}+\dfrac{11}{21}$ = $\dfrac{6+11}{21}$ = $\dfrac{17}{21}$

Câu trả lời:

Tính:

C = $3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right).2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)$

= $\dfrac{7}{2}.\dfrac{4}{49}-\left[\left(2+0,\left(1\right).4\right).\dfrac{27}{5}\right].\dfrac{-5}{42}$

= $\dfrac{1.2}{1.7}-\left[\left(2+\dfrac{1.4}{9}\right).\dfrac{27}{5}\right].\dfrac{-5}{42}$

= $\dfrac{2}{7}-\left[\dfrac{18+4}{9}.\dfrac{27}{5}\right].\dfrac{-5}{42}$

= $\dfrac{2}{7}-\left[\dfrac{22.9}{3.5}\right].\dfrac{-5}{42}$

= $\dfrac{2}{7}-\dfrac{198}{15}.\dfrac{-5}{42}=\dfrac{2}{7}-\dfrac{11}{3}.\dfrac{-1}{7}$

= $\dfrac{2}{7}+\dfrac{11}{21}$ = $\dfrac{6+11}{21}$ = $\dfrac{17}{21}$

Jenny_2690 · Answer

Bạn cóa thể giải giúp mình câu B ko?

Câu trả lời:

Bạn cóa thể giải giúp mình câu B ko?

Nguyễn Lê Phước Thịnh · Answer

$B=\left[-1.5+4\cdot2.15-9\cdot\dfrac{7}{6}\right]\cdot\dfrac{5}{4}$

$=-\dfrac{17}{5}\cdot\dfrac{5}{4}=-\dfrac{17}{4}$

Câu trả lời:

$B=\left[-1.5+4\cdot2.15-9\cdot\dfrac{7}{6}\right]\cdot\dfrac{5}{4}$

$=-\dfrac{17}{5}\cdot\dfrac{5}{4}=-\dfrac{17}{4}$