Bài 1: Tính hợp lý
\(\frac{5}{6}\)- 2 . \(\sqrt{\frac{4}{9}}\)+ \(\sqrt{\left(-2\right)^2}\)
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a) \(\left|-\frac{6}{7}\right|\div\left(-2\right)^3-\sqrt{\frac{9}{16}}\)
\(=\frac{6}{7}\div\left(-8\right)-\frac{3}{4}\)
\(=\frac{-3}{28}-\frac{3}{4}\)
\(=\frac{-6}{7}\)
b) \(-5\frac{5}{9}\div\left(-1\frac{3}{4}\right)+4\frac{5}{9}\div\left(-1\frac{3}{4}\right)\)
\(=\left(-5\frac{5}{9}+4\frac{5}{9}\right)\div\left(-1\frac{3}{4}\right)\)
\(=\left(-5+\frac{5}{9}+4+\frac{5}{9}\right)\div\frac{-7}{4}\)
\(=\left(-1+\frac{10}{9}\right).\frac{-4}{7}\)
\(=\frac{1}{9}.\frac{-4}{7}\)
\(=\frac{-4}{63}\)
c) \(-63,99-\left(\frac{4}{9}-63,99\right)-\left(-1\frac{2}{3}\right)^2\)
\(=-63,99-\frac{4}{9}+63,99-\left(\frac{-5}{3}\right)^2\)
\(=-63,99-\frac{4}{9}+63,99-\frac{25}{9}\)
\(=\left(-63,99+63,99\right)-\left(\frac{4}{9}+\frac{25}{9}\right)\)
\(=-\frac{29}{9}\)
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
\(\frac{5}{6}-2.\sqrt{\frac{4}{9}}+\sqrt{\left(-2\right)^2}\)= \(\frac{5}{6}-2.\frac{2}{3}+2\)
\(=\frac{5}{6}-\frac{8}{6}+\frac{12}{6}=\frac{9}{6}=\frac{3}{2}\)
\(\frac{5}{6}\)-2.\(\sqrt{\frac{4}{9}}\)+\(\sqrt{\left(-2\right)^2}\)
=\(\frac{5}{6}\)-2.\(\frac{2}{3}\)+2
=\(\frac{5}{6}\)-\(\frac{4}{3}\)+2
=\(\frac{5}{6}\)+\(\frac{-8}{6}\)+2
=\(\frac{-1}{2}\)+2=\(\frac{3}{2}\)
Hok tốt!
@Kaito Kid