TL

1,0 nha bn

HT 

nhớ k

cũng có thể kb nếu muốn

0,952 \(\approx\)1,0

2,1,0,1     học tốt nhé

= 2;=1;=0;=1

tTL

=-262,672 nha bn

ht

nhớ ko

= -252672

⇔\orbr{x⋅(x−3)=xx⋅(x−3)=−x⇔\orbr{x−3=xxx−3=−xx⇔\orbr{x−3=1x−3=−1⇔\orbr{x=1+3x=−1+3⇔\orbr{x⋅(x−3)=xx⋅(x−3)=−x⇔\orbr{x−3=xxx−3=−xx⇔\orbr{x−3=1x−3=−1⇔\orbr{x=1+3x=−1+3

⇔\orbr{x=4x=2⇔\orbr{x=4x=2

Vậy x=2 hoặc x=4

\(\frac{5}{6}-2.\sqrt{\frac{4}{9}}+\sqrt{\left(-2\right)^2}\)= \(\frac{5}{6}-2.\frac{2}{3}+2\)

\(=\frac{5}{6}-\frac{8}{6}+\frac{12}{6}=\frac{9}{6}=\frac{3}{2}\)

\(\frac{5}{6}\)-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

\(a,\frac{5}{6}-2\sqrt{\frac{4}{9}}+\sqrt{\left(-2\right)^2}\)

\(=\frac{5}{6}-2.\frac{2}{3}+2\)

\(=\frac{5}{6}-\frac{4}{6}+\frac{12}{6}\)

\(=\frac{5-4+12}{6}=\frac{13}{6}\)

\(b,\left(-3\right)^2.\left(\frac{1}{3}\right)^3:\left[\left(-\frac{2}{3}\right)^3-1\frac{1}{3}\right]-\left(-200\right)^0\)

\(=9.\frac{1}{27}:\left(-\frac{8}{27}-\frac{5}{3}\right)-1\)

\(=\frac{1}{3}:\left(-\frac{8}{27}-\frac{45}{27}\right)-1\)

\(=\frac{1}{3}:\left(-\frac{53}{27}\right)-1\)

\(=\frac{1}{3}.\left(-\frac{27}{53}\right)-1\)

\(=-\frac{9}{53}-1=-\frac{9}{53}-\frac{53}{53}\)

\(=-\frac{62}{53}\)

\(c,\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):2\)

\(=\left(-\frac{1}{2}-\frac{3}{5}\right).\frac{1}{3}+\frac{1}{3}-\left(-\frac{1}{6}\right).\left(-\frac{1}{2}\right)\)

\(=\left(-\frac{5}{10}-\frac{6}{10}\right).\frac{1}{3}+\frac{1}{3}-\frac{1}{12}\)

\(=-\frac{11}{10}.\frac{1}{3}+\frac{1}{3}-\frac{1}{12}\)

\(=\frac{1}{3}\left(-\frac{11}{10}-\frac{1}{12}\right)\)

\(=\frac{1}{3}\left(-\frac{66}{60}-\frac{5}{60}\right)\)

\(=\frac{1}{3}.\left(-\frac{71}{60}\right)\)

\(=-\frac{71}{180}\)