\(a,=\left(-\dfrac{1}{3}\right)^3=-\dfrac{1}{27}\\ b,=\left(-2\right)^6=2^6=64\\ c,=5^6:5^2=5^4=625\)

\(a,\left(-\dfrac{1}{3}\right)^2.\left(-\dfrac{1}{3}\right)=-\dfrac{1}{3}\)

\(b,\left(-2\right)^2.\left(-2\right)^4=\left(-2\right)^6\)

\(c,25^3:5^2=5^6:5^2=5^3\)

(5/6+1/6)x4/5

= 1 x 4/5

=4/5

\(\dfrac{4}{5}\) x ( \(\dfrac{5}{6}\) + \(\dfrac{1}{6}\))

\(\dfrac{4}{5}\) x  1

\(\dfrac{4}{5}\)

\(a,\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{9}{7}=\left(\dfrac{-4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}-\dfrac{9}{7}\right)=-1+\left(-1\right)=-2\)

\(b,\dfrac{1}{12}-\left(\dfrac{1}{2}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\left(\dfrac{6}{12}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\dfrac{13}{12}=\dfrac{-12}{12}=-1\)

\(c,\dfrac{-12}{5}-\left(\dfrac{3}{5}-\dfrac{1}{8}\right)=\dfrac{-12}{5}-\left(\dfrac{24}{40}-\dfrac{5}{40}\right)=\dfrac{-12}{5}-\dfrac{19}{40}=\dfrac{-96}{40}-\dfrac{16}{40}=\dfrac{-100}{40}=\dfrac{-5}{2}\)

\(d,\dfrac{-5}{7}-\left(\dfrac{-7}{6}+\dfrac{2}{7}\right)+\dfrac{7}{-6}=\dfrac{-5}{7}+\dfrac{7}{6}-\dfrac{2}{7}+\dfrac{-7}{6}=\left(\dfrac{-5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{7}{6}+\dfrac{-7}{6}\right)=-1\)

\(e,\dfrac{-6}{5}-\dfrac{2}{7}+\dfrac{1}{5}-\dfrac{5}{7}=\left(\dfrac{-6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{-2}{7}-\dfrac{5}{7}\right)=-1+\left(-1\right)=-2\)

a: \(=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)+1+\dfrac{10}{13}=\dfrac{3}{13}+\dfrac{10}{13}+1=2\)

b: \(=4\left(2.86+3.14\right)-25\cdot6.01+9\cdot0.75\)

=4*6-25*6,01+9*0,75

=-119,5

c: \(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}+\dfrac{15}{4}=\dfrac{7}{8}+\dfrac{15}{4}=\dfrac{7+30}{8}=\dfrac{37}{8}\)

d: \(=\dfrac{-3}{31}-\dfrac{28}{31}-\dfrac{6}{17}-\dfrac{11}{17}+\dfrac{1}{25}-\dfrac{5}{25}=\dfrac{-4}{25}-2=-\dfrac{54}{25}\)

\(1.678\simeq1.68;1.734\simeq1.73\)
\(2.014\simeq2.01;2.536\simeq2.54\)

\(2.834\simeq2.83;2.927\simeq2.93\)

đề bài ko có số điện tiêu thụ thì làm sao tính đc ơi

Bài 1:

a) \(\dfrac{5x-6}{2x-1}=\dfrac{3x}{x-2}\)      (1)

ĐK:\(\left\{{}\begin{matrix}2x-1\ne0\\x-2\ne0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x\ne\dfrac{1}{2}\\x\ne2\end{matrix}\right.\)

(1) ⇒\(\dfrac{\left(5x-6\right)\left(x-2\right)}{\left(2x-1\right)\left(x-2\right)}=\dfrac{3x\left(2x-1\right)}{\left(2x-1\right)\left(x-2\right)}\)

⇔\(5x^2-10x-6x+12=6x^2-3x\)

⇔\(5x^2-10x-6x+12-6x^2+3x=0\)

⇔\(-x^2-13+12=0\)

⇔\(\left[{}\begin{matrix}x=\dfrac{-13+\sqrt{217}}{2}\\x=\dfrac{-13-\sqrt{217}}{2}\end{matrix}\right.\left(TM\right)\)

KL:

sorry! câu này mih làm sai mất rồi