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

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

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

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

a.(-1/3)^2.(-1/3)=(-1/3)^3

b.(-2)^2.(-2)^4=(-2)^6

c.25^3:5^2=5^6:5^2=5^4

ht

(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

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

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

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

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

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

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

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

\(=\dfrac{2}{7}-\dfrac{9}{7}-\dfrac{4}{5}-\dfrac{1}{5}\)

\(=-\dfrac{7}{7}-\dfrac{5}{5}\)

\(=-1-1=-2\)

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

\(=\dfrac{1}{12}-\dfrac{7}{12}-\dfrac{1}{2}\)

\(=-\dfrac{6}{12}-\dfrac{1}{2}=-1\)

\(c,-\dfrac{12}{5}-\left(\dfrac{3}{5}-\dfrac{1}{8}\right)\)

\(=-\dfrac{12}{5}-\dfrac{3}{5}+\dfrac{1}{8}\)

\(=-\dfrac{15}{5}+\dfrac{1}{8}\)

\(=-3+\dfrac{1}{8}=-\dfrac{23}{8}\)

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

=\(-\dfrac{5}{7}+\dfrac{7}{6}-\dfrac{2}{7}-\dfrac{7}{6}\)

\(=-\dfrac{5}{7}-\dfrac{2}{7}\)

\(=-\dfrac{7}{7}=-1\)

\(e.-\dfrac{6}{5}-\dfrac{2}{7}+\dfrac{1}{5}-\dfrac{5}{7}\)

\(=-\dfrac{6}{5}+\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{5}{7}\)

\(=\dfrac{5}{5}-\dfrac{7}{7}\)

\(=1-1=0\)

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

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

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

Điện trở tương đương: \(R=R1+R2=20+40=60\Omega\)

Chỉ số của Ampe kế: \(I=U:R=12:60=0,2A\)

Giải:
điện trở tương đương của đoạn mạch AB là :

$R=R1+R2=20+40=60\Omega$

Chỉ số của Ampe kế là :

 $I=U:R=12:60=0,2A$

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:

Bạn nên đăng vào lớp 9 để nhận được nhiều sự trợ giúp hơn nhé !

Bài 1: b) \(\dfrac{7x}{2x+2}=\dfrac{3x+1}{x-5}\)     (2)

ĐK: \(\left\{{}\begin{matrix}2x+2\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\)

(2)⇒ \(\dfrac{7x\left(x-5\right)}{\left(2x+2\right)\left(x-5\right)}=\dfrac{\left(3x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(x-5\right)}\)

\(\Leftrightarrow7x^2-35x=6x^2+6x+2x+2\)

\(\Leftrightarrow7x^2-35x-6x^2-6x-2x-2=0\)

\(\Leftrightarrow x^2+27x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-27+\sqrt{737}}{2}\\x=\dfrac{-27+\sqrt{737}}{2}\end{matrix}\right.\left(TM\right)\)

KL:

\(Cl_2+H_2\rightarrow2HCl\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnCl_2+Ba\left(OH\right)_2\rightarrow Zn\left(OH\right)_2+BaCl_2\)

\(Zn\left(OH\right)_2+2HCl\rightarrow ZnCl_2+2H_2O\)