Giúppp
K
Khách
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Những câu hỏi liên quan
4 tháng 10 2021
\(a,=\left(-\dfrac{1}{3}\right)^3=-\dfrac{1}{27}\\ b,=\left(-2\right)^6=2^6=64\\ c,=5^6:5^2=5^4=625\)
giúppp
TT
25 tháng 3 2022
\(\dfrac{4}{5}\) x ( \(\dfrac{5}{6}\) + \(\dfrac{1}{6}\))
\(\dfrac{4}{5}\) x 1
\(\dfrac{4}{5}\)
8 tháng 3 2023
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: \(=\dfrac{-3}{31}-\dfrac{28}{31}-\dfrac{6}{17}-\dfrac{11}{17}+\dfrac{1}{25}-\dfrac{5}{25}=\dfrac{-4}{25}-2=-\dfrac{54}{25}\)
23 tháng 2 2023
\(1.678\simeq1.68;1.734\simeq1.73\)
\(2.014\simeq2.01;2.536\simeq2.54\)
\(2.834\simeq2.83;2.927\simeq2.93\)
giúppp mình vs m.n
T
5 tháng 12 2021
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:
3 tháng 3 2022
\(Cl_2+H_2\rightarrow2HCl\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnCl_2+Ba\left(OH\right)_2\rightarrow Zn\left(OH\right)_2+BaCl_2\)
\(Zn\left(OH\right)_2+2HCl\rightarrow ZnCl_2+2H_2O\)
3 tháng 3 2022
Cl2+H2→2HClCl2+H2→2HCl
Zn+2HCl→ZnCl2+H2Zn+2HCl→ZnCl2+H2
ZnCl2+Ba(OH)2→Zn(OH)2+BaCl2ZnCl2+Ba(OH)2→Zn(OH)2+BaCl2
Zn(OH)2+2HCl→ZnCl2+2H2O
\(a,\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{9}{7}=\left(\dfrac{-4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}-\dfrac{9}{7}\right)=-1+\left(-1\right)=-2\)
\(b,\dfrac{1}{12}-\left(\dfrac{1}{2}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\left(\dfrac{6}{12}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\dfrac{13}{12}=\dfrac{-12}{12}=-1\)
\(c,\dfrac{-12}{5}-\left(\dfrac{3}{5}-\dfrac{1}{8}\right)=\dfrac{-12}{5}-\left(\dfrac{24}{40}-\dfrac{5}{40}\right)=\dfrac{-12}{5}-\dfrac{19}{40}=\dfrac{-96}{40}-\dfrac{16}{40}=\dfrac{-100}{40}=\dfrac{-5}{2}\)
\(d,\dfrac{-5}{7}-\left(\dfrac{-7}{6}+\dfrac{2}{7}\right)+\dfrac{7}{-6}=\dfrac{-5}{7}+\dfrac{7}{6}-\dfrac{2}{7}+\dfrac{-7}{6}=\left(\dfrac{-5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{7}{6}+\dfrac{-7}{6}\right)=-1\)
\(e,\dfrac{-6}{5}-\dfrac{2}{7}+\dfrac{1}{5}-\dfrac{5}{7}=\left(\dfrac{-6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{-2}{7}-\dfrac{5}{7}\right)=-1+\left(-1\right)=-2\)