Giúp mik câu 1,2 ở part 2 nha
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1.47:
a) Ta có: \(\dfrac{3}{7}x-\dfrac{2}{5}x=-\dfrac{17}{35}\)
\(\Leftrightarrow\dfrac{1}{35}x=\dfrac{-17}{35}\)
hay x=-17
Vậy: x=-17
b) Ta có: \(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}:x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{1}{3}+\dfrac{-3}{5}:x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\dfrac{-3}{5}:x=\dfrac{-1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{9}{16}\cdot\dfrac{4}{3}=\dfrac{36}{48}=\dfrac{3}{4}\\x=\dfrac{-3}{5}:\dfrac{-1}{3}=\dfrac{-3}{5}\cdot\dfrac{-3}{1}=\dfrac{9}{5}\end{matrix}\right.\)
Bài 1.48:
a) Ta có: \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< \dfrac{-2}{5}\end{matrix}\right.\)
b) Ta có: \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1>0\\x+\dfrac{3}{5}< 0\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{-3}{5}\)
bạn đặt thừa số chung là 4.32 ra nè, còn những cái kia cho vào trong ngoặc rồi tính á
\(A=1,2\cdot594\cdot4+1,6\cdot235\cdot3+1,2\cdot684\)
\(A=1,2\cdot594\cdot4+1,2\cdot235\cdot4+1,2\cdot171\cdot4\)
\(A=4,8\cdot594+4,8\cdot235+4,8\cdot171\)
\(A=4,8\left(594+235+171\right)\)
\(A=4,8\cdot1000=4800\)
Câu 2:
\(1,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ 2,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ 3Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ 4,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\\ 5,2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ 6,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ 7,4Na+O_2\to 2Na_2O\\ 8,C_nH_{2n-2}+\dfrac{3n-1}{2}O_2\xrightarrow{t^o}nCO_2\uparrow+(n-1)H_2O\)
1 C
2 B
3 A
4 A
5 B
6 A
7 B
8 D
9 C
10 D
11 A
12 A
13 C
14 C
14 D
II
1 less
2 electrician
3 competitions
4 librarian
5 Learing
6 littering
7 behavior
1 less
2 electrician