chỉ có 3 em đi câu cá suy ra em thứ 4 câu được 0 con cá

Em thứ 4 câu được 0 con cá

Vì chỉ có 3em nhỏ đi câu cá 

\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)

\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)

$4.17+4.84-4$

$=4.17+4.84-4.1$

$=4.17+4.84-1.4$

$=4.(17+84-1)$

$=4.100$

$=400$

_________________

$17.18-18+84.18$

$=17.18-18.1+84.18$

$=17.18+18.(-1)+84.18$

$=18.(17-1+84)$

$=18.100$

$=1800$

_____________________

$75.12+75.9$

$75.12+75.9-21.25$

$=75.(12+9)-21.25$

$=75.21-21.25$

$=75.21+21.(-25)$

$=21.(75-25)$

$=21.50$

$=1050$

2.

Áp dụng định lý hàm cosin:

\(b=\sqrt{a^2+c^2-2ac.cosB}=\sqrt{8^2+3^2-2.8.3.cos60^0}=7\)

\(S_{ABC}=\dfrac{1}{2}ac.sinB=\dfrac{1}{2}.8.3.sin60^0=6\sqrt{3}\)

4.

\(\Delta=\left(m+2\right)^2-16>0\Leftrightarrow m^2+4m-12>0\Rightarrow\left[{}\begin{matrix}m>2\\m< -6\end{matrix}\right.\) (1)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m-2\\x_1x_2=4\end{matrix}\right.\)

\(x_1+x_2+x_1x_2>1\)

\(\Leftrightarrow-m-2+4>1\)

\(\Rightarrow m< 1\) (2)

Kết hợp (1); (2) ta được \(m< -6\)

Chia bớt ra đi bạn 

Bài 5:

\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ 2KMnO_4+16HCl_{đặc,nóng}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ n_{Cl_2\left(LT\right)}=\dfrac{5}{2}.n_{KMnO_4}=2,5.0,2=0,5\left(mol\right)\\ n_{Cl_2\left(TT\right)}=0,5.90\%=0,45\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,45.22,4=10,08\left(l\right)\)

Bài 4:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\CuO+2HCl\rightarrow CuCl_2+H_2\\ n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ a,\%m_{Fe}=\dfrac{56.0,1}{13,6}.100\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

7:

A+B

\(=x^4-2xy+y^2+y^2+2xy+x^2+1\)

=x^4+2y^2+x^2+1

A-B

=x^4-2xy+y^2-y^2-2xy-x^2-1

=x^4-4xy-x^2-1

5:

a: =8x^2-4x^2=4x^2

b: =(5-7)*x^2y^3z^3=-2x^2y^3z^3

c: =(3+2-1/3-1/2-1/6)*x^2y^2

=4x^2y^2

a, Độ rượu thu được là: \(\dfrac{30}{30+90}.100=25^o\)

b, Ta có: \(m_{C_2H_5OH}=30.0,8=24\left(g\right)\Rightarrow n_{C_2H_5OH}=\dfrac{24}{46}=\dfrac{12}{23}\left(mol\right)\)

PT: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)

_______12/23____________________6/23 (mol)

\(\Rightarrow V_{H_2}=\dfrac{6}{23}.22,4\approx5,84\left(l\right)\)

Bạn tham khảo nhé!