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2.

a. \(\dfrac{-4}{9}\) . \(\dfrac{7}{15}+\dfrac{4}{-9}.\dfrac{8}{15}\) = \(\dfrac{-4}{9}.\left(\dfrac{7}{15}+\dfrac{8}{15}\right)\) = \(\dfrac{-4}{9}\) . 1 = \(\dfrac{-4}{9}\) 

b. \(\dfrac{5}{-4}.\dfrac{16}{25}+\dfrac{-5}{4}.\dfrac{9}{25}\) = \(\dfrac{-5}{4}.\left(\dfrac{16}{25}+\dfrac{6}{25}\right)\) = \(\dfrac{-5}{4}.1\) = \(\dfrac{-5}{4}\) 

c. \(4\dfrac{11}{23}-\dfrac{9}{14}+2\dfrac{12}{23}-\dfrac{5}{4}\) = \(\left(4\dfrac{11}{23}+2\dfrac{12}{23}\right)\) \(-\dfrac{9}{14}-\dfrac{5}{4}\) = \(\dfrac{68}{23}-\dfrac{9}{14}-\dfrac{5}{4}\) = \(\dfrac{745}{322}\) - \(\dfrac{5}{4}=\dfrac{685}{644}\) 

d. \(2\dfrac{13}{27}-\dfrac{7}{15}+3\dfrac{14}{27}-\dfrac{8}{15}\) = \(\left(2\dfrac{13}{27}+3\dfrac{14}{27}\right)\) - \(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) = \(\dfrac{68}{27}\) - \(\dfrac{-1}{15}\) = 

e. \(11\dfrac{1}{4}-\left(2\dfrac{7}{5}+5\dfrac{1}{4}\right)\) = \(11\dfrac{1}{4}\) - \(\dfrac{81}{20}\) = \(\dfrac{-13}{10}\) 

g. \(\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\) = \(\dfrac{7}{9}.\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\) = \(\dfrac{7}{9}.1+\dfrac{12}{19}\) = \(\dfrac{7}{19}+\dfrac{12}{19}\) = \(1\)

1 the

2 age

3 anniversary

4 ended

5 appeared

6 hands

7 to

8 veterans

9 national

10 during

\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt[]{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

\(=18+3\sqrt{81-80}.x=18+3x\)\(\Rightarrow x^3-3x=18\left(1\right)\)

\(y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

\(\Rightarrow y^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(=6+3\sqrt[3]{9-8}.y=6+3y\)\(\Rightarrow y^3-3y=6\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow P=x^3+y^3-3\left(x+y\right)+1996=x^3-3x+y^3-3y+1996\)

\(=18+6+1996=2020\)

Ok lại là mik đây :)
mà bài này có j đó sai sai ;-;

a) = (B1+B2+B3+B4+B5+B6)/6
|Kết quả|= 10000
b) = AVERAGE(B1:B6)
|Kết quả|= 0
c) = C1+C2+C3+C4+C5+C6
|Kết quả|= 10
d) = SUM(C1:C6)
|Kết quả|= 0
e) = (B1+B2+B3+B4+B5)/5
    = 12000
f) =AVERAGE(B2:B6)
|Kết quả|= 5000
g) = MAX(C1+C6)
|Kết quả|= 0
h) = MIN(C1:C6)
|Kết quả|= 0
=> Đây sẽ là kết quả nếu làm theo đề, nếu có thắc mắc gì thì hỏi mik. Phần phát biểu nhận xét mik nghĩ bạn có thể làm dc nên sẽ ko làm. Còn nếu bạn muốn làm kết quả mà ko theo đề(theo ý của mik) thì bình luận mik sẽ cho bạn lời giải mà mik cho là đúng nhất. ok :Đ

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)

\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)

\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)

\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)

\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)

Bài 3:

a. \(R=R1+R2=15+30=45\Omega\)

b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)

Bài 4:

\(I1=U1:R1=6:3=2A\)

\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)

\(U=R.I=\left(3+15\right).2=36V\)

\(U2=R2.I2=15.2=30V\)