\(R1//R2\)

\(\Rightarrow\left\{{}\begin{matrix}Rtd=\dfrac{R1R2}{R1+R2}=8\Omega\\U2=U1=Ia1.R1=12V\Rightarrow Ia2=\dfrac{U2}{R2}=\dfrac{12}{40}=0,3A\\Ia=Ia1+Ia2=0,3+1,2=1,5A\\Uv=U2=12V\end{matrix}\right.\)

yêu cầu tính nhanh nha

Ta có: \(\dfrac{3}{n\left(n+3\right)}=\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}=\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}=\dfrac{1}{n}-\dfrac{1}{n+3}\)

Do đó:

\(C=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..+\dfrac{1}{37}-\dfrac{1}{40}\)

\(=1-\dfrac{1}{40}=\dfrac{39}{40}\)

cho hướng dẫn giải chi tiết thế rồi mà bạn cũng không biết làm?

a) Áp suất của nước tác dụng lên vật là

\(p=d.h=10000.12=120000\left(Pa\right)\)

b) Áp suất của nước nếu chìm sâu thêm 5m

\(p=d.h=10000.\left(12+5\right)=170000\left(Pa\right)\)

c) Áp lực của nước là

\(p=\dfrac{F}{S}=\dfrac{10000}{0,03}=3333333,33\left(Pa\right)\)

1. Products

2. disappointingly

3. consuming

4. polluting

5. respiratory

6. seriously

7. harmful

8. Conservationists

9. deforestation

10. necessary

11. celebrations

12. ineffectively

13. innovations

14. respiratory

15. respiratory

Câu 3:

a: A=1/2

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{2}\)

=>\(2\sqrt{x}-4=\sqrt{x}\)

=>\(2\sqrt{x}-\sqrt{x}=4\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

b: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{x-1}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(\sqrt{P}< \dfrac{1}{2}\)

=>\(0< =P< \dfrac{1}{4}\)

=>P>=0 và P<1/4

=>P>=0 và P-1/4<0

=>\(\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}-2>=0\\\dfrac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>=2\\3\sqrt{x}-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>=2\\\sqrt{x}< 3\end{matrix}\right.\)

=>\(2< =\sqrt{x}< 3\)

=>4<=x<9

\(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)

(2x-1)⁵=x⁵

2x-1=x

2x-x=1

x=1

\(a,C_5H_{12}O\\ b,C_2H_6O\)