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

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

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

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

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

Đề bài đâu rồi ạ, có đề mới giải được ạ

Bài 5:

Ta có : \(\widehat{A_1}+\widehat{A_3}=180^o\) (kề bù)

            \(100^o+\widehat{A_3}=180^o\)

            \(\widehat{A_3}=80^o\)

Ta có: \(\widehat{A_3}=\widehat{B_1}=80^o\)

            \(\widehat{A_3}\) và \(\widehat{B_1}\) ở vị trí đồng vị 

\(\Rightarrow AC//BD\)

\(\Rightarrow\widehat{C}_1=\widehat{D_1}=135^o\) (đồng vị)

\(x=135^o\)

b)

Ta có: \(\widehat{G_1}+\widehat{B_1}=180^o\left(120^o+60^o=180^o\right)\)

               \(\widehat{G_1}\) và \(\widehat{B_1}\) ở vị trí trong cùng phía

\(\Rightarrow QH//BK\)

\(\Rightarrow\widehat{H_1}=\widehat{K_1}=90^o\)(so le)

\(x=90^o\)

Hệ này sẽ có 1 nghiệm vì 2/1<>-3/1

Từ 0 đến 1 được chia thành 10 phần bằng nhau.

Giá trị của mỗi phần là: \(\dfrac{1}{10}\)

Từ lập luận trên ta có:

Số thích hợp để điền vào các ô trống lần lượt  là: 

10; 5; 6; 9; 10

0 ; 1/10 ; 2/10 ; 3/10 ; 4/10 ; 5/10 ; 6/10 ; 7/10 ; 8/10 ; 9/10 ; 1

Chúc bạn học tốt!

10

5

6

9

10

\(1,Q=\dfrac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+2}\\ 2,x=16\Leftrightarrow Q=\dfrac{1}{4+2}=\dfrac{1}{6}\\ 3,Q=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\Leftrightarrow x=1\left(tm\right)\\ 4,\Leftrightarrow Q-\dfrac{1}{9}=\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{9}>0\\ \Leftrightarrow\dfrac{9-\sqrt{x}-2}{9\left(\sqrt{x}+2\right)}>0\\ \Leftrightarrow7-\sqrt{x}>0\left(\sqrt{x}+2>0\right)\\ \Leftrightarrow\sqrt{x}< 7\Leftrightarrow0\le x< 49;x\ne9\)

\(5,Q=\dfrac{1}{\sqrt{x}+2}\le\dfrac{1}{0+2}=\dfrac{1}{2}\\ Q_{max}=\dfrac{1}{2}\Leftrightarrow x=0\)

\(1.Q=\dfrac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x+2}}\)
       \(=\dfrac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{\left(x-x\right)+\left(2\sqrt{x}-\sqrt{x}\right)-\left(10-4-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{1}{\sqrt{x}+2}\)
\(2.\)Với x=16(TM) thì:\(Q=\dfrac{1}{\sqrt{16}+2}=\dfrac{1}{6}\)
3. Để Q=\(\dfrac{1}{3}\) thì:
\(\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{3}\left(x\ge0,x\ne9\right)\)
\(< =>\sqrt{x}+2=3\)
\(< =>x=1\left(TM\right)\)
Vậy với x=1 thì Q=\(\dfrac{1}{3}\)
4. Để \(Q\ge\dfrac{1}{9}\) thì:
\(\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{9}\)\(\left(x\ge0;x\ne9\right)\)
\(< =>\sqrt{x}+2=9\)
\(< =>x=49\left(TM\right)\)
 

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