Bài 2: 

a: \(23+4\sqrt{15}=\left(2\sqrt{5}+3\right)^2\)

b: \(19-4\sqrt{15}=\left(\sqrt{15}-2\right)^2\)

c: \(10-2\sqrt{21}=\left(\sqrt{7}-\sqrt{3}\right)^2\)

d: \(22+2\sqrt{21}=\left(\sqrt{21}+1\right)^2\)

e: \(19-4\sqrt{21}=\left(2\sqrt{3}-\sqrt{7}\right)^2\)

f: \(31+4\sqrt{21}=\left(2\sqrt{7}+\sqrt{3}\right)^2\)

g: \(11-2\sqrt{30}=\left(\sqrt{6}-\sqrt{5}\right)^2\)

\(23+4\sqrt{15}=\left(2\sqrt{5}+\sqrt{3}\right)^2\)

\(19-4\sqrt{15}=\left(\sqrt{15}-2\right)^2\)

\(10-2\sqrt{21}=\left(\sqrt{7}-\sqrt{3}\right)^2\)

\(22+2\sqrt{21}=\left(\sqrt{21}+1\right)^2\)

\(19-4\sqrt{21}=\left(2\sqrt{3}-\sqrt{7}\right)^2\)

\(31+4\sqrt{21}=\left(2\sqrt{7}-\sqrt{3}\right)^2\)

\(11-2\sqrt{30}=\left(\sqrt{6}-\sqrt{5}\right)^2\)

Bài 1: 

a: ĐKXĐ: \(x\ge\dfrac{5}{2}\)

b: ĐKXĐ: \(x\le3\)

c: ĐKXĐ: \(x>\dfrac{3}{4}\)

d: ĐKXĐ: \(x< 0\)

e: ĐKXĐ: \(x\ge\dfrac{1}{2}\)

f: ĐKXĐ: \(x\ge\dfrac{7}{2}\)

g: ĐKXĐ: \(\left\{{}\begin{matrix}x\le3\\x\ne0\end{matrix}\right.\)

h: ĐKXĐ: \(\left\{{}\begin{matrix}x< 0\\x\ne-1\end{matrix}\right.\)

Bài 21: 

a) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

b) Ta có: \(\left|-5x\right|=3x-16\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x=3x-16\left(x\le0\right)\\5x=3x-16\left(x>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x-3x=-16\\5x-3x=-16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-8x=-16\\2x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-8\left(loại\right)\end{matrix}\right.\)

c) Ta có: \(\left|x-4\right|=-3x+5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=-3x+5\left(x\ge4\right)\\x-4=3x-5\left(x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3x=5+4\\x-3x=-5+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=9\\-2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(loại\right)\\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

d) Ta có: \(\left|3x-1\right|-x=2\)

\(\Leftrightarrow\left|3x-1\right|=2-x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2-x\left(x\ge\dfrac{1}{3}\right)\\3x-1=x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+x=2+1\\3x-x=-2+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(nhận\right)\\x=\dfrac{-1}{2}\left(nhận\right)\end{matrix}\right.\)

D

C

D

Câu 1: 

a,MCD: R1//R2

\(R_{12}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{30\cdot20}{30+20}=12\left(\Omega\right)\)

b, MCD: R3nt(R1//R2)

\(R_{tđ}=R_3+R_{12}=30+12=42\left(\Omega\right)\)

Câu 2

a Điện trở và cường độ dòng điện tối đa mà biến trở đó có thể có

b,\(S=\dfrac{l\cdot\rho}{R}=\dfrac{100\cdot1,1\cdot10^{-6}}{200}=5,5\cdot10^{-7}\)

\(R=\sqrt{\dfrac{S}{\pi}}=\sqrt{\dfrac{5,5\cdot10^{-7}}{\pi}}=4,18\cdot10^{-4}\left(m\right)=0,418\left(mm\right)\)

1

a)tôi(CN)đem xác Dế Choắt chôn vào một vùng cỏ um tùm(VN).Tôi(CN)đắp thành nấm mộ to(VN).tôi(CN)đứng lặng giờ lâu nghĩ về bài học đường đời đầu tiên(VN)

b)Dế Mèn đã nghĩ:"Trong cuộc sóng mà ngông cuồng bậy bạ,có óc mà ko biết nghĩ sẽ trút họa vào thân

Câu 2

BPTT:so sánh

công cha như núi Thái Sơn

nghĩa mẹ như nước trong nguồn chảy ra

5. The teacher advised Nam to do morning exercises regularly.

6. My mother told me to clean up the bedroom.

7. The doctor advised Lan not to eat too much meat.

8, My mother asked me to wait for her there.

9. Miss Jackson said Tim should work harder on English pronunciation.

10. The manager invited me to drink some coffee.

1.Lan asked me if i went to school by bike

2.They said that they wanted to meet Mr Nam the following week

3. Mrs Hoa asked Ba not to stay up too late

4.He told me to carry heavy bag for him

5. The doctor advised my father not to drink too much coffee in the evening

a) \(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

b) \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\\ \Leftrightarrow-2\left(2x-5\right)=0\\ \Leftrightarrow2x-5=0\\ \Leftrightarrow x=\dfrac{5}{2}\)

c) \(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a) 3x(x-1)+(x-1).1=0

(x-1)(3x+1)=0

=> x-1=0 hoặc 3x+1 =0 

=> x=1 hoặc x= -1/3