1. On the ocean 

2. Yes

3. do house work , cleaning the floor , cooking meals , washing clothes , feeding his pets .

Bài 3: 

b: \(B_1=-\left|2x-3\right|+2\le2\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

\(B_2=-\left|x+4\right|+5\le5\forall x\)

Dấu '=' xảy ra khi x=-4

Bài 3:

b) Xét số \(-B_3=6+\left|x+4\right|\ge6\Rightarrow B_3\le-6\)

Dấu '=' xảy ra \(\Leftrightarrow x=-4\)

\(3,\\ b,B_1=2-\left|3-2x\right|\le2\)

Dấu \("="\Leftrightarrow x=\dfrac{3}{2}\)

\(B_2=5-\left|x+4\right|\le5\)

Dấu \("="\Leftrightarrow x=-4\)

\(B_3=-6-\left|x+4\right|\le-6\)

Dấu \("="\Leftrightarrow x=-4\)

\(c,15-\left(x^2-4\right)^2-\left|x+2\right|\le15\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x^2-4=0\\x+2=0\end{matrix}\right.\Leftrightarrow x=-2\)

\(C_2=10-\left|2x-1\right|-\left|y+x\right|\le10\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Does your father get to work by bicycle?

Does your father get to work by bicycle?

3. Does your father cycle to work?

Does your father get to work by bike?

Bài 4: 

Nhóm 1: x;1/3x; 8x

Nhóm 2: \(x^2;5x^2;-3x^2\)

Lỗi r, xem lại nhé

1 Yes, there are

2 The city changed a lot

3 It's always sunny

1, Yes, there are

1.

a, \(\left(C\right)x^2+y^2-6x-2y+6=0\)

\(\Leftrightarrow\left(C\right)\left(x-3\right)^2+\left(y-1\right)^2=4\)

\(\Rightarrow\) Tâm \(I=\left(3;1\right)\), bán kính \(R=2\)

b, Tiếp tuyến đi qua A có dạng: \(\left(\Delta\right)ax+by-5a-7b=0\left(a^2+b^2\ne0\right)\)

Ta có: \(d\left(I;\Delta\right)=\dfrac{\left|3a+b-5a-7b\right|}{\sqrt{a^2+b^2}}=2\)

\(\Leftrightarrow\left|a+3b\right|=\sqrt{a^2+b^2}\)

\(\Leftrightarrow6ab+8b^2=0\)

\(\Leftrightarrow2b\left(3a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=0\\3a+4b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\Delta_1:x=5\\\Delta_2:4x-3y+1=0\end{matrix}\right.\)

TH1: \(\Delta_1:x=5\)

Tiếp điểm có tọa độ là nghiệm hệ: \(\left\{{}\begin{matrix}x=5\\x^2+y^2-6x-2y+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y^2-2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\Rightarrow\left(5;1\right)\)

TH2: \(\Delta_2:4x-3y+1=0\)

Tiếp điểm có tọa độ là nghiệm hệ: \(\left\{{}\begin{matrix}4x-3y+1=0\\x^2+y^2-6x-2y+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}\\y=\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left(\dfrac{7}{5};\dfrac{11}{5}\right)\)

Kết luận: Phương trình tiếp tuyến: \(\left\{{}\begin{matrix}\Delta_1:x=5\\\Delta_2:4x-3y+1=0\end{matrix}\right.\)

Tọa độ tiếp điểm: \(\left\{{}\begin{matrix}\left(5;1\right)\\\left(\dfrac{7}{5};\dfrac{11}{5}\right)\end{matrix}\right.\)

5.

\(n_{Na}=a\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

\(a...............a.....0.5a\)

\(m_{\text{dung dịch sau phản ứng}}=23a+500-0.5a\cdot2=22a+500\left(g\right)\)

\(m_{NaOH}=40a\left(g\right)\)

\(C\%_{NaOH}=\dfrac{40a}{22a+500}\cdot100\%=20\%\)

\(\Rightarrow a=2.8\)

\(b.\)

\(n_{H^+}=10^{-3}\cdot V\cdot\left(1+0.5\cdot2\right)=2\cdot10^{-3}V\left(mol\right)\)

\(\Rightarrow n_{NaOH}=2\cdot10^{-3}V=2.8\left(mol\right)\)

\(\Rightarrow V=1400\)

Câu 5:

\(a.PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\\ \dfrac{m}{23}........................\dfrac{m}{23}.........\dfrac{m}{46}\left(mol\right)\\ Có:\dfrac{\dfrac{m}{23}.40}{500+m}.100=20\%\\ \Leftrightarrow m\approx64,972\left(g\right)\\ b.n_{NaOH}=\dfrac{64,972}{23}=2,825\left(mol\right)\\ n_{HCl}=1.0,5=0,5\left(mol\right)\\ Đặt:V=a\left(l\right)\\ n_{H^+}=1.a+0,5.2.a=2a\left(mol\right)\\ n_{NaOH}=n_{H^+}\\ \Leftrightarrow2a=2,825\\ \Leftrightarrow a=1,4125\left(l\right)=1412,5\left(ml\right)\)

1. \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2}{\sqrt{x}+3}\right)\)

\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{3\sqrt{x}+1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}.\sqrt{x}-1\)

P=\(\sqrt{x}+4\)

b)  \(P=\dfrac{x}{4}+5\)

⇔\(\sqrt{x}+4=\dfrac{x}{4}+5\)

⇔\(\dfrac{x}{4}-\sqrt{x}+1=0\)

⇔\(x-4\sqrt{x}+4=0\)

⇔\(\left(\sqrt{x}-2\right)^2=0\)

⇔\(\sqrt{x}-2=0\)

⇔\(\sqrt{x}=2\)

⇔\(x=4\) 

Vậy x=4 thì P=\(\dfrac{x}{4}+5\)

Bài 1: 

a) Ta có: \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\right)\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b) Ta có: \(P=\dfrac{x}{4}+5\)

\(\Leftrightarrow\sqrt{x}+4=\dfrac{1}{4}x+5\)

\(\Leftrightarrow\dfrac{1}{4}x-\sqrt{x}+1=0\)

\(\Leftrightarrow x-4\sqrt{x}+4=0\)

\(\Leftrightarrow x=4\)