A

C

C

B

B

a: \(=\dfrac{1}{3}-\dfrac{17}{6}+\dfrac{4}{3}=\dfrac{5}{3}-\dfrac{17}{6}=\dfrac{10-17}{6}=\dfrac{-7}{7}\)

b: \(=\dfrac{5+6}{12}=\dfrac{11}{12}\)

c: \(=\dfrac{-12+7}{28}\cdot\dfrac{28}{15}=\dfrac{-5}{15}=\dfrac{-1}{3}\)

d: \(=\dfrac{2}{3}+\dfrac{1}{5}-\dfrac{4}{15}=\dfrac{10+3-4}{15}=\dfrac{9}{15}=\dfrac{3}{5}\)

e: \(=\dfrac{-3}{16}\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{5}{16}=\dfrac{-3-5}{16}=\dfrac{-1}{2}\)

f: \(=\dfrac{-20}{23}-\dfrac{2}{23}+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{10+6+7}{15}=\dfrac{-15+23}{15}=\dfrac{8}{15}\)

g: =5/7(5/11+2/11-14/11)

=-7/11*5/7=-5/11

h: =-5/7(10/13+3/13)+1+5/7

=-5/7+1+5/7

=1

i: \(=\dfrac{7}{4}\left(\dfrac{29}{5}-\dfrac{9}{5}\right)+3+\dfrac{2}{13}=7+3+\dfrac{2}{13}=10+\dfrac{2}{13}=\dfrac{132}{13}\)

P6:

1. big

2. been

3. bought

4. people

P7:

1. T

2. F

3. F

4. F

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:

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

ĐK: \(\left\{{}\begin{matrix}x\ne-y\\y\ge\dfrac{3}{2}\end{matrix}\right.\).

\(\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}=1\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-1=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-\dfrac{x+y}{x+y}=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y+3-x-y=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y+3=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\left(2y-3\right)=0\\2x-\sqrt{2y-3}=0\end{matrix}\right..\)

Đặt a = x, b = \(\sqrt{2y-3}\).

Hệ phương trình trở thành: \(\left\{{}\begin{matrix}a-b^2=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\2b^2-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\b\left(2b-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}a=0\\a=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y-3=\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y=\dfrac{13}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\y=\dfrac{13}{8}\end{matrix}\right.\end{matrix}\right..\)

Vậy hệ phương trình có nghiệm (x;y) \(\in\) \(\left\{\left(0;\dfrac{3}{2}\right),\left(\dfrac{1}{4};\dfrac{13}{8}\right)\right\}\).

a, \(2\sqrt{3}-\sqrt{4+x^2}=0\Leftrightarrow\sqrt{4+x^2}=2\sqrt{3}\)

\(\Leftrightarrow x^2+4=12\Leftrightarrow x^2=8\Leftrightarrow x=\pm2\sqrt{2}\)

b, \(\sqrt{16x+16}-\sqrt{9x+9}=0\)ĐK : x >= -1 

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=0\Leftrightarrow\sqrt{x+1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

c, \(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow2\left|x+2\right|=8\Leftrightarrow\left|x+2\right|=4\)

TH1 : \(x+2=4\Leftrightarrow x=2\)

TH2 : \(x+2=-4\Leftrightarrow x=-6\)

c: Ta có: \(\sqrt{4\left(x+2\right)^2}=8\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)