Cường có số thời gian rảnh rỗi là:   \(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{12}-\frac{1}{8}-\frac{1}{24}=\frac{1}{4}\)

các cậu diễn giải ra đc k

25/9-2=25/9-18/9=7/9

4-5/7=28/7-2/7=26/7

189/45-2=189/45-90/45=99/45

6-1515/1818=6-15/18=6-5/6=36/6-5/6=31/6

1 A

2 A

3 A

4 D

9. A

10. B

11. A

12. D

1. A

2. A

3. A

4. D

a) Thay m=3 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}x+3y=3\\3x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+9y=9\\3x+4y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=3\\x+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{5}\\x=3-3y=3-3\cdot\dfrac{3}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy: Khi m=3 thì hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{6}{5};\dfrac{3}{5}\right)\)

a) Thay m=3 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}x+3y=3\\3x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+9y=9\\3x+4y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=3\\x+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{5}\\x=3-3\cdot\dfrac{3}{5}=\dfrac{15}{5}-\dfrac{9}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy: \(\left(x,y\right)=\left(\dfrac{6}{5};\dfrac{3}{5}\right)\)

2. ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

\(P=\dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1}{2}\Rightarrow2\sqrt{x}-2=\sqrt{x}+2\Rightarrow\sqrt{x}=4\Rightarrow x=16\)

b) Ta có: \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

Ta có: \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\Rightarrow1-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{1}{2}\)

\(\Rightarrow P_{min}=-\dfrac{1}{2}\) khi \(x=0\)

9C

10A

11B

12B

1. B /ð/ ; còn lại /θ/

2. D /t/; còn lại /d/

3. B /ʌ/; còn lại /ʊ/

4. B /z/; còn lại /s/

B= 1.99+2.98+2.97+...98.2+99.1

=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)

=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99

=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+97.98+98.99)

=99.(1+2+3+...+98+99)-(1.2+2.3+3.4+...+97.98+98.99)

=99.4950-(1.2+2.3+3.4+...+97.98+98.99)

=490050-(1.2+2.3+3.4+...+97.98+98.99)

Đặt C=1.2+2.3+3.4+...+97.98+98.99

=> 3C=1.2.3+2.3.3+3.4.3+...+97.98.3+98.99.3

=1.2.3+2.3.(4-1)+...+98.99.(100-97)

=1.2.3+2.3.4-1.2.3+...+98.99.100-97.98.99

=98.99.100

=> A=(98.99.100):3=323400

Vậy B=490050-323400=166650

=1.99+2.(99-1)+3.(99-2)+4.(99-3)+......+99.(99-98)

=99.(1+2+3+.......+99)-(2+2.3+3.4+........+98.99)

=99.(1+99).99:2-98.99.100:3

=99.50.99-98.33.100

=490050-323400=166650