5a.

Pt có 2 nghiệm pb lhi:

\(\Delta=9+4m>0\Leftrightarrow m>-\dfrac{9}{4}\)

b. Phương trình có 2 nghiệm khi:

\(\Delta=1+4\left(-2m+1\right)\ge0\Rightarrow m\le\dfrac{5}{8}\)

6.

a. Pt có 2 nghiệm khi: 

\(\Delta'=1-\left(m+2\right)\ge0\Leftrightarrow m\le-1\)

6b

Khi \(m\le-1\), theo hệ thức Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m+2\end{matrix}\right.\)

\(x^2_1+x^2_2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)

\(\Leftrightarrow4-2\left(m+2\right)=10\)

\(\Leftrightarrow m=-5\)

B.

\(x^2_1+x_2^2+4x_1x_2=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+2x_1x_2=0\)

\(\Leftrightarrow4+2\left(m+2\right)=0\)

\(\Leftrightarrow m=-4\)

C.

Để biểu thức đề bài xác định \(\Rightarrow x_1x_2\ne0\Rightarrow m\ne-2\)

Khi đó:

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-3\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=-3\)

\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=-3x_1x_2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow4+m+2=0\)

\(\Leftrightarrow m=-6\)

D.

\(x_1^3+x_2^3=15\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=15\)

\(\Leftrightarrow8-6\left(m+2\right)=15\)

\(\Leftrightarrow m=-\dfrac{19}{6}\)

Bài 5:

Bài 5

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl --> 2AlCl₃ + 3H₂O

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,5}{6}\) => Al dư, HCl hết

PTHH: Al₂O₃ + 6HCl --> 2AlCl₃ + 3H₂O

            \(\dfrac{1}{12}\)<----0,5------->\(\dfrac{1}{6}\)----->0,25

=> \(\left\{{}\begin{matrix}m_{Al\left(dư\right)}=10,2-\dfrac{1}{12}.102=1,7\left(g\right)\\m_{AlCl_3}=\dfrac{1}{6}.133,5=22,25\left(g\right)\\m_{H_2}=0,25.18=4,5\left(g\right)\end{matrix}\right.\)

Bài 6

a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

           0,1<-------------0,1<----0,1

=> \(n_{Mg\left(pư\right)}=0,1\left(mol\right)< 0,2\)

=> Mg dư => HCl hết

b) \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

\(m_{Mg\left(dư\right)}=\left(0,2-0,1\right).24=2,4\left(g\right)\)

h tới mai mà k cs ng lm thì t lm hết h đi ngủ :>

câu 6 thể tích 1kg khí ở 0 độ 1,29 m³ 

áp dụng đẳng quá trình \(\dfrac{P_0V_0}{T_0}=\dfrac{PV}{T}\Leftrightarrow\dfrac{1,01.10^5.1,29}{0+273}=\dfrac{3.10^5.V}{70+273}\Rightarrow V\approx0,5456\left(m^3\right)\)

khối lượng riêng 0,5456kg/m³

Bài 5: 

a. \(\sqrt{x^2+2x+1}=\sqrt{9x^2}\)

<=> \(\sqrt{\left(x+1\right)^2}=\sqrt{\left(3x\right)^2}\)

<=> \(\left|x+1\right|=\left|3x\right|\)

<=> \(\left[{}\begin{matrix}x+1=3x\\x+1=-3x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0,5\\x=-0,25\end{matrix}\right.\)

b. \(\sqrt{x^2-\dfrac{2}{5}x+\dfrac{1}{25}}=\sqrt{\left(2x+1\right)^2}\)

<=> \(\sqrt{\left(x-\dfrac{1}{5}\right)^2}=\sqrt{\left(2x+1\right)^2}\)

<=> \(\left|x-\dfrac{1}{5}\right|=\left|2x+1\right|\)

<=> \(\left[{}\begin{matrix}x-\dfrac{1}{5}=2x+1\\x-\dfrac{1}{5}=-2x-1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1,2\\x=-\dfrac{4}{15}\end{matrix}\right.\)

c. \(\sqrt{25x^2}=\sqrt{x^4}\)

<=> \(\sqrt{\left(5x\right)^2}=\sqrt{\left(x^2\right)^2}\)

<=> \(\left|5x\right|=\left|x^2\right|\)

<=> \(\left[{}\begin{matrix}5x=x^2\\5x=-x^2\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}5x-x^2=0\\5x+x^2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x\left(5-x\right)=0\\x\left(5+x\right)=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\5+x=0\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

Chào em, em tham khảo nhé!

Passage 5:

1. lie

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3. across

4. established

5. for

6. estimated

Passage 6:

1. what

2. smoking

3. listening

4. sign

5. while

6. quit

7. laying

8. would

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