2.

\(R=d\left(I;\Delta\right)=\dfrac{\left|3.2+4.\left(-1\right)-27\right|}{\sqrt{3^2+4^2}}=5\)

Phương trình: \(\left(x-2\right)^2+\left(y+1\right)^2=25\)

3.

- Với \(m=1\) pt trở thành: \(2=0\) (vô nghiệm) \(\Rightarrow\) thỏa mãn

- Với \(m\ne1\) pt đã cho vô nghiệm khi:

\(\Delta'=\left(m-1\right)^2-2m\left(m-1\right)< 0\)

\(\Leftrightarrow\left(m-1\right)\left(-m-1\right)< 0\Rightarrow\left[{}\begin{matrix}m< -1\\m>1\end{matrix}\right.\)

Vậy pt vô nghiệm khi: \(\left[{}\begin{matrix}m< -1\\m\ge1\end{matrix}\right.\)

4. Đặt \(AB=x>0\)

\(tan35^0=\dfrac{BC}{BD}=\dfrac{BC}{x+10}\)

\(tan40^0=\dfrac{BC}{AB}=\dfrac{BC}{x}\)

\(\Rightarrow\dfrac{tan35^0}{tan40^0}=\dfrac{x}{x+10}\Leftrightarrow x.tan35^0+10tan35^0=x.tan40^0\)

\(\Rightarrow x=\dfrac{10.tan35^0}{tan40^0-tan35^0}\Rightarrow BC=x.tan40^0=\dfrac{10.tan35^0.tan40^0}{tan40^0-tan35^0}\approx42,3\left(m\right)\)

Bài 5 :

n Cu = a(mol) ; n Al = b(mol) ; n Mg = c(mol)

=> 64a + 27b + 24c = 11,5(1)

2Al + 6HCl $\to$ 2AlCl3 + 3H2

Mg + 2HCl $\to$ MgCl2 + H2

Theo PTHH :

n H2 = 1,5b + c = 5,6/22,4 = 0,25(2)

Bảo toàn electron :

2n SO2 = 2n Cu 

2.0,1 = 2a (3)

Từ (1)(2)(3) suy ra a = 0,1 ; b = 0,1 ; c = 0,1

Vậy : 

m Cu = 0,1.64 = 6,4 gam

m Al = 0,1.27 = 2,7 gam

m Mg = 0,1.24 = 2,4 gam

\(1,\frac{-x}{15}=\frac{10}{25}\)

\(\Leftrightarrow\frac{-x}{15}=\frac{2}{5}\Leftrightarrow\left(-x\right).5=2.15\)

\(\Leftrightarrow\left(-x\right).5=30\)

\(\Leftrightarrow-x=6\Rightarrow x=-6\)

\(2,\frac{x+1}{14}=\frac{10}{35}\)

\(\Leftrightarrow\frac{x+1}{14}=\frac{2}{7}\)

\(\Leftrightarrow\left(x+1\right).7=2.14\)

\(\Leftrightarrow7x+7=28\)

\(\Leftrightarrow7x=21\)

\(\Rightarrow x=3\)

\(3,\frac{\left|x+2\right|}{26}=\frac{12}{39}\)

\(\Leftrightarrow\frac{\left|x+2\right|}{26}=\frac{4}{13}\)

\(\Leftrightarrow\left|x+2\right|.13=4.26\)

\(\Leftrightarrow\left|x+2\right|.13=104\)

\(\Leftrightarrow\left|x+2\right|=8\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=8\\x+2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)

Vậy x = 6 hoặc x = -10

\(4,x-\left(1\frac{1}{3}-\frac{2}{3}\right)=2\)

\(\Leftrightarrow x-\left(\frac{4}{3}-\frac{2}{3}\right)=2\)

\(\Leftrightarrow x-\frac{2}{3}=2\)

\(\Leftrightarrow x=2+\frac{2}{3}\)

\(\Leftrightarrow x=\frac{8}{3}\)

\(5,\frac{2}{5}:x-\frac{1}{5}=\frac{3}{5}\)

\(\Leftrightarrow\frac{2}{5}:x=\frac{3}{5}+\frac{1}{5}\)

\(\Leftrightarrow\frac{2}{5}:x=\frac{4}{5}\)

\(\Leftrightarrow x=\frac{2}{5}:\frac{4}{5}\)

\(\Rightarrow x=\frac{1}{2}\)

\(6,\frac{1}{2}.\left(3x-\frac{2}{5}\right)=\left(-3\right)^2\)

\(\Leftrightarrow\frac{1}{2}.\left(3x-\frac{2}{5}\right)=9\)

\(\Leftrightarrow3x-\frac{2}{5}=9:\frac{1}{2}\)

\(\Leftrightarrow3x-\frac{2}{5}=18\)

\(\Leftrightarrow3x=\frac{92}{5}\)

\(\Leftrightarrow x=\frac{92}{5}:3\)

\(\Rightarrow x=\frac{92}{15}\)

\(\frac{-x}{15}=\frac{10}{25}\)

\(\Leftrightarrow25\times\left(-x\right)=10\times15\)

\(-25x=150\)

\(x=150\div\left(-25\right)\)

\(x=-6\)

Câu 01:

A.

chemistry

B.

church

C.

choose

D.

cheerful

Câu 02:

A.

finished

B.

provided

C.

watched

D.

cooked
Choose the word that has a different stress pattern from the others.

Câu 03:

A.

changing

B.

arranging

C.

boring

D.

interesting

Câu 04:

A.

difficult

B.

beautiful

C.

volunteer

D.

interesting

oxit bazơ

Đáp án : D

Make somebody/something + adj = khiến cho cái gì trở nên thế nào. More + adj/adv là dạng so sánh hơn. Make her mother much more beautiful = khiến mẹ cô ấy xinh đẹp hơn nhiều

Đáp án : D

Câu điều kiện loai 3 dạng đảo, giả định một điều trái với quá khứ: Had + S + past participle, S + would have + past participle