a: BC=căn 6^2+8^2=10cm

AM=BC/2=5cm

b:

ΔAEH vuông tại A có AI là trung tuyến

nên IH=IA

=>góc IHA=góc IAH

góc IAH+góc MAB

=góc MBA+góc IHA=90 độ

=>góc IAM=90 độ

=>AI vuông góc AM

III

1 catches => caught

2 is => was

3 enough experienced= > experienced

4 annoyed => annoying

5 American character => American's character

IV

1 A

2 C

3 C

4 A

5 D

2: 

a: 

b: PTHĐGĐ là:

-x^2=x-2

=>x^2=-x+2

=>x^2+x-2=0

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

=>y=-4 hoặc y=-1

1:

a: 

b: PTHĐGĐ là;

x^2+x-2=0

=>(x+2)(x-1)=0

=>x=1 hoặc x=-2

=>y=1 hoặc y=4

x=0 

\(1,\Rightarrow3:\dfrac{9}{4}=\dfrac{3}{4}:x\\ \Rightarrow\dfrac{4}{3}=\dfrac{3}{4}:x\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{4}{3}=\dfrac{9}{16}\\ 2,\)

Nửa chu vi là \(50:2=25\left(cm\right)\)

Gọi cd là a, cr là b (cm)(a,b>0)

Ta có \(a:b=3:2\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\) và \(a+b=25\left(cm\right)\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{a+b}{3+2}=\dfrac{25}{5}=5\\ \Rightarrow\left\{{}\begin{matrix}a=15\\b=10\end{matrix}\right.\)

Vậy ...

B2:chiều dài là 15

Chiều rộng là 10

Từ hình vẽ ta thấy : 

\(x< -1\)

\(\Leftrightarrow x+4< 4+\left(-1\right)\)

\(\Leftrightarrow x+4< 3\)

\(\Rightarrow C\)

1. ĐKXĐ: \(x\ne5\)

\(\dfrac{3x+1}{x-5}+\dfrac{-2x-6}{x-5}\)

\(=\dfrac{3x+1-2x-6}{x-5}\)

\(=\dfrac{x-5}{x-5}\)

\(=1\)

2. ĐKXĐ:  \(x\ne\pm1\)

\(\dfrac{x+1}{2x-2}-\dfrac{x^2+3}{2x^2-2}\)

\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x^2-1\right)}\)

\(=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)

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

\(=\dfrac{1}{x+1}\)