x=0 

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:  \(x\ne\pm1\)

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

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

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

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

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

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

Trắc nghiệm
Câu 1:B

Câu 2:C

Câu 3:A

Câu 4:C

Câu 5:A 

Câu 6:B

Câu 7:D

Câu 8:D

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

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

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

1, D

2, B

3, D

4, A

5, D 

6, B

7, A

8, thiếu dữ kiện kìa 

9, D

11C

12D

13B

14A

15 thiếu hình

16B

2. It is thought that he stole the money.

He is thought to have stolen the money.

3. It is said that she is going to study abroad.

She is said to be going to study abroad.

4. It is thought that the house was broken into last night.

The house was thought to be broken into last night.

r) \(100x^2-\left(x^2-25\right)^2\)

\(=\left(10x\right)^2-\left(x^2+25\right)^2\)

\(=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)

\(=\left(-x^2+10x-25\right)\left(x^2+10x+25\right)\)

\(=-\left(x-5\right)^2\left(x+5\right)^2\)

v) \(\left(x+y\right)^2-2\left(x+y\right)+1\)

\(=\left(x+y\right)^2-2\left(x+y\right)\cdot1+1^2\)

\(=\left(x+y-1\right)^2\)

y) \(12y-36-y^2\)

\(=-y^2+12x-36\)

\(=-\left(y^2-12x+36\right)\)

\(=-\left(y-6\right)^2\)

r: =(10x)^2-(x^2+25)^2

=(10x-x^2-25)(10x+x^2+25)

=-(x^2-10x+25)(x+5)^2

=-(x-5)^2(x+5)^2

t: =(2x-1)^2-(x+1)^2

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

=3x*(x-2)

v: =(x+y)^2-2(x+y)*1+1^2

=(x+y-1)^2

u: =(x-y+5)^2-2(x-y+5)*1+1^2

=(x-y+5-1)^2

=(x-y+4)^2

x: =-(x^2+2xy+y^2)

=-(x+y)^2

y: =-(y^2-12y+36)

=-(y-6)^2

t) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)

\(=\left(2x-1\right)^2-\left(x+1\right)^2\)

\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)

\(=3x\left(x-2\right)\)

u) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)

\(=\left(x-y+5\right)^2-2\left(x-y+5\right)\cdot1+1^2\)

\(=\left(x-y+5-1\right)^2\)

\(=\left(x-y+4\right)^2\)

x) \(-x^2-2xy-y^2\)

\(=-\left(x^2+2xy+y^2\right)\)

\(=-\left(x+y\right)^2\)