1. my sister is talking .................................. her friend

a. to b. for c. at d. in

2. she is .................................... than her brother

a. tall b. short c. taller d. much tall

3. he lives .................. 19 Nguyen Du Street

a. in b. on c. at d. from

4. they ...................... in Ho Chi Minh City now

a. live b. living c. are living d. lives

5.............................. is it from your house to shool?

a. how b. how many c. how much d. how far

6. Nam can swim. .................... can his sister

a. So b. too c. and d. but

7. his mother is a nurse. His sister is a nurse, ...................................

a. so b. too c. like d. or

8. Our new school is ....................................... from our old school

a.larger b. smaller c. newer d. different

9.what .................................. at the moment?

a. does your father do b. does your father doing

c. is your father doing d. is your father do

10.I......................... a bicycle to school every morning

a. ride b. am riding c.riding d.rides

11. where .................................... she come from ? - From Nha trang

a. do b. is c. does d. are

12.I am from Ha Noi. - Oh, ............................

a. I am, too b.so am I c. I am so d. A&B are correct

she often helps my mum

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

Nếu \(x^2\ge0\) thì \(\left(x+1\right)^2>0\)

Ngược lại \(\left(x+1\right)^2\ge0\) thì \(x^2>0\)

=> x² + (x + 1)² > 0 \(\forall x\)

hay \(2x^2+2x+1>0\forall x\)

--> đpcm

thêm ĐKXĐ là: x > 1

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

a. \(\left(2x-3\right)\left(x+1\right)+\left(2x-3\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1+3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x-6=0\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}\)

b. \(\left(x-4\right)\left(3x-2\right)+x^2-16=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-2\right)+\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-2+x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Tên: Lê Minh Hoàng

Lớp: 1

\(x^2\left(x-2\right)-2x^2+8x-8=0\)

\(\Leftrightarrow x^3-2x^2-2x^2+8x-8=0\)

\(\Leftrightarrow x^3-4x^2+8x-8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2-2x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\\left(x-1\right)^2+3=0\left(voly\right)\end{matrix}\right.\)

Vậy pt có nghiệm là x = 2

Có N nhỏ nhất khi \(\dfrac{12}{3\left|x+5\right|+4}\) lớn nhất

\(\Rightarrow3\left|x+5\right|+4\) nhỏ nhất

Có: \(3\left|x+5\right|\ge0\forall x\Rightarrow3\left|x+5\right|+4\ge4\)

\(\Rightarrow\dfrac{12}{3\left|x+5\right|+4}\ge\dfrac{12}{4}=3\)

Dấu ''='' xảy ra \(\Leftrightarrow x=-5\)

\(\Rightarrow N_{MAX}=2+3=5\)

Vậy \(N_{MAX}=5\) khi x = -5