4.2:

a: x^2-x+1=x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4>0 với mọi x

=>x^2-x+1 ko có nghiệm

b: 3x-x^2-4

=-(x^2-3x+4)

=-(x^2-3x+9/4+7/4)

=-(x-3/2)^2-7/4<=-7/4<0 với mọi x

=>3x-x^2-4 ko có nghiệm

5:

a: x^2+y^2=25

x^2-y^2=7

=>x^2=(25+7)/2=16 và y^2=16-7=9

x^4+y^4=(x^2)^2+(y^2)^2

=16^2+9^2

=256+81

=337

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

=1^2-2*(-6)

=1+12=13

x^3+y^3=(x+y)^3-3xy(x+y)

=1^3-3*1*(-6)

=1+18=19

1 interested in travelling by plane

2 15 minutes riding his bike to school

3 is made by my mother everymorning

4 take care of her little brother

5 able to speak 2 languages when he was young

6 able to play the piano

7 is played all over VN

8 to play voleyball very well when he was young

9 to go fishing when he was young

10 way to the shopping mall

over

camping

was

couldn't

percussion

takes

look after

go around 

go through

wind

how long

for

do

Bài cuối mình không thấy rõ đề nhưng mình đoán là thế này bạn nhé.

a.

$ax-ay+bx-by=(ax-ay)+(bx-by)=a(x-y)+b(x-y)$

$=(a+b)(x-y)$

b. Trùng phần a

c. 

$x^2+xy+ax+ay=(x^2+xy)+(ax+ay)=x(x+y)+a(x+y)$

$=(x+a)(x+y)$

d.

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

$=xy(x+y)-(x+y)=(xy-1)(x+y)$

e.

$x^2y-xy^2-5x+5y=(x^2y-xy^2)-(5x-5y)$

$=xy(x-y)-5(x-y)=(x-y)(xy-5)$

f. Biểu thức không phân tích được thành nhân tử

g.

$x^2y-4x+xy^2-4y$

$=(x^2y+xy^2)-(4x+4y)=xy(x+y)-4(x+y)=(x+y)(xy-4)$
h.

$x^4+x^3+x+1=(x^4+x^3)+(x+1)$
$=x^3(x+1)+(x+1)=(x^3+1)(x+1)=(x+1)(x^2-x+1)(x+1)$

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

Bài 5:

a: \(=\dfrac{a+2\sqrt{a}+a-2\sqrt{a}}{a-4}\cdot\dfrac{a-4}{2\sqrt{a}}=\dfrac{2a}{2\sqrt{a}}=\sqrt{a}\)

b: Để A-2>0 thì căn a-2>0

=>căn a>2

=>a>4

c: Để 4/A+1 là số nguyên thì \(\sqrt{a}+1\inƯ\left(4\right)\)

=>\(\sqrt{a}+1\in\left\{1;2;4\right\}\)

=>\(a\in\left\{1;9\right\}\)

1 does this red dress cost

2 is this shirt

does this short cost

3 do these shoes cost

4 has brown backpack

5 how to get to the bus stop

6 to stay in bed

7 are cleaned by Nga everyday

8 is played with a bow

9 worked for that company for 10 years

10 15 minutes walking to school

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

\(ĐK:x\le\dfrac{3}{2}\)

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

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

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)