a: Đặt P(x)=0

=>12-5x=0

=>5x=12

=>x=2,5

b: Đặt Q(y)=0

=>4y-3-5y=0

=>-y-3=0

=>y=-3

c: Đặt E(x)=0

=>\(4x^2-4=0\)

=>\(x^2=1\)

=>\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

d: Đặt H(x)=0

=>\(x^2+9=0\)

mà \(x^2+9>=9>0\forall x\)

nên \(x\in\varnothing\)

3 A lot of people go to Cusco in Peru to attend the Festival of the Sun

4 Children eat moon cakes and sing tradidional songs at Mid-Autumn Festival

5 I saw a parade of flower floats when I was attending the festival last year

6 Critics said that the film would be fantastic and gripping despite several small faults

7 I couldn't enjoy the film the last time because two people in front of me didn't stop using phones

8 Mai can't go to the cinema at present because she is studying for the exam

9 There will be less air pollution if we stop using fossil fuel

10 Last year they placed solar panels on the roofs of houses to produce hot water

a: \(A\left(x\right)=4x^2+4x+1\)

bậc là 2

Hạng tử tự do là 1

Hạng tử cao nhất là 4x²

b: A(x)+B(x)=5x²+5x+1

=>\(B\left(x\right)=5x^2+5x+1-A\left(x\right)\)

=>\(B\left(x\right)=5x^2+5x+1-4x^2-4x-1=x^2+x\)

c: \(\dfrac{A\left(x\right)}{2x+1}=\dfrac{4x^2+4x+1}{2x+1}=\dfrac{\left(2x+1\right)^2}{2x+1}=2x+1\)

a: Xét ΔEBC vuông tại E và ΔDCB vuông tại D có

BC chung

\(\widehat{EBC}=\widehat{DCB}\)

Do đó: ΔEBC=ΔDCB

b: Ta có: ΔEBC=ΔDCB

=>\(\widehat{ECB}=\widehat{DBC}\)

=>\(\widehat{HBC}=\widehat{HCB}\)

=>ΔHBC cân tại H

Xét ΔBHC có HB+HC>BC

=>BC<2BH

=>\(BH>\dfrac{BC}{2}\)

1.i think i can build a new road to reduce the traffic jam in the future

2. It will look like a ball with a golden shoe on it

3. It will cost one billion vietnam dong

4. it will use water 

5, when there is a traffic jam it will be a robot police to reduce the traffic jam

6. 400 kilometers per hour

7. It will not make the smoke and noise

5.

a: \(A\left(x\right)=2x^3-6x^2-5\left(x^2-2x-5\right)\)

\(=2x^3-6x^2-5x^2+10x+25\)

\(=2x^3-11x^2+10x+25\)

\(B\left(x\right)=x^3-3\left(x^3-2x^2-5x\right)\)

\(=x^3-3x^3+6x^2+15x\)

\(=-2x^3+6x^2+15x\)

b: \(A\left(x\right)=2x^3-11x^2+10x+25\)

Bậc là 2

Hệ số cao nhất là 2

Hệ số tự do là 25

c: A(x)-C(x)=B(x)

=>C(x)=A(x)-B(x)

\(=2x^3-11x^2+10x+25+2x^3-6x^2-15x\)

\(=4x^3-17x^2-5x+25\)

d: Đặt P(x)=0

=>B(x)+2x³=0

=>\(-2x^3+6x^2+15x+2x^3=0\)

=>\(6x^2+15x=0\)

=>3x(2x+5)=0

=>x(2x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a: \(A\left(x\right)=2x^4+4x^3-3x^2-4x+1\)

bậc là 4

Hạng tử tự do là 1

Hạng tử cao nhất là \(2x^4\)

b: \(A\left(x\right)+B\left(x\right)=2x^3-x^2+5\)

=>\(B\left(x\right)=2x^3-x^2+5-A\left(x\right)\)

\(=2x^3-x^2+5-2x^4-4x^3+3x^2+4x-1\)

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