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**Solve the exercise Joint exercise page 34 (Chapter 7 Math 7 Connect)**

### Exercise 7.18 page 35 math 7 episode 2 KNTT

For monomials: $2x^6$; $-5x^3$; $-3x^5$; $x^3$; $\frac{3}{5}x^2$;$\frac{-1}{2}x^2$; $84; $-3x$. Let $A$ be the sum of the given monomers.

a) Simplify the sum $A$ and arrange the terms to get a polynomial.

b) Find the highest coefficient, the coefficient of freedom and the coefficient of $x^2$ of the obtained polynomial.

Solution guide:

a) $A = 2x^6 – 5x^3 – 3x^5 + x^3 + \frac{3}{5}x^2 – \frac{1}{2}x^2 + 8 -3x$

$= 2x^6 – 3x^5 + (-5x^3 + x^3) + (\frac{3}{5}x^2 – \frac{1}{2}x^2) – 3x + 8 $

$= 2x^6 – 3x^5 – 4x^3 + \frac{1}{10}x^2 – 3x + 8$

b)

– In $A$, the term $2x^6$ has the highest degree.

=> The highest coefficient in $A$ is: 2

– Factor of freedom: 8

– The coefficient of $x^2$ is: $\frac{1}{10}$

### Exercise 7.19 page 35 math 7 volume 2 KNTT

A water tank in the shape of a rectangular box is designed with proportional dimensions:

Height : width : length = 1 : 2 : 3.

There is 0.7 m$^3$ of water left in the tank. Let the height of the tank be x (meters).

Write a polynomial that represents the number of cubic meters of water that must be added to the tank to fill the tank. Determine the degree of that polynomial.

Solution guide:

– According to the problem, we have: The height of the tank is: $x$ (meters)

– The size of the tank according to the ratio: height: width: length = 1 : 2 : 3.

Should:

The width of the tank is: $2x$ (meters)

The length of the tank is: $3x$ (meters)

=> The polynomial expressing the volume of the pool is: $V = x . 2x . 3x = 6x^3$ (m$^3$)

* So: the polynomial expressing the number of cubic meters of water that needs to be added to the tank to fill the tank is:

$A = 6x^3 – 0.7$ (m$^3$)

### Exercise 7.20 page 35 math 7 volume 2 KNTT

In addition to the Celsius (degrees Celsius) temperature scale, many countries also use the Fahrenheit temperature scale, called degrees Fahrenheit, to measure temperature in weather forecasts. To calculate how many degrees F corresponds to $x^o$C, we use the formula:

$T(x) = 1.8x + 32$

For example, 0$^o$C corresponds to $T(0) = 32$ ($^o$F).

a) Ask $0^o$F for how many degrees Celsius?

b) The temperature on a summer day in Hanoi is $35^o$C. How many degrees Fahrenheit does that correspond to?

c) The temperature on a winter day in New York (USA) is $41^o$F. That temperature corresponds to how many degrees Celsius?

Solution guide:

a) To know how many degrees Celsius corresponds to $0^o$F, we have:

$T(x) = 0$

<=> $0 = 1.8x + 32$

=> $x \approx -17.78 $

So 0$^o$F corresponds to minus 17.78 degrees Celsius.

b) To calculate how many degrees F corresponds to $35^o$C, we substitute $x = 35$ into the expression T(x):

$T(35) = 1.8 . 35 + 32 = 95 $

So the temperature on a summer day in Hanoi is $35^o$C, then that temperature corresponds to $95$ degrees F.

c) To know how many degrees Celsius is $41^o$F, we have:

$T(x) = 41$

<=> $41 = 1.8x + 32$

=> $x = 5$

So the temperature on a winter day in New York (USA) is $41^o$F, then that temperature corresponds to $5$ degrees Celsius.

### Exercise 7.21 page 35 math 7 volume 2 knowledge skills

Given two polynomials:

$P = -5x^4 + 3x^3 + 7x^2 + x – 3$

$Q = 5x^4 – 4x^3 – x^2 + 3x + 3$

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a) Determine the degree of the polynomials $P + Q$ and $P – Q$.

b) Calculate the value of each polynomial $P + Q$ and $P – Q$ at $x= 1$; $x= -1$.

c) Which of the two polynomials $P + Q$ and $P – Q$ has a solution of $x = 0$?

Solution guide:

a)

$P + Q$

$= -5x^4 + 3x^3 + 7x^2 + x – 3 + (5x^4 – 4x^3 – x^2 + 3x + 3)$

$= -5x^4 + 3x^3 + 7x^2 + x – 3 + 5x^4 – 4x^3 – x^2 + 3x + 3$

$= (-5x^4 + 5x^4) + (3x^3 – 4x^3) + (7x^2 – x^2) + (x + 3x) + (-3 + 3)$

$= -x^3 + 6x^2 + 4x$

$P – Q$

$= -5x^4 + 3x^3 + 7x^2 + x – 3 – (5x^4 – 4x^3 – x^2 + 3x + 3)$

$= -5x^4 + 3x^3 + 7x^2 + x – 3 – 5x^4 + 4x^3 + x^2 – 3x – 3$

$= (-5x^4 – 5x^4) + (3x^3 + 4x^3) + (7x^2 + x^2) + (x – 3x) + (-3 – 3)$

$= -10x^4 + 7x^3 + 8x^2 – 2x – 6$

b)

Substituting $x = 1$ into the polynomial $P + Q$, we get:

$P + Q = -1^3 + 6.1^2 + 4.1 = 9$

Substituting $x = -1$ into the polynomial $P + Q$, we get:

$P + Q = -(-1)^3 + 6.(-1)^2 + 4.(-1) = 3$

Substituting $x = 1$ into the polynomial $P – Q$, we get:

$P – Q = -10 . 1^4 + 7.1^3 + 8.1^2 – 2.1 – 6 = -3$

Substituting $x = -1$ into the polynomial $P – Q$, we get:

$P – Q = -10 . (-1)^4 + 7.(-1)^3 + 8.(-1)^2 – 2.(-1) – 6 = -13$

c) We see:

The expression $P + Q$ has a coefficient of freedom of 0

=> Substituting $x = 0$ into the polynomial $P + Q$, we get: $P + Q = 0$

The expression $P + Q$ has a coefficient of freedom of -6

=> Substituting $x = 0$ into the polynomial $P – Q$, we get: $P – Q = -6$

* So: Polynomial $P + Q$ has a solution of $x = 0$.

### Exercise 7.22 page 35 math 7 episode 2 KNTT

A bus goes from Hanoi to Yen Bai (on the Hanoi – Lao Cai highway) with a speed of 60 km/h. After 25 minutes, a tourist car also goes from Hanoi to Yen Bai (on the same road as the bus) with a speed of 85 km/h. Both cars did not stop along the way.

a) Let $D(x) $ be the polynomial representing the distance traveled by the bus and $K(x) $ be the polynomial representing the distance traveled by the bus from the start to the time the bus stops. go x hours. Find $D(x)$ and $K(x) $.

b) Show that the polynomial $f(x) = K(x) – D(x) $ has a solution of $x = 1$. Explain the meaning of the solution $x = 1$ of the polynomial $f(x) $.

Solution guide:

a)

25 minutes = $\frac{5}{12}$ hours

According to the topic, we have:

$D(x) = 85x$

$K(x) = 60.\frac{5}{12} + 60x = 60x + 25$

b) $f(x) = K(x) – D(x)$

$= 60x + 25 – 85x $

$= 25 – 25x$

Substituting $x = 1$ into $f(x)$, we get:

$f(1) = 25 – 25.0 = 0$

So: Polynomial $f(x) = K(x) – D(x)$ has a solution of $x = 1$

* Meaning: When two cars have traveled for 1 hour, they will meet.