Task 6. Dùng các hình thức so sánh với tính từ trong ngoặc
1. Mary is 10 years old. Julie is 8 years old. Mary is (old) ................... .........Julie.
2. The Alps are very high. They are (high)..........................mountains in Europe.
3. An ocean is (large)................................ a sea.
4. A Rolls Royce costs a lot of money. A Twingo costs less money. A Rolls Royce is (expensive)...............................a Twingo.
5. John's results were good. Fred's results were very poor. Fred's results were (bad)................................John's.
6. This exercise is not difficult. It's (easy)...............................I expected. 7. The weather is not good today - it's raining. I hope the weather will be (good)........................next week.
8. People are not friendly in big cities. They are usually (friendly)......................in small towns.
9. In the government of a country, the President is (important).....................person.
10. People say that Chinese is (difficult)...................to learn than English.
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1. If you read many books, you will have much knowledge. ⇒ The more books you read, the more knowledge you have.
2. He speaks too much and people feel bored. ⇒ The more speaking he does, the more bored people feel.
3. The growth in the economy makes people’s living condition better. ⇒ The more growth the economy makes, the better the condition is.
4. He learned a lot of things as he traveled far. ⇒ The farther he traveled, the more things he learned.

Để rút gọn biểu thức, ta sẽ thực hiện các phép tính và kết hợp các thành phần tương tự: P(2x-1).4x^2 + 2x + 1 + (x+1)x^2 - x + 1 = P(8x^3 - 4x^2) + 2x + 1 + x^3 + x^2 - x + 1 = P(8x^3) - P(4x^2) + x^3 + (2x-x) +(1+1) = **8Px^3 - 4Px^2**+ x^3 **+ x**+ **2** Vậy biểu thức đã được rút gọn thành: **8Px³ - 4Px²+x³+x+2**

a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)
\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)
\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
b) \(27x^3-54x^2+36x=9\)
\(\Rightarrow27x^3-54x^2+36x-9=0\)
\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)
\(\Rightarrow\left(3x-2\right)^3-1=0\)
\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)
mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)
\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)
(\(x-5\))2 = (3 +2\(x\))2 ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}
27\(x^3\) - 54\(x^2\) + 36\(x\) = 9
27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1
(3\(x\) - 2)3 = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1

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

\(1,\left(x-3\right)^2+3-x=0\)
\(\Leftrightarrow x^2-6x+9+3-x=0\)
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow x^2-3x-4x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy.........
\(\left(x-3\right)^2+3-x=0\\ \Leftrightarrow x^2-6x+9+3-x=0\\ \Leftrightarrow x^2-7x+12=0\\ \Leftrightarrow x^2-3x-4x+12=0\\ \Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{3;4\right\}\)

B1:
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cout << "Nhap so n: ";
cin >> n;
cout << int(sqrt(n));
}
B2:
#include <bits/stdc++.h>
using namespace std;
int main() {
float a,b;
cout << "Nhap vao 2 so phan biet: ";
cin >> a >> b;
if (a>b) cout << a;
else cout << b;
}

\(\left\{{}\begin{matrix}3x-6y+2z=-4\\3x-y-3z=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x-6y+2z=-4\\3x-y-3z=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x-6y=-4-2z\\3x-y=1+3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5y=1+3z+4+2z\\3x-y=1+3z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5y=5+5z\\3x=y+1+3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=1+z\\3x=1+z+1+3z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=1+z\\x=\dfrac{4z+6}{3}\end{matrix}\right.\)
\(S=9x^2-8\left(y^2+z^2\right)\)
\(S=9\left(\dfrac{4z+2}{3}\right)^2-8\left[\left(1+z\right)^2+z^2\right]\)
\(S=9.\dfrac{16z^2+16z+4}{9}-8\left[1+2z+z^2+z^2\right]\)
\(S=16z^2+16z+4-8-16z-16z^2\)
\(S=-4\)
Đính chính \(x=\dfrac{4z+2}{3}\) không phải \(x=\dfrac{4z+6}{3}\)

Bài 9A:
\(a,\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\\ \Leftrightarrow\left(x^2+10x+25\right)-\left(x^2-10x+25\right)-2x+1=0\\ \Leftrightarrow x^2-x^2+10x+10x-2x=-1-25+25\\ \Leftrightarrow18x=-1\\ \Leftrightarrow x=-\dfrac{1}{18}\\ b,\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\\ \Leftrightarrow4x^2-28x+49-x^2-6x-9-3x^2-6=0\\ \Leftrightarrow4x^2-x^2-3x^2-28x-6x=6+9-49\\ \Leftrightarrow22x=-34\\ \Leftrightarrow x=-\dfrac{17}{11}\\ c,\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\\ \Leftrightarrow9x^2+12x+4-9\left(x^2-25\right)=225-5x\\ \Leftrightarrow9x^2-9x^2+12x+5x=225-4+9.25\\ \Leftrightarrow17x=446\\ \Leftrightarrow x=\dfrac{446}{17}\)
Sao bài này câu nào x cũng k nguyên ta, hơi xấu hi
9B
\(a,\left(4x-1\right)^2-4\left(2x-3\right)^2-x-4=0\\ \Leftrightarrow16x^2-8x+1-4\left(4x^2-12x+9\right)-x-4=0\\ \Leftrightarrow16x^2-16x^2-8x+48x-x=4+36-1\\ \Leftrightarrow39x=39\\ \Leftrightarrow x=1\\ b,x\left(x-5\right)-\left(4-x\right)^2=7x+1\\ \Leftrightarrow x^2-5x-\left(16-8x+x^2\right)-7x-1=0\\ \Leftrightarrow x^2-x^2-5x+8x-7x=1+16\\ \Leftrightarrow-4x=17\\ \Leftrightarrow x=\dfrac{-17}{4}\\ c,\left(2x-6\right)\left(x+3\right)=2\left(x-3\right)^2\\ \Leftrightarrow2x^2-6x+6x-18=2\left(x^2-6x+9\right)\\ \Leftrightarrow2x^2-2x^2-6x+6x+12x=18+18\\ \Leftrightarrow12x=36\\ \Leftrightarrow x=\dfrac{36}{12}=3\)
1. Mary is older than Julie.
2. The Alps are higher than any other mountains in Europe.
3. An ocean is larger than a sea.
4. A Rolls Royce is more expensive than a Twingo.
5. Fred's results were worse than John's.
6. This exercise is easier than I expected.
7. I hope the weather will be better next week.
8. People are usually friendlier in small towns than in big cities.
9. In the government of a country, the President is the most important person.
10. People say that Chinese is more difficult to learn than English.
(compare more).