Bài 11: 

a: x/5=2/10

=>x/5=1/5

hay x=1

b: x/-3=-5/15

=>x/3=1/3

hay x=1

Bài 6:

b: PTHĐGĐ là:

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-7\\y=-2\end{matrix}\right.\)

Câu 4: 

Số đo các góc còn lại là \(47^0;133^0;133^0\)

Bài 3: 

a: Ta có: \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(=6x^2+9x+14x+21-6x^2-33x+10x+55\)

=76

b: Ta có: \(B=\left(x-3\right)\left(x+2\right)-\left(x-5\right)\left(x+4\right)\)

\(=x^2+2x-3x-6-x^2-4x+5x+20\)

=14

Bài 1:

a) \(\left(x+2\right)\left(x-3\right)=x^2-x-6\)

b) \(\left(x^2-1\right)\left(x+2\right)=x^3+2x^2-x-2\)

c) \(\left(x^2+2x\right)\left(x^2-1\right)=x^4+2x^3-x^2-2x\)

d) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)

e) \(\left(x^2-2x+3\right)\left(x-4\right)=x^3-6x^2+11x-12\)

3 uncomfortable

4 decision

5 solution

3. uncomfortable

4. decision

5. solution

Chúc bạn học tốt

uncomfortable

decision

solution

\(7,x^4+x^3+x^2-1=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)=\left(x^3+x-1\right)\left(x+1\right)\)

\(8,x^2y^2+1-x^2-y^2=\left(x^2y^2-y^2\right)-\left(x^2-1\right)\\ =y^2\left(x^2-1\right)-\left(x^2-1\right)=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)

\(10,x^4-x^2+2x-1=x^4-\left(x-1\right)^2=\left(x^2-x+1\right)\left(x^2+x-1\right)\\ 11,3a-3b+a^2-2ab+b^2=3\left(a-b\right)+\left(a-b\right)^2=\left(3+a-b\right)\left(a-b\right)\\ 12,a^2+2ab+b^2-2a-2b+1=\left(a+b\right)^2-2\left(a+b\right)+1=\left(a+b-1\right)^2\\ 13,a^2-b^2-4a+4b=\left(a-b\right)\left(a+b\right)-4\left(a-b\right)=\left(a+b-4\right)\left(a-b\right)\\ 14,a^3-b^3-3a+3b=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 15,x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+4x+1\right)\)

1)
=0,25y.(64x³+z³)
2)
=x²(x²-4x+4)
=x²(x-2)²
5)
=x²(x+1)-4(x+1)
=(x²-4)(x+1)
=(x-2)(x+2)(x+1)
6)
=x²(x-1)-(x-1)
=(x²-1)(x-1)
=(x-1)(x+1)(x-1)
=(x-1)²(x+1)
 

1 B

2 D

3 A

4 A

5 D

6 C

7 A

8 C

9 C

10 C

11 C

12 B

13 A

7A

9C

10C

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