pạn ơi pạn đã lm đk chưa? nếu lm đk oy cho mk xem cách lm bài 2 nhé. cảm ơn pạn nhìu lắm

Câu 1:

a/ (-5x³)(2x²+3x-5)

=-10x⁵-15x⁴+25x³

b/(2x-1)x

=2x²-x

c/(x-y)(3x²+4xy)

=3x³+4x²y-3x²y-4xy²

=3x³ +x²y-4xy²

Câu 2:

a/ x³-2x²+x

=x(x²-2x+1)

=x(x-1)²

b/x²-x-12

=x² +3x-4x-12

=(x² +3x)+(-4x-12)

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

=(x+3)(x-4)

c/ 2x-6

=2(x-3)

e/ x²+4x+4-y²

=(x²+4x+4)-y²

=(x+2)²-y²

=(x+2-y)(x+2+y)

d/ x²-2xy+y²-16

=(x²-2xy+y²⁾-16

=(x-y)²-16

=(x-y-4)(x-y+4)

Câu 3:

a: \(=\dfrac{5xy-4+3xy+4}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)

b: \(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)

c: \(=\dfrac{3x+1-2x+3}{x+y}=\dfrac{x+4}{x+y}\)

d: \(=\dfrac{4x+7+5x+7}{9}=\dfrac{9x+14}{9}\)

e: \(=\dfrac{5\left(x+2\right)}{2\left(2x-1\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-5\left(x-2\right)}{2x-1}\)

1) 4x\(^2\).(5x³+2x-1)

= 20x\(^5\)+8x\(^3\)-4x\(^2\).

2) 4x\(^3\): x²

= 4x

3) ( 15x²y³-10x³y³+6xy): 5xy

= 3xy²-2x²y²+\(\dfrac{6}{5}\)

4) (5x³+14x²+12x+8 ): (x+2)

= 5x²+4x+4

5)\(\dfrac{7}{2x}\)+\(\dfrac{11}{3y^2}\)

=\(\dfrac{7.3y^2+11.2x}{6xy^2}\) =\(\dfrac{21y^2+22x}{6xy^2}\) = \(\dfrac{21+22}{6}\) =\(\dfrac{43}{6}\)

6) \(\dfrac{x}{x+2}\) +\(\dfrac{3}{\left(x+2\right)\left(4x-7\right)}\)

7)\(\dfrac{3}{x-y}\)-\(\dfrac{2x^2}{x+y}\)

= \(\dfrac{3\left(x+y\right)-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{3x+3y-2x-2y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{1}{x-y}\).

8)\(\dfrac{1}{2}\)x²y².(2x+y)(2x-y)

= \(\dfrac{1}{2}\)x²y².(4x²-2xy+2xy-y²)

= \(\dfrac{1}{2}\)x²y².(4x²-y²)

= 2x⁴y²-\(\dfrac{1}{2}\)x²y⁴

9) (x-\(\dfrac{1}{2}\)).(x+\(\dfrac{1}{2}\)).(4x-1)

= x².(4x-1)

= 4x³-x²

10)\(\dfrac{3x}{2x+6}\)+\(\dfrac{6-x}{2x^2+6x}\)

= \(\dfrac{3x}{2\left(x+3\right)}\)+\(\dfrac{6-x}{2x\left(x+3\right)}\)= \(\dfrac{3x^2+6-x}{2x\left(x+3\right)}\)=\(\dfrac{3-x}{3}\)= -x

11) x²-\(\dfrac{1}{2x-2}\)+3x+\(\dfrac{3}{1-x^2}\)

12)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)

= \(\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)= \(\dfrac{x}{x+y}\)

cảm ơn bạn nhé ^^

1: =(4x-1)^2-3(4x-1)

=(4x-1)(4x-1-3)

=4(x-1)(4x-1)

2: =-8x^4y^5(2y+3x)

3: =(a-5)^2-4b^2

=(a-5-2b)(a-5+2b)

5: =x^2-mx-nx+mn

=x(x-m)-n(x-m)

=(x-m)(x-n)

6: =(4a^2-3a-18-4a^2-3a)(4a^2-3a-18+4a^2+3a)

=(-6a-18)(8a^2-18)

=-6(2a-3)(2x+3)(a+3)

1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)

2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)

4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

5, \(2x^3+3x^2+2x+3\)

\(=x^2\left(2x+3\right)+2x+3\)

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

6, \(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^3z-x^2z^2+x^2yz-xy^2\)

\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)

\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)

\(=xz\left(x+y\right)\left(x-z\right)\)

8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)

9, \(x^2-6x+8\)

\(=x^2-4x-2x+8\)

\(=x\left(x-4\right)-2\left(x-4\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

10, \(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-2\right)\left(x-6\right)\)

Chỗ còn lại mai làm nốt nha.

Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha

11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)

\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)

\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)

\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)

\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

12, \(x^3-7x-6\)

\(=x^3-3x^2+3x^2-9x+2x-6\)

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

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

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

\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

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

13, \(x^4+4\)

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

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

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

14, \(a^4+64\)

\(=a^4+16a^2+64-16a^2\)

\(=\left(a^2+8\right)^2-16a^2\)

\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)

15, \(x^5+x+1\)

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

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

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

\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

16, \(x^5+x-1\)

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

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

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

17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)

19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)

Đặt \(x^2+8x+7=a\) ta có:

(*) \(\Leftrightarrow a\left(a+8\right)+15\)

\(\Leftrightarrow a^2+8a+15\)

\(\Leftrightarrow a^2+3a+5a+15\)

\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)

\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)

Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)

Đặt \(x^2+3x+1=a\) ta có:

(*) \(\Leftrightarrow a\left(a+1\right)-6\)

\(\Leftrightarrow a^2+a-6\)

\(\Leftrightarrow a^2+3a-2a-6\)

\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)

\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)

Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)

\(A=\left(x+1\right)^2+\left(x+2\right)^2=\left(x+1\right)^2+\left(-2-x\right)^2\ge\frac{1}{2}\left(x+1-2-x\right)^2=\frac{1}{2}.1^2=\frac{1}{2}\Rightarrow A_{min}=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)

\(B=-2x^2-4\le0-4=-4\Rightarrow B_{max}=-4\Leftrightarrow x=0\)

\(C=-5x^2+10x-7=-5x^2+10x-5-2=-5\left(x-1\right)^2-2\le0-2=-2\Rightarrow C_{min}=-2\Leftrightarrow x-1=0\Leftrightarrow x=1\)