26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

c) Ta có: \(C=4\left(3x-2\right)^2+\left(4-x\right)^2-\left(6x-4\right)\left(8-2x\right)\)

\(=4\left(9x^2-12x+4\right)+x^2-8x+16-\left(48x-12x^2-32+8x\right)\)

\(=36x^2-48x+16+x^2-8x+16-48x+12x^2+32-8x\)

\(=49x^2-112x+64\)

\(=\left(7x-8\right)^2\)

\(=\left(7\cdot149-8\right)^2\)

\(=1071225\)

d) \(\left(3x-4\right)^2-9\left(x-2\right)\left(x+2\right)\)

\(=9x^2-24x+16-9\left(x^2-4\right)\)

\(=9x^2-24x+16-9x^2+36\)

\(=-24x+52\)

\(=-24\cdot\left(-2\right)+52\)

=48+52=100

e) Ta có: \(x\left(x-3\right)^2-\left(x-1\right)\left(x+5\right)-x\left(x-2\right)\left(x+2\right)\)

\(=x\left(x^2-6x+9\right)-\left(x^2+4x-5\right)-x\left(x^2-4\right)\)

\(=x^3-6x^2+9x-x^2-4x+5-x^3+4x\)

\(=-7x^2+9x+5\)

\(=-7\cdot\left(-1\right)^2+9\cdot\left(-1\right)+5\)

\(=-7-9+5\)

=-16+5=-11

mn giúp e sớm đc ko ạ

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

Với \(x\ge-1\) thì \(\left|x+1\right|=x+1\)\(\Rightarrow A=\frac{x+1+2x}{3x^2-2x+1}=\frac{3x+1}{3x^2-2x+1}\)

         Thay \(x=\frac{3}{4}>-1\) vào ta được:\(A=\frac{3.\frac{3}{4}+1}{3.\left(\frac{3}{4}\right)^2-2.\frac{3}{4}+1}=\frac{52}{19}\)

Với \(x< -1\) thì \(\left|x+1\right|=-\left(x+1\right)=-x-1\)\(\Rightarrow A=\frac{-x-1+2x}{3x^2-2x+1}=\frac{x-1}{3x^2-2x+1}\)

      Thay \(x=-2< -1\) vào ta được \(A=\frac{-2-1}{3.\left(-2\right)^2-2.\left(-2\right)+1}=-\frac{3}{17}\)

câu 1:

x³-1+3x²-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)

Câu 2 :

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

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

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

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

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

b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)

\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)

Câu 3 :

Sửa đề :

\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)

a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)

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

\(=\dfrac{3x-1}{3x+1}\)

\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)

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

\(=\dfrac{x-3}{3x}\)

\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)

c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

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

\(=\dfrac{x-2}{2x}\)

\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

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

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

Bài 2: 

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

\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)

\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)

\(=-117x\)

b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)

\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)

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

=4

c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)

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

=-6