1) `2x(3x-1)-(2x+1)(x-3)`

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

\(\dfrac{x^2-4x+4}{x^2+2x+1}.\dfrac{x+1}{x^2-2x}.\dfrac{6x}{2x+4}=\dfrac{\left(x-2\right)^2}{\left(x+1\right)^2}.\dfrac{x+1}{x\left(x-2\right)}.\dfrac{6x}{2\left(x+2\right)}\)

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

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

(x-2)(x²+2x+4)-(2+x)(4-2x+x²)

=x³-8-x³-8

=-16

(2x+1)²-(4-x)-(5+x)(2-x)+(2+x)³ 

(sai đề)

(x-1)³-(x+1)³+6(x+1)(x-1)

=[(x-1)-(x+1)]*[(x-1)²+(x-1)(x+1)+(x+1)²]+6(x+1)(x-1)

=-2[x²-2x+1+x²-1+x²+2x+1]+6(x²-1)

=-2[3x²+1]+6x²-6

=-6x²-2+6x²-6

=-8

Bài 2:

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

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

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

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

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

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

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

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

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

a: ĐKXĐ: \(x\ne-2\)

\(\left(\dfrac{-2x-1}{x+2}+\dfrac{3x+4}{x+2}\right)\cdot\left(x^2-4\right)\)

\(=\dfrac{-2x-1+3x+4}{x+2}\cdot\left(x-2\right)\left(x+2\right)\)

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

b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

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

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

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

\(P\left(-1\right)=\left(-1\right)^4+2\cdot\left(-1\right)^2+1=1+2+1=4\)

\(P\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^4+2\cdot\left(\dfrac{1}{2}\right)^2+1=\dfrac{1}{16}+\dfrac{1}{2}+1=\dfrac{9}{16}\)

\(Q\left(-2\right)=\left(-2\right)^4+4\cdot\left(-2\right)^3+2\cdot\left(-2\right)^2-4\cdot\left(-2\right)+1=16-32+8+8+1=1\)

\(Q\left(1\right)=1^4+4\cdot1^3+2\cdot1^2-4\cdot1+1=1+4+2-4+1=5+3-4=2-4=-2\)

a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)

b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)

c: \(=\dfrac{x}{2x+1}-\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(x-2\right)}{2x-1}\)

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

\(=\dfrac{2x^2-x-1-2x^2-x+4x+2}{\left(2x+1\right)\left(2x-1\right)}\)

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

d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)

\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)