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

\(=9x^2+12x+4-\left(9x^2-4\right)-6x=6x+8\)

Vậy bth phụ thuộc biến x, ko có đpcm 

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

\(=8x^3+1-\left(8x^3-1\right)=8x^3+1-8x^3+1=2\)

1: \(x^2-25=\left(x-5\right)\left(x+5\right)\)

2: \(9x^2-\dfrac{1}{16}y^2=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2\)

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

3: \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

4: \(\left(2x-5\right)^2-64=\left(2x-5-8\right)\left(2x-5+8\right)\)

\(=\left(2x-13\right)\left(2x+3\right)\)

5: \(81-\left(3x+2\right)^2\)

\(=\left(9-3x-2\right)\left(9+3x+2\right)\)

\(=\left(-3x+7\right)\left(3x+11\right)\)

6: \(9\left(x-5y\right)^2-16\left(x+y\right)^2\)

\(=\left(3x-15y\right)^2-\left(4x+4y\right)^2\)

\(=\left(3x-15y-4x-4y\right)\left(3x-15y+4x+4y\right)\)

\(=\left(-x-19y\right)\left(7x-11y\right)\)

7: \(x^3-8=x^3-2^3=\left(x-2\right)\left(x^2+2x+4\right)\)

8: \(27x^3+125y^3=\left(3x\right)^3+\left(5y\right)^3\)

\(=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)

9: \(x^6+216=\left(x^2\right)^3+6^3\)

\(=\left(x^2+6\right)\left(x^4-6x^2+36\right)\)

10: \(x^2+8x+16=x^2+2\cdot x\cdot4+4^2=\left(x+4\right)^2\)

11: \(9x^2-12xy+4y^2\)

\(=\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\)

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

12: \(-25x^2y^2+10xy-1\)

\(=-\left[\left(5xy\right)^2-2\cdot5xy\cdot1+1^2\right]\)

\(=-\left(5xy-1\right)^2\)

13: \(x^3-6x^2+12x-8\)

\(=x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3\)

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

14: \(8x^3+12x^2y+6xy^2+y^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

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

Dựng \(AH\perp CD;BK\perp CD\left(H;K\in CD\right)\)

Xét tg vuông ADH có

\(\widehat{DAH}=90^o-\widehat{D}=30^o\)

\(\Rightarrow DH=\dfrac{AD}{2}=\dfrac{4}{2}=2cm\) (trong tg vuông cạnh đối diện góc \(30^o\) băng nửa cạnh huyền)

\(\Rightarrow AH=\sqrt{AD^2-DH^2}=\sqrt{16-4}=\sqrt{12}=2\sqrt{3}cm\)

\(\Rightarrow AH=BK=2\sqrt{3}cm\) (đường cao của hình thang)

Xét tg vuông BCK có

\(\widehat{KBC}=90^o-\widehat{C}=45^o\)

=> tg BCK vuông cân tại K \(\Rightarrow CK=BK=2\sqrt{3}cm\)

\(\Rightarrow BC=\sqrt{BK^2+CK^2}=\sqrt{12+12}=2\sqrt{6}cm\)

Xét HCN ABKH có

\(AB=KH=CD-DH-CK=8-2\sqrt{3}-2\sqrt{3}=8-4\sqrt{3}=4\left(2-\sqrt{3}\right)cm\)

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

\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\)

\(=6ab^2\)

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

\(=x^3-8-x^3-3x^2-3x-1+3\left(x^2-1\right)\)

\(=-3x^2-3x-9+3x^2-3=-3x-12\)

\(\left(x+3\right)\left(x^2-3x+9\right)=28\)

=>\(x^3+27=28\)

=>\(x^3=1=1^3\)

=>x=1

\(\dfrac{2020^3+1}{2020^2-2019}=\dfrac{\left(2020+1\right)\left(2020^2-2020\cdot1+1\right)}{2020^2-2019}\)

\(=\dfrac{2021\cdot\left(2020^2-2019\right)}{2020^2-2019}\)

=2021

Sửa đề: \(\dfrac{2020^3-1}{2020^2+2021}\)

\(=\dfrac{\left(2020-1\right)\left(2020^2+2020+1\right)}{2020^2+2020+1}\)

=2020-1=2019

ĐKXĐ: \(x\ne2\)

\(P=\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{x^4+4x^2-4x^3-16x+4x^2+16}\)

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

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

Để P nguyên thì \(x+2⋮x-2\)

=>\(x-2+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

a: \(P=27-27x+9x^2-x^3\)

\(=3^3-3\cdot3^2\cdot x+3\cdot3\cdot x^2-x^3=\left(3-x\right)^3\)

Khi x=-17 thì \(P=\left[3-\left(-17\right)\right]^3=\left(3+17\right)^3=20^3=8000\)

b: \(Q=x^3+3x^2+3x\)

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

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

Khi x=99 thì \(Q=\left(99+1\right)^3-1=100^3-1=1000000-1=999999\)

giúp,có tick ạ