Trả lời:

Bài 1:

a, \(9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\)

b, \(x^3+27=x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)

c, \(8-y^3=2^3-y^3=\left(2-y\right)\left(4+2y+y^2\right)\)

d, \(x^4-81=\left(x^2\right)^2-9^2=\left(x^2-9\right)\left(x^2+9\right)\)\(=\left(x^2-3^2\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

e, \(64x^3-1=\left(4x\right)^3-1^3=\left(4x-1\right)\left(16x^2+4x+1\right)\)

f, \(x^6+8y^3=\left(x^2\right)^3+\left(2y\right)^3=\left(x^2+2y\right)\left(x^4-2x^2y+4y^2\right)\)

Bài 3 : 

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

Thay x = 6 ta được : \(-\left(6-1\right)^3=-\left(5\right)^3=-125\)

b, \(8-12x+6x^2-x^3=\left(2-x\right)^3\)

Thay x = 12 ta được : \(\left(2-12\right)^3=\left(-10\right)^3=-1000000\)

Bài 4 : 

a, \(A=163^2+74.163+37^2=163^2+2.37.163+37^2\)

\(=\left(163+37\right)^2=\left(200\right)^2=40000\)

Trả lời:

Bài 3: 

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

Thay x = 6 vào biểu thức trên, ta có:

\(-\left(6-1\right)^3=-5^3=-125\)

b, \(8-12x+6x^2-x^3=2^3-3.2^2.x+3.2.x^2-x^3=\left(2-x\right)^3\)

Thay x = 12 vào biểu thức trên, ta có:

\(\left(2-12\right)^3=\left(-10\right)^3=-1000\)

Bài 4:

a, Ta có: \(A=\) \(163^2+74.163+37^2=163^2+2.163.37+37^2=\left(163+37\right)^2=200^2\)

            \(B=\)\(147^2-94.147+47^2=147^2-2.147.47+47^2=\left(147-47\right)^2=100^2\)

Vì \(200^2>100^2\)

nên \(A>B\)

b, Ta có: \(C=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)

\(=2^2+4^2+...+100^2-1^2-3^2-...-99^2\)

\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)

\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)

\(=1.3+1.7+...+1.199\)

\(=3+7+...+199\)

\(=\frac{\left(199+3\right).50}{2}=5050\)  (50 là số số hạng)

\(D=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)

\(=\left(3.7\right)^8-\left[\left(21^4\right)^2-1\right]=21^8-21^8+1=1\)

Vì \(5050>1\)

nên \(C>D\)

a)\(\left(-a+\frac{2}{3}\right)\left(a+\frac{2}{3}\right)=\left(\frac{2}{3}-a\right)\left(\frac{2}{3}+a\right)=\left(\frac{2}{3}\right)^2-a^2=\frac{4}{9}-a^2\)

b)\(\left(x+5\right)\left(x^2-5x+25\right)=x^3+5^3=x^3+125\)

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

d)\(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=\left(a^2+3\right)^2-\left(2a\right)^2=\left(a^2+3\right)^2-4a^2\)

e)\(\left(x+3y\right)\left(9y^2-3xy+x^2\right)=x^3+\left(3y\right)^3=x^3+9y^3\)

f)\(2\left(x-\frac{1}{2}\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x\right)^3-1=8x^3-1\)

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

\(\Leftrightarrow B=16-x^4\)

Do \(x^4\ge0\forall x\Leftrightarrow-x^4\le0\Leftrightarrow16-x^4\le16\)

Dau '' = '' xay ra khi \(\Leftrightarrow x=0\)

\(\Rightarrow MaxB=16\Leftrightarrow x=0\)

bạn chia 1 góc 45 độ thành 3 góc 15 độ ik

(mik nghĩ v ko bt đúng ko)

Trả lời:

Bài 2: 

a, \(\left(-a+\frac{2}{3}\right)\left(a+\frac{2}{3}\right)=\left(\frac{2}{3}-a\right)\left(\frac{2}{3}+a\right)=\left(\frac{2}{3}\right)^2-a^2=\frac{4}{9}-a^2\)

b, \(\left(x+5\right)\left(x^2-5x+25\right)=x^3+5^3=x^3+125\)

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

d, \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=\left[\left(a^2+3\right)-2a\right]\left[\left(a^2+3\right)+2a\right]\)

\(=\left(a^2+3\right)^2-\left(2a\right)^2=\left(a^2+3\right)^2-4a^2\)

e, \(\left(x+3y\right)\left(9y^2-3xy+x^2\right)=x^3+\left(3y\right)^3=x^3+27y^3\)

f, \(2\left(x-\frac{1}{2}\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x\right)^3-1^3=8x^3-1\)

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

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

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

Trả lời:

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

\(=1-4x+4x^2+2+4x-4x-8x^2+1+4x+4x^2\)

\(=4\)

(1-2x)^2+(2-4x)*(1+2x)+(1+2x)^2

= 2^2