này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

2x y 2  +  x 2 y 4 + 1 = x y 2 2 + 2.x y 2 .1 + 1 2  = x y 2 + 1 2

Ta có hằng đẳng thức: 

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Ta thấy \(\left(x-1\right)+\left(x-2\right)+\left(3-2x\right)=0\)

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

suy ra \(\left(x-1\right)\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x_1=1\\x_2=2\\x_3=\frac{3}{2}\end{cases}}\)

\(S=\frac{29}{4}\).

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

a) ( x   +   1 ) 2 .                     b) ( x   –   4 ) 2 .

c) x 2 4 + x + 1 ;                   d) ( 2 x   –   2 y ) 2 .

a) \(=\left(x-2\right)^2\)

b) \(=\left(3x-2\right)^2\)

c) \(=\left(x-3y\right)^2\)

d) \(=\left(\dfrac{x}{2}+1\right)^2\)

e) \(=\left(x-4\right)^2\)

f) \(=\left(\dfrac{1}{2}xy^2+1\right)^2\)

g) \(=\left(x-1\right)\left(x+1\right)\)

h) \(=\left(5x-4\right)\left(5x+4\right)\)

a) x²-4x+4=\(\left(X-2\right)^2\)

b)9x²-12x+4=\(\left(3x-2\right)^2\)

c)x²-6xy+9y^{2=\(\left(x-3y\right)^2\)}

d)x²/4+x+1=\(\left(\dfrac{x}{2}+1\right)^2\)

e)-8x+16+x² =\(X^2-8x+16=\left(x-4\right)^2\)

f)xy²+1/4x²y⁴+1 => sai đề

g)x²-1=\(\left(X-1\right)\left(x+1\right)\)

h)25x²-16=\(\left(5x-4\right)\left(5x+4\right)\)

a. (x + y)² = x² + 2xy + y²

b. (x - 2y)² = x² - 4xy - 4x²

c. (xy² + 1)(xy² - 1) = x²y⁴ - 1

d. (x + y)²(x - y)² = (x² + 2xy + y²)(x² - 2xy + y²) = x⁴ - (2xy + y²)² = x⁴ - (4x²y² + y⁴) = x⁴ - 4x²y² - y⁴

^{Chucs hocj toots}

Câu 2: 

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(x^2+10x+25=\left(x+5\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)

e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)

Câu 7:

a: Ta có: \(A=x^2-2x+7\)

\(=x^2-2x+1+6\)

\(=\left(x-1\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x=1

b: Ta có: \(B=5x^2-20x\)

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

\(=5\left(x-2\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x=2

x 2  + x + 1/4 =  x 2  + 2.x.1/2 +  1 / 2 2  =  x + 1 / 2 2