16ab + 4b² - 9 + 16a²

= (16a² + 16ab + 4b²) - 9

= (4a+2b)² - 3²

= (4a+2b-3)(4a+2b+3)

\(16a^2b-16ab+4b=4b\left(4a^2-4a+1\right)=4b\left(2a-1\right)^2\\ 5a^3-10a=5a\left(a^2-2\right)=5a\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\\ 3x-3z+x^2-2xz+z^2\\ =3\left(x-z\right)+\left(x-z\right)^2=\left(x-z\right)\left(3+x-z\right)\)

\(16a^2b-16ab+4b=4b\left[\left(4a^2\right)-4a+1\right]=4b\left[\left(2a\right)^2-4a+1\right]=4b\left(2a-1\right)^2\)

\(5a^3-10a=5a\left(a^2-2\right)\)

\(3x-3z+x^2-2xz+z^2=\left(3x-3z\right)+\left(x^2-2xz+z^2\right)=3\left(x-z\right)+\left(x-z\right)^2=\left(x-z\right)\left[3+\left(x-z\right)\right]=\left(x-z\right)\left(3+x-z\right)\)

Lời giải:

a.

\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)

\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)

\(=-a^4b^4(3a+4b)^2\)

b.

$x^3-6x^2y+12xy^2-8x^3$

$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$

c.

$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$

$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$

$=(x+\frac{1}{2})^3$

a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

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

b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)

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

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

c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

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

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

a) \(x^6+x^4+x^2y^2+y^4-y^6\)

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

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

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

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

\(=\left(x-y\right)\left(x+y\right)\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]+\left(x^2+y^2\right)^2-x^2y^2\)

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

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

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

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

c) \(\left(2a+b\right)^3+6a+3b-4\)

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

Đặt 2a + b = t.

Ta có: \(t^3+3t-4\)

\(=t^3-t^2+t^2-t+4t-4\)

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

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

Thay t = 2a + b vào biểu thức:

\(\left(t-1\right)\left(t^2+t+4\right)=\left(2a+b-1\right)\left(4a^2+4ab+b^2+2a+b+4\right)\)

a. 16a² - 49.( b - c )²

= ( 4a )² - 7².( b - c )²

= ( 4a )² - [ 7.( b - c ) ]²

= ( 4a )² - ( 7b - 7c )²

= ( 4a - 7b + 7c ).( 4a + 7b - 7c )

b. ( ax + by )² - ( ax - by )²

=( ax + by + ax - by ).( ax + by - ax + by )

= 2ax . 2by

= 2.( ax + by )

c.a⁶ - 1

= ( a³ )² - 1

= ( a³ - 1 ).( a³ + 1 )

= ( a - 1 ).( a² + a + 1 ).( a + 1 ).( a² - a + 1 )

d. a⁸ - b⁸

= ( a⁴ )² - ( b⁴ )²

= ( a⁴ - b⁴ ).( a⁴ + b⁴ )

= [ ( a² )² - ( b² )² ].( a⁴ + b⁴ )

= ( a² - b² ).( a² + b² ).( a⁴ + b⁴ )

= ( a - b ).( a + b ).( a² + b² ).( a⁴ + b⁴ )

B₂

( x - 4 )² - 36 = 0

\(\Leftrightarrow\) ( x - 4 )² = 36

\(\Leftrightarrow\) ( x - 4 )² = 6²

\(\Leftrightarrow\) x + 4 = \(\pm\) 6

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

Vậy x = 10 , x = -2

b. ( x - 8 )² = 121

\(\Leftrightarrow\) ( x - 8 )² = 11²

\(\Leftrightarrow\) x - 8 = \(\pm\)11

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=11\\x-8=-11\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=19\\x=-3\end{cases}}\)

Vậy x = 19 , x = -3

c. x² + 8x + 16 = 0

\(\Leftrightarrow\)x² + 2.4x + 4² = 0

\(\Leftrightarrow\) ( x + 4 )² = 0

\(\Leftrightarrow\) x + 4 = 0

\(\Leftrightarrow\) x = -4

Vậy x = -4

d. 4x² - 12x = - 9

\(\Leftrightarrow\)( 2x )² - 2.2.x.3 + 3² = 0

\(\Leftrightarrow\) ( 2x - 3 )² = 0

\(\Leftrightarrow\) 2x - 3 = 0

\(\Leftrightarrow\) 2x = 3

\(\Leftrightarrow\) \(x=\frac{3}{2}\)

Vậy x = \(\frac{3}{2}\)