Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

a) \(x^3-0,25x=0\\ < =>x\left(x^2-0,25\right)=0\\ =>\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\sqrt{0,25}\end{matrix}\right.\)

b) \(x^2-10x=-25\\ < =>x^2-10x+25=0\\ < =>\left(x-5\right)^2=0\\ < =>x-5=0\\=>x=5\)

Ta có: \(4x^2-9y^2\\ =\left(2x\right)^2-\left(3y\right)^2\\ =\left(2x-3y\right)\left(2x+3y\right)\)

Vậy: Chọn D

a) (a-3)(a+3)

b)(2x-5)(2x+5)

c)(x³-y³)(x³+y³)

a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)\)

= 400

b) \(87^2+73^2-27^2-13^2\)

\(\Leftrightarrow\left(87^2-13^2\right)\)+\(\left(73^2-27^2\right)\)

\(\Leftrightarrow\left(87+13\right)\left(87-13\right)+\left(73+27\right)\left(73-27\right)\)

\(\Leftrightarrow7400+4600=12000\)

a, \(9x^2+6xy+y^2=\left(3x\right)^2+2\times3xy+y^2=\left(3x+y\right)^2\)

b, \(6x-9-x^2=-\left(x^2-2\times3x+3^2\right)=-\left(x-3\right)^2\)

c, \(x^2+4y^2+4xy=x^2+2\times2xy+\left(2y\right)^2=\left(x+2y\right)^2\)

Bài giải:

a) 73² – 27² = (73 + 27)(73 – 27) = 100 . 46 = 4600

b) 37² - 13² = (37 + 13)(37 – 13) = 50 . 25 = 100 . 12 = 1200

c) 2002² – 2² = (2002 + 2)(2002 – 2) = 2004 . 2000 = 400800

Bài giải:

a) x³ + $\frac{1}{27}$ = x³ + ($\frac{1}{3}$)³ = (x + $\frac{1}{3}$)(x² – x . $\frac{1}{3}$+ ($\frac{1}{3}$)²)

=(x + $\frac{1}{3}$)(x² – $\frac{1}{3}$x + $\frac{1}{9}$)

b) (a + b)³ – (a - b)³

= [(a + b) – (a – b)][(a + b)² + (a + b) . (a – b) + (a – b)²]

= (a + b – a + b)(a² + 2ab + b² + a² – b² + a² – 2ab + b²)

= 2b . (3a³ + b²)

c) (a + b)³ + (a – b)³ = [(a + b) + (a – b)][(a + b)² – (a + b)(a – b) + (a – b)²]

= (a + b + a – b)(a² + 2ab + b² – a² +b² + a² – 2ab + b²]

= 2a . (a² + 3b²)

d) 8x³ + 12x²y + 6xy² + y³ = (2x)³ + 3 . (2x)² . y +3 . 2x . y + y³ = (2x + y)³

e) - x³ + 9x² – 27x + 27 = 27 – 27x + 9x² – x³ = 3³ – 3 . 3² . x + 3 . 3 . x² – x³ = (3 – x)³

Bài giải:

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

= -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - $\frac{1}{8}$ = (2x)³ – ($\frac{1}{2}$)³ = (2x - $\frac{1}{2}$)[(2x)² + 2x . $\frac{1}{2}$ + ($\frac{1}{2}$)²]

^{= (2x - $\frac{1}{2}$)(4x2 + x + $\frac{1}{4}$)}

d) $\frac{1}{25}$x² – 64y² = ${\left(\frac{1}{5}x\right)}^2$- (8y)² = ($\frac{1}{5}$x + 8y)($\frac{1}{5}$x - 8y)

Bài giải:

a) 2 – 25x² = 0 => (√2)² – (5x)² = 0

=> (√2 – 5x)( √2 + 5x) = 0

Hoặc √2 – 5x = 0 => 5x = √2 => x = $\frac{\sqrt{2}}{5}$

Hoặc √2 + 5x = 0 => 5x = -√2 => x = - $\frac{\sqrt{2}}{5}$

b) x² - x + $\frac{1}{4}$ = 0 => x² – 2 . x . $\frac{1}{2}$ + ($\frac{1}{2}$)² = 0

=> (x - $\frac{1}{2}$)² = 0 => x - $\frac{1}{2}$ = 0 => x = $\frac{1}{2}$

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

\(\Leftrightarrow\left[\left(x+y\right)+\left(x-y\right)\right]\left[\left(x+y\right)-\left(x-y\right)\right]\)

\(\Leftrightarrow\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(\Leftrightarrow2x.2y=4xy\)

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

\(\Leftrightarrow\left[\left(3x+1\right)+\left(x+1\right)\right]\left[\left(3x+1\right)-\left(x+1\right)\right]\)

\(\Leftrightarrow\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)

\(\Leftrightarrow\left(4x+2\right).2x\)

\(\Leftrightarrow8x^2+4x\)

\(\Leftrightarrow x\left(8x+4\right)\)