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\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
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= (5x-25) + (5x - x2)
= 5(x-5) + x(5-x)
= 5(x-5) - x(x-5)
= (5 - x)(x - 5)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)
= (3x + 1 - x - 1)(3x + 1 + x + 1)
= 2x(4x + 2)
Em áp dụng hđt số 3 trong sgk nhé.
\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)
\(b,x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
\(c,4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)
a) x⁶ + y⁶ = (x²)³ + (y²)³
= (x² + y²)(x⁴ - x²y² + y⁴)
b) x⁶ - y⁶
= (x³)² - (y³)²
= (x³ - y³)(x³ + y³)
= (x - y)(x² + xy + y²)(x + y)(x² - xy + y²)
a) \(=\left(a+2c\right)^2-16=\left(a+2c-4\right)\left(a+2c+4\right)\)
b) \(=3y\left(4-x^2\right)+9\left(4-x^2\right)=3\left(4-x^2\right)\left(y+3\right)\)
\(=3\left(2-x\right)\left(2+x\right)\left(y+3\right)\)
a, a2 + 4ac + 4c2 - 16 = (a + 2c)2 - 42 = (a + 2c -4).(a + 2c +4)
b, 12y - 9x2 + 36 - 3x2y = (12y + 36) - (3x2y + 9x2) = 12.(y+ 3) - 3x2.(y + 3) =(y + 3).(12 - 3x2)
a: =4x^2+8x-3x-6
=4x(x+2)-3(x+2)
=(x+2)(4x-3)
b: =3(3x^2-2x-1)
=3(3x^2-3x+x-1)
=3(x-1)(3x+1)
c: =2x^2-4x+x-2
=2x(x-2)+(x-2)
=(x-2)(2x+1)
d: =3x^2+3x-2x-2
=3x(x+1)-2(x+1)
=(x+1)(3x-2)
e: =3x^2+9x+x+3
=3x(x+3)+(x+3)
=(x+3)(3x+1)
a) \(4x^2+5x-6\)
\(=4x^2+8x-3x-6\)
\(=\left(4x^2+8x\right)-\left(3x+6\right)\)
\(=4x\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(4x-3\right)\)
b) \(9x^2-6x-3\)
\(=3\left(3x^2-2x-1\right)\)
\(=3\left(3x^2-3x+x-1\right)\)
\(=3\left[3x\left(x-1\right)+\left(x-1\right)\right]\)
\(=3\left(x-1\right)\left(3x+1\right)\)
c) \(2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=\left(2x^2-4x\right)+\left(x-2\right)\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(2x+1\right)\left(x-2\right)\)
d) \(3x^2+x-2\)
\(=3x^2+3x-2x-2\)
\(=\left(3x^2+3x\right)-\left(2x+2\right)\)
\(=3x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-2\right)\)
e) \(3x^2+10x+3\)
\(=3x^2+9x+x+3\)
\(=3x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(3x+1\right)\)
a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;
b ) -(y-x-3)*(y+x+3)
\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right).\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)