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Thế x = 3 , y = -5 vào biểu thức ta được :
a.3[ 3 - ( -5 ) ] + ( -5 )4( 3 - 5 )
= a.3.8 + 625.( -2 )
= 24a - 1250
a ) \(24a^2b-36ab^2-60b^3\)
\(=12b\left(2a^2-3ab-5b^2\right)\)
b ) \(2x^2y-3x+3x^2-2xy\)
\(=\left(2x^2y-2xy\right)+\left(3x^2-3x\right)\)
\(=2xy\left(x-1\right)+3x\left(x-1\right)\)
\(=\left(2xy+3x\right)\left(x-1\right)\)
\(=x\left(2y+3\right)\left(x-1\right)\)
c ) \(x^2-y^2-6x+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right)\left(x-3+y\right)\)
:D
\(a,2a^2b-36ab^2-60b^3\)
\(=12b\left(2a^2-3ab-5b^2\right)\)
\(=12b\left(2a^2-5ab+2ab-5b^2\right)\)
\(=12b\left[a\left(2a-5b\right)+b\left(2a-5b\right)\right]\)
\(=12b\left(a+b\right)\left(2a-5b\right)\)
\(b,2x^2y-3x+3x^2-2xy\)
\(=\left(2x^2y-2xy\right)+\left(3x^2-3x\right)\)
\(=2xy\left(x-1\right)+3x\left(x-1\right)\)
\(=\left(2xy+3x\right)\left(x-1\right)\)
\(=x\left(2y+3\right)\left(x-1\right)\)
\(c,x^2-y^2-6x+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x^2-2.x.3+3^2\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right)\left(x-3+y\right)\)
Chúc bạn hok tốt!!!
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: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=\left(x+\dfrac{1}{2}\right)^3\)
