rút gọn rồi tính
3-2x/x^2-9 + 2/x-3 tại x=1
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a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
c) Ta có: \(C=4\left(3x-2\right)^2+\left(4-x\right)^2-\left(6x-4\right)\left(8-2x\right)\)
\(=4\left(9x^2-12x+4\right)+x^2-8x+16-\left(48x-12x^2-32+8x\right)\)
\(=36x^2-48x+16+x^2-8x+16-48x+12x^2+32-8x\)
\(=49x^2-112x+64\)
\(=\left(7x-8\right)^2\)
\(=\left(7\cdot149-8\right)^2\)
\(=1071225\)
d) \(\left(3x-4\right)^2-9\left(x-2\right)\left(x+2\right)\)
\(=9x^2-24x+16-9\left(x^2-4\right)\)
\(=9x^2-24x+16-9x^2+36\)
\(=-24x+52\)
\(=-24\cdot\left(-2\right)+52\)
=48+52=100
e) Ta có: \(x\left(x-3\right)^2-\left(x-1\right)\left(x+5\right)-x\left(x-2\right)\left(x+2\right)\)
\(=x\left(x^2-6x+9\right)-\left(x^2+4x-5\right)-x\left(x^2-4\right)\)
\(=x^3-6x^2+9x-x^2-4x+5-x^3+4x\)
\(=-7x^2+9x+5\)
\(=-7\cdot\left(-1\right)^2+9\cdot\left(-1\right)+5\)
\(=-7-9+5\)
=-16+5=-11
a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)
\(=-4.\dfrac{1}{4}+10=9\)
b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)
\(=\left(-2\right).\left(32-32\right)=0\)
a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)
\(=4x^2-4x+1+9-4x^2\)
\(=-4x+10\)
\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)
Lời giải:
$H=(x^3-3x^2+3x-1)-(x^3+8)+3(x^2-16)$
$=x^3-3x^2+3x-1-x^3-8+3x^2-48$
$=(x^3-x^3)+(-3x^2+3x^2)+3x+(-1-8-48)$
$=3x-57=3.\frac{-1}{2}-57=\frac{-117}{2}$
Lời giải:
a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$
$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.
$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$
$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$
d.
$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$
$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$
$=-x^2y+4x^2-2xy^2-10x$
$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$
\(\left(x+3y\right)^3-\left(x+3y\right)\left(x^2-3xy+9y^2\right)-2x\left(x-2\right)^2=\left(x+3y\right)^3-\left(x^3+27y^3\right)-2x\left(x-2\right)^2\)
Thay x=1 y=2 ta có:
\(\left(1+3.2\right)^3-\left(1^3+27.2^3\right)-2.1.\left(1-2\right)^2=7^3-\left(1+216\right)-2=343-217-2=124\)