Rút gọn phân thức sau:
a) \(\dfrac{12xy^3z⁴}{24x^2y^3z^3}\)
b)\(\dfrac{3x-6}{6x^2-12x}\)
a: \(=\dfrac{12xy^3z^4}{24x^2y^3z^3}=\dfrac{1}{2}\cdot\dfrac{1}{x}\cdot z=\dfrac{z}{2x}\)
b: \(=\dfrac{3\left(x-2\right)}{6x\left(x-2\right)}=\dfrac{1}{2x}\)
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tính tổng S = x + 2y + 3z biết rằng 1/(x+ 2y) + 1/(2y+3z)+1/(x+3z)= 12x/(2y+3z)+24y/(x+3z)+ 36z/(x+2y)=2016
BT9: Thực hiện phép tính
a, xy^2+x^2y+(-2xy^2)
b, 12x^2y^3z^4+(-7x^2y^3z^4)
c, -6xy^3-(-6xy^3)+6x^3
d, -x^2/2+7/2x^2+x
e, 2x^3+3x^3-1/3x^3
f, 5xy^2+1/2xy^2+1/4xy^2
a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
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1/2.(6x-2y).(3x+y)
(2/3z-2/5x).(1/3z+1/5x).1/2
(5y-3x).1/4.(12x+20y)
(3/4y-1/2x).(x+3/2y).2
(a+b+c).(a+b-c)
(x-y+z).(x+y-z)
mng giúp mình vs ạ
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
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a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)
b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)
\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)
\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)
\(=\left(5y-3x\right)\left(5y+3x\right)\)
\(=25y^2-9x^2\)
d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)
\(=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)\)
\(=\dfrac{9}{4}y^2-x^2\)
e: \(\left(a+b+c\right)\left(a+b-c\right)\)
\(=\left(a+b\right)^2-c^2\)
\(=a^2+2ab+b^2-c^2\)
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cho 2 đa thức A= \(-4x^5y^3+x^4y^3-3x^2y^3z^2-x^4y^3+x^2y^3z^2-2y^4\)
a) thu gọn rồi tìm bậc đa thức A
b) tìm đa thức B biết rằng B\(-2x^2y^3z^2+\dfrac{2}{3}y^4-\dfrac{1}{5}x^4y^3=A\)
a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)
b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)
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1. Tính tổng: S = x+ 2y+3z, biết rằng:
\(\dfrac{1}{x+2y}+\dfrac{1}{2y+3z}+\dfrac{1}{3z+x}=\dfrac{12x}{2y+3z}+\dfrac{24y}{3z+x}+\dfrac{36z}{x+2y}=2016\)
bạn chịu khó suy nghĩ chút sẽ ra bài này dễ mà
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Tính tổng \(S=x+2y+3z\), biết rằng:
\(\dfrac{1}{x+2y}+\dfrac{1}{2y+3z}+\dfrac{1}{3z+x}=\dfrac{12x}{2y+3z}+\dfrac{24y}{3z+x}-\dfrac{36z}{x+2y}=2016\)
Tinh tong : S= x+2y +3z, biet rang : \(\frac{1}{x+2y}+\frac{1}{2y+3z}+\frac{1}{3z+z}=\frac{12x}{2y+3z}+\frac{24y}{3z+x}-\frac{36z}{x+2y}=2016\)
cho 3x=4y; 4y=5z
Tính S=x-2y+3z/ x+2y-3z
tính tổng S= x+2y+3z biết rằng:
\(\frac{1}{x+2y}\)+\(\frac{1}{2y+3z}\)+\(\frac{1}{3z+x}\)= \(\frac{12x}{2y+3z}\)+\(\frac{24y}{3z+x}\)+\(\frac{36z}{x+2y}\)=2016
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