2/3 x 3/4 + 3/4 x 2/5
tính nhanh nhé
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Lời giải:
$a=2+\sqrt{5}$
$a-2=\sqrt{5}$
$a^2-4a+4=5\Leftrightarrow a^2-4a-1=0$
$p(a)=a^5-13a^4+7a^3-4a^2-6a$
$=a^3(a^2-4a-1)-9a^2(a^2-4a-1)-28a(a^2-4a-1)-125a^2-34a$
$=-125a^2-34a=-125(a^2-4a-1)-534a-125$
$=-534a-125=-534(2+\sqrt{5})-125=-1193-534\sqrt{5}$
Có \(x+y=7+4\sqrt{3}+7-4\sqrt{3}=14\)
\(xy=\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)=1\)
\(x^2+y^2=\left(x+y\right)^2-2xy=14^2-2=194\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=14^3-3.1.14=2702\)
\(x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^3y^3\left(x+y\right)\)\(=2702\left[\left(x^2+y^2\right)^2-2x^2y^2\right]-14\)
\(=2702\left(194^2-2\right)-14=101687054\)
Vậy...
\(x+y=14\) ; \(xy=\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=1\)
\(x^2+y^2=\left(x+y\right)^2-2xy=14^2-2.1=194\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=14^3-3.1.14=2702\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=194^2-2.1^2=37634\)
\(x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-\left(xy\right)^3\left(x+y\right)=2702.37634-1^3.14=...\)
\(f\left(\dfrac{1}{2}\right)=2.\left(\dfrac{1}{2}\right)^2-5=\dfrac{-9}{2}\)
\(f\left(-1\right)=2.\left(-1\right)^2-5=-3\)
\(f\left(3\right)=2.3^2-5=13\)
(x)+Q(x)=(x3-2x+1)+(2x2 -2x3+x-5)
=x3-2x+1+2x2-2x3+x-5 = -x3+2x2-x-4
P(x)-Q(x)=(x3-2x+1)+(2x2-2x3+x-5)
=x3-2x+1-2x2+2x3-x+5
=3x3-2x2-3x+6
a) \(2\left|x\right|-5=3\)
=> \(2\left|x\right|=3+5\)
=> \(2\left|x\right|=8\)
=> \(\left|x\right|=4\)
=> x = 4 hoặc x = -4
b) \(x-5=\left(-14\right)+2^3\)
=> \(x-5=\left(-14\right)+8\)
=> \(x-5=-6\)
=> \(x=-6+5=-1\)
c) \(10+2x=4^5:4^3\)
=> \(10+2x=4^{5-3}\)
=> \(10+2x=4^2\)
=> \(2x=4^2-10=16-10=6\)
=> \(2x=6\)
=> \(x=3\)
d) (x + 7) - 13 = 4
=> x + 7 = 17
=> x = 17 - 7 = 10
e) \(2x-10=2^4:2^2\)
=> \(2x-10=2^2\)
=> \(2x-10=4\)
=> \(2x=14\)
=> \(x=7\)
\(\dfrac{2}{3}\times\dfrac{3}{4}+\dfrac{3}{4}\times\dfrac{2}{5}=\dfrac{3}{4}\times\left(\dfrac{2}{3}+\dfrac{2}{5}\right)=\dfrac{3}{4}\times\dfrac{16}{15}=\dfrac{4}{5}\)
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