cho A=2+2/2+2/3+...+2/10
a A :3 c 2/3b2/3.2
b A:31 d 2/3[A+2]
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a) \(ab-ac-b^2+bc=\left(ab-ac\right)-\left(b^2-bc\right)\)( Phương pháp nhóm các hạng tử )
\(=a.\left(b-c\right)-b.\left(b-c\right)\) ( Phương pháp đặt nhân tử chung )
\(=\left(a-b\right)\left(b-c\right)\) ( Phương pháp đặt nhân tử chung )
b) \(10a^3-10a=10a.\left(a^2-1\right)=10a.\left(a+1\right)\left(a-1\right)\)
c) \(2a^2xy-18b^2xy=2xy.\left(a^2-9b^2\right)=2xy.\left(a+3y\right)\left(a-3y\right)\)
d) \(\left(a-b\right)\left(a+b\right)+3\left(a+b\right)=\left(a+b\right)\left(a-b+3\right)\)
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=64\\b^2=144\\c^2=256\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\pm8\\b=\pm12\\c=\pm16\end{matrix}\right.\)
Vậy \(\left(a;b;c\right)\in\left\{\left(8;12;16\right),\left(-8;-12;-16\right)\right\}\)
Cách khác:
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)
\(\Leftrightarrow k^2=16\)
Trường hợp 1: k=4
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k=8\\b=3k=12\\c=4k=16\end{matrix}\right.\)
Trường hợp 2: k=-4
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k=-8\\b=3k=-12\\c=4k=-16\end{matrix}\right.\)
a: \(=\left(a+b\right)^2-\left(c+d\right)^2\)
b: \(=\left(a-d\right)^2-\left(b-c\right)^2\)
c: \(=\left(x+3z\right)^2-4y^2\)
d: \(=\left(a^2-9\right)\left(a^2+9\right)=a^4-81\)
e: \(=\left(a-5\right)^2\cdot\left(a+5\right)^2=\left(a^2-25\right)^2\)
a,
A = 2 + 22 + 23 +...+210
A = (2 + 22 ) + (23 +24 ) + ...+ (29 + 210 )
A = 2 ( 1+2 ) + 23(1+2 ) + ...+ 29(1+2)
A = 2 .3 + 23 .3 + ...+29.3
A = 3 ( 2+ 23 + ...+ 29 ) \(⋮\) 3 3
Vậy A \(⋮\) 3
b, A = 2 + 22 + 23 +...+210
A = ( 2 + 22 + 23 + 24 + 25 ) + ( 26 + 27 + 28 + 29 + 210 )
A = 2 ( 1+2+22 + 23 + 24 ) + 26(1+2+22 + 23 + 24)
A = 2 . 31 + 26 .31
A = 31(2+26 ) \(⋮\) 31
vậy A \(⋮\) 31
d , A = 2 + 22 + 23 +...+210
Đặt a/3=b/5=k
=>a=3.k
=>a2=9.k2
=>b=5.k
=>b2=25.k2
Ta có: C= 5a2+3b2/10a2-3b2
=> c= 5.9.k2+3.25.k2/10.9.k2-3.25.k2
=> C= k2.(5.9+3.25) / k2.(9.10-3.25)
=> C= 120/15
=> C=8
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