Rút gọn các biểu thức sau: a 4 3 . a - 1 3 + a 2 3 a 1 4 . a 3 4 + a - 1 4
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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
A = 1 + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
3\(\times\) A = 3 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)
3A - A = 3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\)
2A = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)
A = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2
A = \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2
A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2B = 2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2B + B = 2 - \(\dfrac{1}{2^{100}}\)
3B = 2 - \(\dfrac{1}{2^{100}}\)
B = ( 2 - \(\dfrac{1}{2^{100}}\)): 3
B = \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3
B = \(\dfrac{2^{101}-1}{3.2^{100}}\)
a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)
b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)
c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1