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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a. \(x-\left(1,5-7\right)=0,35\\ \Rightarrow x+5,5=0,35\\ \Rightarrow x=-5,14\)
b. \(\left(x-1\right)^5=32\\ \Rightarrow\left(x-1\right)^5=2^5\\ \Rightarrow x-1=2\\ \Rightarrow x=3\)
a)\(\frac{-11}{12}.x+0,25=5\)
\(\Rightarrow-\frac{11}{12}.x=5-0,25=\frac{19}{4}\)
\(\Rightarrow-\frac{11}{12}.x=\frac{19}{4}\)
\(\Rightarrow x=\frac{-57}{11}\)
b)\(\left(x-1\right)^5=-32=-2^5\)
\(\Rightarrow\left(x-1\right)=-2\)
\(\Rightarrow x=-2+1=-1\)
a/|x|-2,5=27,5
=>|x|=27,5+2,5=30
=>x=30 hoặc x=-30
b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)
=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)
c/(x-1)\(^5\)=-32
=>x-1=-2 vì (-2)\(^5\)=-32
=>x=-2+1=-1
d/\(\dfrac{4}{5}.x+0,5=4.5\)
=>\(\dfrac{4}{5}.x+0,5=20\)
=>\(\dfrac{4}{5}.x=20-0,5=19,5\)
=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)
Lời giải:
a. Vì $x^2\geq 0$ với mọi $x\in\mathbb{R}$ nên $x^2+2\geq 2$
$\Rightarrow A=\frac{32}{x^2+2}\leq \frac{32}{2}=16$
Vậy $A_{\max}=16$ khi $x^2=0\Leftrightarrow x=0$
b.
$(x+1)^2\geq 0$ với mọi $x\in\mathbb{R}$
$\Rightarrow 2(x+1)^2+3\geq 3$
$\Rightarrow B=\frac{5}{2(x+1)^2+3}\leq \frac{5}{3}$
Vậy $B_{\max}=\frac{5}{3}$ khi $x+1=0\Leftrightarrow x=-1$
Vậy \(( - 32{x^5} + 1):( - 2x + 1) = 16{x^4} + 8{x^3} + 4{x^2} + 2x + 1\).
a) \(\left(\frac{1}{2}\right)^x=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
=> x = 5
b) \(\left(\frac{5}{7}\right)^x=\frac{125}{343}\)
\(\left(\frac{5}{7}\right)^x=\left(\frac{5}{7}\right)^3\)
=> x = 3
`(x+1)^5=-32`
`=>(x+1)^5=(-2)^5`
`=>x+1=-2`
`=>x=-2-1=-3`
b, (X + 1) 5 = -32
(x+1)5 = (-2)5
x + 1 = -2
x = -2 -1
X = -3