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Đầu tiên ta phải biết: Bán kính quả bóng bàn.
Ta có:
Chu vi hình tròn = Đường kính x 3,14
\(=35.9.65-8.100\)
\(=35.9.65-800\)
\(=20475-800=19675\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(5^x+5^{x+2}=650\)
`\Rightarrow 5^x + 5^x . 5^2 = 650`
`\Rightarrow 5^x . (1 + 5^2) = 650`
`\Rightarrow 5^x . 26 = 650`
`\Rightarrow 5^x = 650 \div 26`
`\Rightarrow 5^x = 25`
`\Rightarrow 5^x = 5^2`
`\Rightarrow x = 2`
Vậy, `x = 2`
`b)`
`(4x + 1)^2 = 25.9`
`\Rightarrow (4x + 1)^2 = 225`
`\Rightarrow (4x + 1)^2 = (+-15^2)`
`\Rightarrow`\(\left[{}\begin{matrix}4x-1=15\\4x-1=-15\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}4x=16\\4x=-14\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-\dfrac{7}{2};4\right\}\)
`c)`
\(2^x+2^{x+3}=144\)
`\Rightarrow 2^x + 2^x . 2^3 = 144`
`\Rightarrow 2^x . (1 + 2^3) = 144`
`\Rightarrow 2^x . 9 = 144`
`\Rightarrow 2^x = 144 \div 9`
`\Rightarrow 2^x = 16`
`\Rightarrow 2^x = 2^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
`d)`
\(3^{2x+2}=9^{x+3}\)
`\Rightarrow `\(3^{2x+2}=\left(3^2\right)^{x+3}\)
`\Rightarrow `\(3^{2x+2}=3^{2x+6}\)
`\Rightarrow 2x + 2 = 2x + 6`
`\Rightarrow 2x - 2x = 6 - 2`
`\Rightarrow 0 = 4 (\text {vô lý})`
Vậy, `x` không có giá trị nào thỏa mãn.
`e)`
\(x^{15}=x^2\)
`\Rightarrow `\(x^{15}-x^2=0\)
`\Rightarrow `\(x^2\cdot\left(x^{13}-1\right)=0\)
`\Rightarrow `\(\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x^{13}=1\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy, `x \in`\(\left\{0;1\right\}.\)
\(\frac{n^2+3n+6}{n+3}=\frac{n^2+3n}{n+3}+\frac{6}{n+3}\)
\(=\frac{n\left(n+3\right)}{n+3}+\frac{6}{n+3}\)
\(=n+\frac{6}{n+3}\)
Để thỏa đề bài thì 6 phải chia hết cho n + 3
\(\Rightarrow n+3\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
n + 3 = 1
n = -2 ( loại )
n + 3 = 2
n = -1 ( loại )
n + 3 = 3
n = 0 ( loại )
n + 3 = 6
n + 3 ( nhận )
Vậy n = 3 thì thỏa đề
\(B=1.3+3.5+5.7+....+27.29+29.31\)
\(6B=1.3.6+3.5.6+5.7.6+....+27.29.6+29.31.6\)
\(6B=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+27.29.\left(31-25\right)+29.31.\left(33-27\right)\)
\(6B=1.3.5+3+3.5.7-1.3.5+5.7.9-3.5.7+...+27.29.31-25.27.29+29.31.33-27.29.31\)
\(6B=3+29.31.33\)
\(B=\frac{3+29.21.33}{6}=4945\)
\(a,5^x+5^{x+2}=650\\ \Rightarrow5^x+5^x.5^2=650\\ \Rightarrow5^x\left(1+5^2\right)=650\\ \Rightarrow5^x.26=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,\left(4x+1\right)^2=25.9\\\Rightarrow\left(4x+1\right)^2=225\\ \Rightarrow\left[{}\begin{matrix}4x+1=15\\4x+1=-15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-4\end{matrix}\right.\)
\(c,2^x+2^{x+3}=144\\ \Rightarrow2^x+2^x.2^3=144\\ \Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:\left(1+2^3\right)\\ \Rightarrow2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\)
\(d,3^{x+2}=9^{x+3}\\ \Rightarrow3^{x+2}=\left(3^2\right)^{x+3}\\ \Rightarrow3^{x+2}=3^{2x+6}\\ \Rightarrow x+2=2x+6\\ \Rightarrow x-2x=6-2\\ \Rightarrow-x=4\\ \Rightarrow x=-4\)
\(e,x^{15}=x^2\\ \Rightarrow x^{15}-x^2=0\\ \Rightarrow x^2\left(x^{13}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a: =>5^x+5^x*25=650
=>5^x*26=650
=>5^x=25
=>x=2
b: =>4x+1=15 hoặc 4x+1=-15
=>4x=-16 hoặc 4x=14
=>x=7/2 hoặc x=-8
c: =>2^x*9=144
=>2^x=16
=>x=4
d: =>2x+2=2x+6
=>2=6(loại)
e: =>x^2(x^13-1)=0
=>x=0 hoặc x=1
\(f,\left(2x+1\right)^3=343\\\Rightarrow \left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow2x=6\\ \Rightarrow x=3\\ g,\left(x-1\right)^3=125\\ \Rightarrow\left(x-1\right)^3=5^3\\ \Rightarrow x-1=5\\ \Rightarrow x=6\\ h,2^{x+2}-2^x=96\\ \Rightarrow2^x.2^2-2^x=96\\ \Rightarrow2^x.\left(2^2-1\right)=96\\ \Rightarrow2^x=96:\left(2^2-1\right)\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)
\(i,\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow\left(x-5\right)^4\left(1-\left(x-5\right)^2\right)=0\\\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{matrix}\right. \\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(t/m\right)\\x=6\left(t|m\right)\\x=4\left(loại\right)\end{matrix}\right.\)
\(j,720:\left[41-\left(2x-5\right)\right]=2^3.5\\ \Rightarrow720:\left(41-2x+5\right)=40\\ \Rightarrow\left(46-2x\right)=720:40\\ \Rightarrow46-2x=18\\ \Rightarrow2x=46-18=28\\ \Rightarrow x=28:2=14\)
@seven
f: =>2x+1=7
=>2x=6
=>x=3
g: =>x-1=5
=>x=6
h: =>2^x*3=96
=>2^x=32
=>x=5
i: =>(x-5)^4*[(x-5)^2-1]=0
=>(x-5)(x-4)(x-6)=0
=>x=5;x=4;x=6
j: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
a: \(\dfrac{25\cdot9-25\cdot17}{-8\cdot80-8\cdot10}=\dfrac{25\cdot\left(-8\right)}{-8\cdot90}=\dfrac{25}{90}=\dfrac{5}{18}=\dfrac{125}{450}\)
\(\dfrac{48\cdot12-48\cdot15}{-3\cdot270-3\cdot30}=\dfrac{48\left(12-15\right)}{-3\left(270+30\right)}=\dfrac{48}{300}=\dfrac{4}{25}=\dfrac{72}{450}\)
b: \(\dfrac{2^5\cdot7+2^5}{2^5\cdot5^2-2^5\cdot3}=\dfrac{2^5\cdot8}{2^5\cdot22}=\dfrac{8}{22}=\dfrac{4}{11}=\dfrac{28}{77}\)
\(\dfrac{3^4\cdot5-3^6}{3^4\cdot13+3^4}=\dfrac{5-3^2}{13+1}=\dfrac{-4}{14}=\dfrac{-2}{7}=\dfrac{-22}{77}\)