x/2=(x-1)/3. hãy tìm x
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`(x+1)+(x+2)+(x+3)+...+(x+9)=945`
`x+1+x+2+x+3+...+x+9=945`
`9x+(1+9)+(2+8)+(3+7)+(4+6)+5=945`
`9x+10+10+10+10+5=945`
`9x+45=945`
`9x=945-45`
`9x=900`
`x=900:9`
`x=100`
`3/4 : x =1 1/2`
`=> 3/4 : x= 3/2`
`=>x= 3/4 : 3/2`
`=>x= 3/4 xx 2/3`
`=> x= 6/12`
`=> x= 1/2`
Vậy `x=1/2`
Thay `5/8-x` thì
`5/8-1/2= 5/8- 4/8=1/8`
\(\dfrac{3}{4}:x=1\dfrac{1}{2}\)
\(\dfrac{3}{4}:x=\dfrac{3}{2}\)
\(x=\dfrac{3}{4}:\dfrac{3}{2}\)
\(x=\dfrac{1}{2}\)
Vậy: \(\dfrac{5}{8}-x=\dfrac{5}{8}-\dfrac{1}{2}=\dfrac{1}{8}\)
`1. P = x/(sqrt x-1)`
`= (x-1+1)/(sqrtx-1)`
`= ((sqrt x+1)(sqrt x-1))/(sqrt x-1) +1/(sqrt x-1)`
`= sqrt x+1 + 1/(sqrt x-1)`
`= sqrtx-1 + 1/(sqrt x-1) + 2 >= 4`.
ĐTXR `<=> (sqrtx-1)^2 = 1`.
`<=> x =4` hoặc `x = 0 ( ktm)`.
Vậy Min A `= 4 <=> x= 4`.
1) \(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{(x-\sqrt{x})+(\sqrt{x}-1)+1}{\sqrt{x}-1}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}+1\)
\(=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)
Với x>1\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1>0\\\dfrac{1}{\sqrt{x}-1}>0\end{matrix}\right.\)
Áp dụng BĐT AM-GM cho 2 số dương \(\sqrt{x}-1\) và \(\dfrac{1}{\sqrt{x}-1}\), ta có:
\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}=2\)
\(\Rightarrow P\ge2+2=4\)
Dấu = xảy ra khi: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
KL;....
Ta có
x+3/5=7/2
=>x=7/2-3/5=29/10
x-1/4=1/5
=> x=9/20
2-x=4/7
=>x=2-4/7=10/7
KL
k mik vs
x + \(\frac{3}{5}\)= \(\frac{7}{2}\)
x = \(\frac{7}{2}\)- \(\frac{3}{5}\)
x = \(\frac{29}{10}\)
x - \(\frac{1}{4}\)= \(\frac{1}{5}\)
x = \(\frac{1}{5}\)+ \(\frac{1}{4}\)
x = \(\frac{9}{20}\)
2 - x = \(\frac{4}{7}\)
x = 2 - \(\frac{4}{7}\)
x = \(\frac{10}{7}\)
x + 1 + x + 2 + x + 3 + x +4 = 5
x + ( 1 + 2 + 3 + 4 ) = 5
x + 10 = 5
x = 5 - 10
x = -5
1) \(3^x+3^{x+1}+3^{x+2}=351\)
\(\Rightarrow3^x\left(1+3^1+3^2\right)=351\)
\(\Rightarrow3^x.13=351\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
2) \(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(\Rightarrow C=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)...+2^{96}\left(2+2^2+2^3+2^4\right)\)
\(\Rightarrow C=30+2^4.30...+2^{96}.30\)
\(\Rightarrow C=\left(1+2^4+...+2^{96}\right).30⋮30\)
mà \(30=5.6\)
\(\Rightarrow C⋮5\left(dpcm\right)\)
1,
Có \(3^x\)+ \(3^{x+1}\) + \(3^{x+2}\) = \(351\)
=> \(3^x\) + \(3^x\).\(3\) + \(3^x\).\(9\) = \(351\)
=> \(3^x\).\(13\) = \(351\)
=> \(3^x\) = \(27\)
=> \(x\) = \(3\)
2,
C = \(2\) + \(2^2\) + \(2^3\) + ... + \(2^{100}\)
2C = \(2^2\) + \(2^3\) + \(2^4\) + ... + \(2^{101}\)
2C - C = \(2^{101}\) - \(2\)
C = \(2^{101}\) - \(2\)
C = \(2\).\(\left(2^{100}-1\right)\)
C = 2.\(\left(\left(2^5\right)^{20}-1^{20}\right)\)
Có \(2^5\) \(-1\) \(⋮\) 5
=> \(\left(\left(2^5\right)^{20}-1^{20}\right)\) \(⋮\) 5
=> C \(⋮\) 5
3,
Xét \(\overline{abcdeg}\)
= \(\overline{ab}\).\(10000\) + \(\overline{cd}\).\(100\) + \(\overline{eg}\)
= \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\) + \(9.\left(1111.\overline{ab}+11.\overline{cd}\right)\)
Có\(\left\{{}\begin{matrix}9.\left(1111.\overline{ab}+11.\overline{cd}\right)⋮9\left(1111.\overline{ab}+11.\overline{cd}\inℕ^∗\right)\\\overline{ab}+\overline{cd}+\overline{eg}⋮9\end{matrix}\right.\)
=> \(\overline{abcdeg}⋮9\)
4,
S = \(3^0+3^2+3^4+...+3^{2002}\)
9S = \(3^2+3^4+3^6+...+3^{2004}\)
9S - S = \(3^2+3^4+3^6+...+3^{2004}\) - (\(3^0+3^2+3^4+...+3^{2002}\))
8S = \(3^{2004}-1\)
=> 8S \(< 3^{2004}\)
nhớ cho tớ cái like nha
(- 2 )bài này mik làm rồi