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AN
2 tháng 1 2018
a, Theo đề ta có:
\(2.3^x-405=3^{x-1}\)
=> \(2.3^x-405=3^x:3\)
=> \(405=(2.3^x)-(3^x:3)\)
=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)
=> \(405=3^x(2-\dfrac{1}{3})\)
=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)
=> \(405=3^x.\dfrac{5}{3}\)
=> \(3^x=405:\dfrac{5}{3}\)
=>\(3^x=405.\dfrac{3}{5}\)
=> \(3^x=81.3\)
=> \(3^x=243\)
=> \(3^x=3^5\)
=> x=5
Vậy:..............................
NK
4
12 tháng 2 2020
(x-99) (1/30 + 1/32 + 1/34 - 1/36 - 1/38) = 0
SUy ra x - 99 = 0
VẬy x =99
NK
1
LN
19 tháng 9 2017
\(\)Bạn viết thiếu 1 vế đúng không?
\(\dfrac{x-69}{30}+\dfrac{x-67}{32}+\dfrac{x-65}{34}=\dfrac{x-63}{36}+\dfrac{x-61}{38}+\dfrac{x-59}{40}\)\(\Rightarrow\left(\dfrac{x-69}{30}-1\right)+\left(\dfrac{x-67}{32}-1\right)+\left(\dfrac{x-65}{34}-1\right)=\left(\dfrac{x-63}{36}-1\right)+\left(\dfrac{x-61}{38}-1\right)+\left(\dfrac{x-59}{40}-1\right)\)
\(\Rightarrow\dfrac{x-99}{30}+\dfrac{x-99}{32}+\dfrac{x-99}{34}=\dfrac{x-99}{36}+\dfrac{x-99}{38}+\dfrac{x-99}{40}\)
\(\Rightarrow\dfrac{x-99}{30}+\dfrac{x-99}{32}+\dfrac{x-99}{34}-\dfrac{x-99}{36}-\dfrac{x-99}{38}-\dfrac{x-99}{40}=0\)\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{34}-\dfrac{1}{36}-\dfrac{1}{38}-\dfrac{1}{40}\right)=0\)
Vì \(\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{34}-\dfrac{1}{36}-\dfrac{1}{38}-\dfrac{1}{40}\ne0\)
Nên:
\(x-99=0\Rightarrow x=99\)
19 tháng 7 2021
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)
\(\Leftrightarrow\dfrac{x+35}{65}+1+\dfrac{x+39}{61}+1=\dfrac{x+43}{57}+1+\dfrac{x+47}{53}+1\)
\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\right)=0\Leftrightarrow x=-100\)
19 tháng 7 2021
Ta có:
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)
Ta thấy:
\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)
Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)
\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)
Vậy \(x=-100\)
NX
30 tháng 8 2023
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)
Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
Vậy \(x\in\left\{-66\right\}\)
3 tháng 8 2023
2: =>(3x-7)^2=25/144=(5/12)^2
=>3x-7=5/12 hoặc 3x-7=-5/12
=>3x=5/12+7=89/12 hoặc 3x=7-5/12=79/12
=>x=89/36 hoặc x=79/36
3:Sửa đề: |2x-3|=|x+1|
=>2x-3=x+1 hoặc 2x-3=-x-1
=>x=4 hoặc 3x=2
=>x=2/3 hoặc x=4
4: =>3x+1=5 hoặc 3x+1=-5
=>3x=4 hoặc 3x=-6
=>x=-2 hoặc x=4/3
1: =>\(2x-7=\sqrt[3]{\dfrac{26}{63}}\)
=>\(2x=\sqrt[3]{\dfrac{26}{63}}+7\)
=>\(x=\dfrac{1}{2}\cdot\left(\sqrt[3]{\dfrac{26}{63}}+7\right)\)
6 tháng 12 2015
mỗi phân số cộng thêm 1 ta có
\(\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
rồi đặt x+66 làm thừa số chung sau đó giải tiếp
\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\dfrac{x-1}{65}-1+\dfrac{x-3}{63}-1=\dfrac{x-5}{61}-1+\dfrac{x-7}{59}-1\)
\(\Leftrightarrow\dfrac{x-66}{65}+\dfrac{x-66}{63}=\dfrac{x-66}{61}+\dfrac{x-66}{59}\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
\(\Leftrightarrow x-66=0\)
\(\Leftrightarrow x=66\)
Vậy x=66.