tìm x
a) x: 27=3,6
b)\(\dfrac{2x+1}{-27}\) =\(\dfrac{-3}{2x+1}\)
K
Khách
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4 tháng 10 2021
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
18 tháng 12 2023
Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)
ĐKXĐ: x<>1/2
\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)
=>\(\left(2x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
25 tháng 2 2021
`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
25 tháng 2 2021
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
BM
2
TC
2
S
20 tháng 2 2021
a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
=> 6(2x-7) = 9(x-3)
=> 12x - 42 = 9x - 27
=> 12x - 9x = -27 + 42
=> 3x = 15
=> x = 5
Vậy x = 5
b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
=> -7(x + 27) = 6(x + 1)
=> -7x - 189 = 6x + 6
=> -7x - 6x = 6 + 189
=> -13x = 195
=> x = -15
Vậy x = -15
20 tháng 2 2021
a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)
\(\Leftrightarrow12x-42=9x-27\)
\(\Leftrightarrow12x-9x=-27+42\)
\(\Leftrightarrow3x=15\)
hay x=5
Vậy: x=5
b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)
\(\Leftrightarrow6x+6=-7x+189\)
\(\Leftrightarrow6x+7x=189-6\)
\(\Leftrightarrow13x=183\)
hay \(x=\dfrac{183}{13}\)
Vậy: \(x=\dfrac{183}{13}\)
VN
Võ Ngọc Phương
VIP
1 tháng 10 2023
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
1 tháng 10 2023
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
24 tháng 11 2021
\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)
\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\dfrac{x^2-x}{x^2-1}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2}\\ \dfrac{3x}{x^3+2x^2+x}=\dfrac{3x}{x\left(x^2+2x+1\right)}=\dfrac{3}{\left(x+1\right)^2}\\ 2x=\dfrac{2x\left(x+1\right)^2}{\left(x+1\right)^2}\)
AH
Akai Haruma
Giáo viên
25 tháng 2 2021
Lời giải:
a)
$3^{2x+1}.7^y=9.21^x=3^2.(3.7)^x=3^{2+x}.7^x$
Vì $x,y$ là số tự nhiên nên suy ra $2x+1=2+x$ và $y=x$
$\Rightarrow x=y=1$
b) \(\frac{27^x}{3^{2x-y}}=\frac{3^{3x}}{3^{2x-y}}=3^{x+y}=243=3^5\Rightarrow x+y=5(1)\)
\(\frac{25^x}{5^{x+y}}=\frac{5^{2x}}{5^{x+y}}=5^{x-y}=125=5^3\Rightarrow x-y=3\) $(2)$
Từ $(1);(2)\Rightarrow x=4; y=1$
a/\(x:27=3,6\)
\(\Rightarrow x=97,2\)
b/\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Rightarrow\left(2x+1\right)^2=81\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)
a) x=3,6\(\times\) 27=97,2