(x-5)^5=(x-5)^3
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SI
1 tháng 7 2021
30% . x - x - \(\dfrac{5}{6}\) = \(\dfrac{1}{3}\)
\(\Rightarrow\) (\(\dfrac{3}{10}\) - 1) . x = \(\dfrac{1}{3}+\dfrac{5}{6}\)
\(\Rightarrow\) \(-\dfrac{7}{10}\) . x = \(\dfrac{7}{6}\)
\(\Rightarrow\) x = \(-\dfrac{3}{5}\)
DD
1 tháng 7 2021
KO ghi đề nhé
\(\dfrac{3}{10}.x-x=\dfrac{1}{3}+\dfrac{5}{6}\)
\(x\left(\dfrac{3}{10}-1\right)=\dfrac{2}{6}+\dfrac{5}{6}\)
\(x.\dfrac{-7}{10}=\dfrac{7}{6}\)
\(x=\dfrac{7}{10}:\dfrac{-7}{6}\)
\(x=\dfrac{7}{10}.\dfrac{6}{-7}\)
\(x=\dfrac{42}{-70}\)
\(x=\dfrac{6}{-10}\)
\(x=\dfrac{3}{-5}\)
Vậy \(x=\dfrac{3}{-5}\)
Y
4 tháng 4 2022
\(\dfrac{9}{7}x+\dfrac{5}{7}x=\dfrac{2}{3}\)
\(\Leftrightarrow x\left(\dfrac{9}{7}+\dfrac{5}{7}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{14}{7}x=\dfrac{2}{3}\)
\(\Leftrightarrow2x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{2}{3}:2\)
\(\Leftrightarrow x=\dfrac{2}{3}\times\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{2}{6}\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
9 tháng 1 2022
\(\Leftrightarrow2.5:x=2+\dfrac{2}{3}-2-\dfrac{1}{5}=\dfrac{7}{15}\)
hay \(x=\dfrac{5}{2}:\dfrac{7}{15}=\dfrac{5}{2}\cdot\dfrac{15}{7}=\dfrac{75}{14}\)
AN
3
DC
19 tháng 4 2022
= 3/4 + 1/3 = 13/12
5/4 x X = 5/8
X = 5/8 : 5/4
X = 1/2
vậy X = ...
DD
3
24 tháng 1
ĐKXĐ: x<>0
2x-y=3
=>\(y=2x-3\)
\(\dfrac{2}{x}=\dfrac{y}{5}\)
=>\(\dfrac{2}{x}=\dfrac{2x-3}{5}\)
=>x(2x-3)=10
=>\(2x^2-3x-10=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{89}}{4}\left(nhận\right)\\x=\dfrac{3-\sqrt{89}}{4}\left(nhận\right)\end{matrix}\right.\)
Khi \(x=\dfrac{3+\sqrt{89}}{4}\) thì \(y=2\cdot\dfrac{3+\sqrt{89}}{4}-3=\dfrac{-3+\sqrt{89}}{2}\)
Khi \(x=\dfrac{3-\sqrt{89}}{4}\) thì \(y=2\cdot\dfrac{3-\sqrt{89}}{4}-3=\dfrac{-3-\sqrt{89}}{2}\)
UN
0
TC
2 tháng 5 2022
a)\(f\left(1\right)=2.1^2+5.1-3=2+5-3=4\)
\(f\left(0\right)=0+0-3=-3\)
\(f\left(1,5\right)=2.\left(1,5\right)^2-5.1,5-3=4,5-7,5-3=-6\)
VD
20 tháng 3 2022
a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne\pm1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)
\(b,A=3\\ \Leftrightarrow\dfrac{x+2}{2x+1}=3\\ \Leftrightarrow6x+3=x+2\\ \Leftrightarrow5x+1=0\\ \Leftrightarrow x=-\dfrac{1}{5}\left(tm\right)\)
\(c,\dfrac{1}{A}=\dfrac{2x+1}{x+2}=\dfrac{2x+4-3}{x+2}=\dfrac{2\left(x+2\right)-3}{x+2}=2-\dfrac{3}{x+2}\)
Để `1/A` là số nguyên thì `3/(x+2)` nguyên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Ta có bảng:
| x+2 | -3 | -1 | 1 | 3 |
| x | -5 | -3 | -1(ktm) | 1(ktm) |
Vậy \(x\in\left\{-5;-3\right\}\)
2
17 tháng 5 2022
`a)`
`@f(1)=2.1^2+5.1-3=2.1+5-3=2+5-3=4`
`@f(0)=2.0^2+5.0-3=-3`
`@f(1,5)=2.(1,5)^2+5.1,5-3=4,5+7,5-3=9`
_____________________________________________________
`b)`
`***f(3)=9`
`=>3a-3=9`
`=>3a=12=>a=4`
`***f(5)=11`
`=>5a-3=11`
`=>5a=14=>a=14/5`
`***f(-1)=6`
`=>-a-3=6`
`=>-a=9=>a=-9`
17 tháng 5 2022
a: f(1)=2+5-3=4
f(0)=-3
f(1,5)=4,5+7,5-3=9
b: f(3)=9 nên 3a-3=9
hay a=4
f(5)=11 nên 5a-3=11
hay a=14/5
f(-1)=6 nên -a-3=6
=>-a=9
hay a=-9
\(\left(x-5\right)^5=\left(x-5\right)^3\)
chỉ có 2 TH là cả hai vế bằng 1 hoặc bằng 0
\(TH1:\left(x-5\right)^5=\left(x-5\right)^3=0\)
\(=>x-5=0\)
\(=>x=-5\)
\(TH2:\left(x-5\right)^5=\left(x-5\right)^3=1\)
\(=>x-5=1\)
\(=>x=-4\)
Có vẻ đúng !!!!!
\(\left(x-5\right)^5=\left(x-5\right)^3\)
\(\left(x-5\right)^5-\left(x-5\right)^3=0\)
\(\left(x-5\right)^3\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^3=0\\\left(x-5\right)^2-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=0+1=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0+5=5\\\left(x-5\right)^2=\left(-1\right)^2\text{ Hoặc }\left(x-5\right)^2=1^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=-1\text{ Hoặc }x-5=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=5\\x=-1+5=4\text{ Hoặc }x=1+5=6\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{5\text{ ; }4\text{ ; }6\right\}\)