Tìm x :
(x - 7)^4 = (x-7)^5
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1) <=> 8x-16-21-7x=-4
<=> x-37=-4
<=> x=33
2) <=> 12x-48+6x-12-16x-48=-28
<=> 2x-108=-28
<=> 2x = 80
<=> x = 40
3)<=> 4x-20-35+7x+50-10x=-3
<=> x-5=-3
<=> x=2
4) <=> -3x-15=-45-21
<=> 51=3x
<=> 17=x
5) 4x-28+15=-2x+14-10
<=> 6x-13=4
<=> 6x=17
<=> x = \(\frac{17}{6}\)
Chúc bạn học tốt
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
a: =>7/9:x=1/18-2/9=1/18-4/18=-3/18=-1/6
=>x=-7/9:1/6=-7/9*6=-42/9=-14/3
b: =>x*7/5=2/15+2/5=8/15
=>x=8/15:7/5=8/21
c: =>x-1/2=3/14:4/7=3/8
=>x=3/8+4/8=7/8
d: =>0,4x+0,3x-0,2x=0,7
=>0,5x=0,7
=>x=1,4
/x+2/=0\(\Rightarrow x+2=0\Rightarrow x=0-2=-2\)
/x-3/=7-(-2)=9\(\Rightarrow x=9+3=12ho\text{ặc}x=-9+3=-6\)
(7-x)-(25+7)=-25\(\Rightarrow\left(7-x\right)-32=-25\Rightarrow7-x=-25+32=7\Rightarrow x=0\)
/x-3/=/5/+/-7/=5+7=12\(\Rightarrow x=15ho\text{ặc}x=-9\)
/x-5/=/-7/=7\(\Rightarrow x=12ho\text{ặc}x=-2\)
4-(7-x)=x-(13-4)\(\Rightarrow x-3=x-9\Rightarrow x-x=-9+3\Rightarrow0=-6\)(vô lí)
Vậy không có x thoả mãn 4-(7-x)=x-(13-4)
|x+2|=0
Vì |0|=0,suy ra x+2=0
x=0-2
x=-2
Vậy x = -2
|x-3|=7-(-2)
|x-3|=7+2
|x-3|=9
Vì |9|=|-9|=9,suy ra x-3 thuộc{9;-9}
*x-3=9
x =9+3
x =12
*x-3=-9
x=-9+3
x=-6
Vậy x thuộc {12;-6}
a)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{47}{10}\)
\(\Rightarrow x=\dfrac{13}{10}.\dfrac{10}{47}\)
\(\Rightarrow x=\dfrac{13}{47}\)
b)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{5}{30}-\dfrac{24}{30}\)
\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{19}{30}\)
\(\Rightarrow x=\dfrac{5}{7}:\left(-\dfrac{19}{30}\right)\)
\(\Rightarrow x=\dfrac{5}{7}.\left(-\dfrac{30}{19}\right)\)
\(\Rightarrow x=-\dfrac{150}{133}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{25}{28}\)
\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)
\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)
\(\Rightarrow x=-\dfrac{37}{30}\)
\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)
\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)
\(\Rightarrow35x=-70\)
\(\Rightarrow x=-2\)
\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)
\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)
\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)
\(\Rightarrow x=-\dfrac{8}{9}\)
(x-1/2).5/3=7/4-1/2
(x-1/2).5/3=5/4
x-1/2=5/4:5/3
x=3/4+1/2
x=5/4
(x-4/3).7/4=5-7/6
(x+4/3).7/4=23/6
x+4/3=23/6:7/4
x+4/3=46/21
x=46/21-4/3
x=6/7
Vì (x-7)4=(x-7)5 => x-7=1 hoặc x-7=0
(+) x-7=0 => x=7
(+) x-7=1 => x=8