Tìm x:
4 7 + x = 2
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a) \(\left(x-4\right)^2-\left(x-4\right)=0\)
\(\left(x-4\right)\left(x-4-1\right)=0\)
\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)
\(\left(x-7\right)\left(5x^2+7\right)=0\)
\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)
\(x=7\)
c) \(x^2\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-1\right)=0\)
\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)
a) (x - 4)^2=(x - 4)
(x - 4) (x -4)=(x -4 )
(x - 4) (x - 4)-(x - 4)=0
(x-4) (x-4-1)=0
(x-4) (x-5)=0
TH1:x-4=0 TH2:x-5=0
x=4 x=5
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
\(\left(x+7\right)\left(3x-15\right)=0\\ \Rightarrow3\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\\ 4x\left(x+7\right)=2\left(x+7\right)\\ \Rightarrow4x\left(x+7\right)-2\left(x+7\right)=0\\ \Rightarrow2\left(2x-1\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-7\end{matrix}\right.\\ \left(x-3\right)^2-x\left(x-4\right)=5\\ \Rightarrow x^2-6x+9-x^2+4x-5=0\\ \Rightarrow-2x+4=0\\ \Rightarrow-2x=-4\Rightarrow x=2\)
hưng phúc đầy đủ chưa bạn nhỉ?
1) \(\left(x+7\right)\left(3x-15\right)=0\)
⇔\(\left[{}\begin{matrix}x+7=0\\3x-15=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
2) \(4x\left(x+7\right)=2\left(x+7\right)\)
\(2\left(2x+1\right)\left(x+7\right)=0\)
⇔\(\left[{}\begin{matrix}2x+1=0\\x+7=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-7\end{matrix}\right.\)
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
(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
Từ đề bài suy ra \(\left(\frac{x}{2}-1\right)^2=\frac{9}{4}=\left(\frac{3}{2}\right)^2\)
Suy ra \(\orbr{\begin{cases}\frac{x}{2}-1=\frac{3}{2}\\\frac{x}{2}-1=-\frac{3}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{x-3}{2}=1\\\frac{x+3}{2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{25}{28}\)
\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)
\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)
\(\Rightarrow x=-\dfrac{37}{30}\)
\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)
\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)
\(\Rightarrow35x=-70\)
\(\Rightarrow x=-2\)
\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)
\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)
\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)
\(\Rightarrow x=-\dfrac{8}{9}\)
4 7 + x = 2 x = 2 - 4 7 x = 10 7