( x - 1) . (x + 7) =0
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7 x 2 = 14 7 x 5 = 35 7 x 6 = 42 0 x 7 = 0
7 x 4 = 28 7 x 3 = 21 7 x 9 = 63 7 x 0 = 0
7 x 8 = 56 7 x 1 = 7 7 x 10 =70 1 x 7 = 7
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
7 x 3 = 21
7 x 8 = 56
7 x 2 = 14
7 x 1 = 7
7 x 4 = 35
7 x 6 = 42
7 x 10 = 70
0 x 7 = 0
7 x 7 = 49
7 x 4 = 28
7 x 9 = 63
7 x 0 = 0
7 x 1 = 7
7 x 8 = 56
7 x 6 = 42
7 x 5 = 35
7x 2 = 14
7 x 9 = 63
7 x 4 = 28
0 x 7 = 0
7x 3 = 21
7 x 7 = 49
7 x 0 = 0
7 x 10 = 70
3 x 7 x ( 9 + 0 + 1 ) x 7 x 3 x ( 1 + 0 + 9 )
= 3 x 7 x 10 x 7 x 3 x 10
= 21 x 10 x 7 x 3 x 10
= 210 x 7 x 3 x 10
= 1470 x 3 x 10
= 4110 x 10
= 41100
k mk nha bn
1) -12+3.(-x+7)=-18
3.(-x+7)=-18+12
3.(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Làm theo công thức: tích bằng 0 thì một trong x thừa số bằng 0 rồi xét các trường hợp
\(1,x.\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)
\(2,\left(x+12\right).\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)
\(3,\left(-x+5\right).\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
4/ \(x.\left(2+x\right).\left(7-x\right)=0\)
\(\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}\)
Vậy \(x=\left\{0,-2,7\right\}\)
5/ \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}}\)=> \(\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}\)
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(x-1).(x+7)=0
=>x-1=0 hoặc x+7=0
TH1:x-1=0
x =0+1
x =1
TH2:x+7=0
x =0-7
x = -7
Vậy x = 1 hoặc x = -7
x=1 hoặc -7 tick na