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\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a) pt => 2x-x=-25+5(chuyển vế đổi dấu) =>x=-20
b)pt=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\)=\(\frac{2016}{2017}\)
=>\(1-\frac{1}{2x+1}=\frac{2016}{2017}\)=>\(\frac{2x}{2x+1}=\frac{2016}{2017}\). Nhân chéo => x=1008
\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
a) \(\left(x-5\right)\left(2x-3^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\2x=9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{9}{2}\end{matrix}\right.\)
b) \(2\left(3x-15\right)\left(5-x\right)=0\)
\(\Rightarrow6\left(x-5\right)\left(5-x\right)=0\Rightarrow x=5\)
(x - 5)(2x - 32) = 0
=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)
\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)
\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)
\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)
hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)
a, (x2+5).(x2-25)=0
trường hợp 1(+):
x2+5=0
x2=0-5
x2=-5
không có giá trị nguyên nào
trường hợp 2
x2-25=0
x2=0+25
x2=25
x=5
vậy x =5
câu b làm tương tự nha. thông cảm