(x^2 - 7)(x^2 - 51) < 0
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tìm x,y thuộc z :
a, (x^2-7) . ( x^2 - 51) < 0
b, ( x-3) . ( 2y + 1) = 7
c, ( 2x+ 1 ) . ( 3y .2) =-55
0
TT
7 tháng 1 2018
a,\(\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy...
b,\(\left(x-5\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)
Vậy...
c,\(\left(x-5\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy...
Câu d bạn viết lại đề nhé
HT
1
4 tháng 7 2016
a) \(60-3.\left(x-2\right)=51\)
\(\Rightarrow3.\left(x-2\right)=60-51=9\)
\(\Rightarrow x-2=9:3=3\)
\(\Rightarrow x=3+2=5\)
b) \(\left(105-x\right):2^5=3^{0+1}=3^1=3\)
\(\Rightarrow\left(105-x\right):32=3\)
\(\Rightarrow105-x=3.32=96\)
\(\Rightarrow x=105-96=9\)
c) \(x+5=20-\left(12-7\right)\)
\(\Rightarrow x+5=20-5\)
\(\Rightarrow x+5=15\)
\(\Rightarrow x=15-5=10\)
NH
0
24 tháng 8 2021
\(x^2=4\rightarrow x=\pm2\)
\(x^2=5\rightarrow x=\pm\sqrt{5}\)
\(x^2=0\rightarrow x=0\)
\(x^2=1\rightarrow x=\pm1\)
\(x^2-9=0\rightarrow x^2=9\rightarrow x=\pm3\)
\(x^2+1=0\rightarrow x^2=-1\rightarrow x\in\varnothing\)
\(x^2=2\rightarrow x=\pm\sqrt{2}\)
\(x^2-3=0\rightarrow x^2=3\rightarrow x=\pm\sqrt{3}\)
\(x^2+1=82\rightarrow x^2=81\rightarrow x=\pm9\)
\(7x^2=63\rightarrow x^2=9\rightarrow x=\pm3\)
\(x^2+\frac{7}{4}=\frac{23}{4}\rightarrow x^2=4\rightarrow x=\pm2\)
Chúc bạn hok tốt!!!
24 tháng 8 2021
x2= 4
=> x = 2 hoặc x = -2
x2= 5
=> x = \(\sqrt{5}\) hoặc \(-\sqrt{5}\)
x2= 0
=> x = 0
24 tháng 8 2021
x2= 1
=> x = 1 hoặc x = -1
x2 - 9 = 0
=> x2 = 9
=> x = 3 hoặc x = -3
x2 + 1 = 0
=> x2 = -1 ( vô lý )( ko có giá trị x nào thoả mãn)
x2 = 2
=> x = \(\sqrt{2}\) hoặc x = \(-\sqrt{2}\)
10 tháng 7 2021
Với \(x>0;x\ne4\)
\(\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right):\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right):\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\frac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right).\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
10 tháng 7 2021
\(A=\left[\frac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right]\times\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\times\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{5\sqrt{x}\left(2\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-7>0\\x^2-51< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-7< 0\\x^2-51>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2>7\\x^2< 51\end{matrix}\right.\\\left\{{}\begin{matrix}x^2< 7\\x^2>51\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\pm\sqrt{7}\\x< \pm\end{matrix}\right.\\\left\{{}\begin{matrix}x< \pm\sqrt{7}\\x>\pm\sqrt{51}\end{matrix}\right.\end{matrix}\right.\sqrt{51}}\)