giai phuong trinh\(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}+\dfrac{1}{x^2+16x+63}=\dfrac{1}{5}\)
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28 tháng 11 2022
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{2}{\left(x+3\right)\left(x+5\right)}+\dfrac{2}{\left(x+5\right)\left(x+7\right)}+\dfrac{2}{\left(x+7\right)\left(x+9\right)}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+9}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{x+9-x-1}{\left(x+1\right)\left(x+9\right)}=\dfrac{2}{5}\)
=>2(x+1)(x+9)=5*8=40
=>x^2+9x+9=20
=>x^2+9x-11=0
hay \(x=\dfrac{-9\pm5\sqrt{5}}{2}\)
=>x^2+9x
DT
0
16 tháng 6 2022
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
TH
0
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 1 2022
1.
\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)
\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)
\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max
2.
\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)
\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)
\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)
\(E_{min}=-1\) khi \(x=0\)
\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)
\(G_{min}=-2\) khi \(x=2\)
6 tháng 2 2018
1) điều kiện xác định : \(x\notin\left\{-1;-3;-5;-7\right\}\)
ta có : \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\) \(\Leftrightarrow\dfrac{\left(x+5\right)\left(x+7\right)+\left(x+1\right)\left(x+7\right)+\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)\(\Leftrightarrow\dfrac{x^2+12x+35+x^2+8x+7+x^2+4x+3}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{3x^2+24x+45}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)
\(\Leftrightarrow9\left(3x^2+24x+45\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)
\(\Leftrightarrow27\left(x^2+8x+15\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)
\(\Leftrightarrow27\left(x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)
\(\Leftrightarrow27=\left(x+1\right)\left(x+7\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow27=x^2+8x+7\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-10\)
15 tháng 8 2018
b , \(\sqrt{1-4x+4x^2}-3=0\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=3\)
\(\Leftrightarrow\left|1-2x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=2\\-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(S=\left\{-1,2\right\}\)
TA
1 tháng 1 2021
- Thay từng giá trị vào, ta thấy A. \(\dfrac{15}{4}\) thỏa mãn.
b) \(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow x^2-6x+9+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow\left(x^2-3x+7\right)-\sqrt{x^2-3x+7}-20=0\)
Đặt \(\sqrt{x^2-3x+7}=t\left(t\ge0\right)\left(1\right)\)
\(\Rightarrow t^2-t-20=0\)
\(\Rightarrow x_1=5\left(TM\right);x_2=-4\left(KTM\right)\)
Thay t=5 vào (1), ta có :
\(\sqrt{x^2-3x+7}=5\)
\(\Leftrightarrow x^2-3x+7=25\)
\(\Leftrightarrow x^2-3x-18=0\)
\(\Rightarrow x_1=6;x_2=-3\)
vậy...
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