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DT
2
7 tháng 3 2021
Ta có: \(\dfrac{x-25}{75}+\dfrac{x-15}{85}+\dfrac{x-5}{95}+\dfrac{x-145}{15}=0\)
\(\Leftrightarrow\dfrac{x-25}{75}-1+\dfrac{x-15}{85}-1+\dfrac{x-5}{95}-1+\dfrac{x-145}{15}+3=0\)
\(\Leftrightarrow\dfrac{x-100}{75}+\dfrac{x-100}{85}+\dfrac{x-100}{95}+\dfrac{x-100}{15}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}\right)=0\)
mà \(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
5 tháng 9 2021
\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\\x+3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)
13 tháng 7 2021
a)Ta có: \(\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)
\(\Leftrightarrow\dfrac{3x+9+x+1}{3\left(x+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1>0\\4x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
b) Ta có: \(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)
\(\Leftrightarrow\dfrac{3x+6+x+3}{3\left(x+3\right)}\le0\)
\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+9>0\\4x+9\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow-3< x\le-\dfrac{9}{4}\)
LT
13 tháng 7 2021
a)\(\dfrac{x+3}{x+1}\ge-\dfrac{1}{3}\left(x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)
\(\Leftrightarrow\dfrac{3x+9+x+1}{3x+3}\ge0\)
\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+10\ge0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+10\le0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{-5}{2}\\x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le\dfrac{-5}{2}\end{matrix}\right.\)
b) \(\dfrac{x+2}{x+3}\le-\dfrac{1}{3}\left(x\ne-3\right)\)
\(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)
\(\Leftrightarrow\dfrac{3x+6+x+3}{3x+9}\le0\)
\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+9\ge0\\3x+9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+9\le0\\3x+9>0\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{9}{4}\\x< -3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{9}{4}\\x>-3\end{matrix}\right.\end{matrix}\right.\)
TH1: loại
TH2: TM
Vậy no của BPT là :\(-\dfrac{9}{4}\ge x>-3\)
chúc bạn học tốt
25 tháng 2 2022
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
Y
9 tháng 3 2023
\(\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)
9 tháng 5 2022
a.
\(A=B\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)
\(\Leftrightarrow8x+16=0\)
\(\Leftrightarrow8\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn A=B
b.
\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{16}>0\)
\(\Leftrightarrow\dfrac{x}{2}>0\)
\(\Leftrightarrow x>0\)
21 tháng 7 2018
\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1-\left(x+1\right)}{x+1}\right)\left(\frac{4x-2\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\left(\frac{x^2+1-x-1}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)=\left(\frac{x^2-x}{x+1}\right)\left(\frac{2\left(x+1\right)}{x\left(x-1\right)}\right)=\frac{2x\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=2\)
Vậy Q = 2
T
21 tháng 7 2018
Hình như đề là rút gọn thì phải.
Giải
\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\left(\frac{x^2}{x}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(x-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\frac{4\left(x-1\right)}{x-1}-\frac{2\left(x-1\right)}{x}=4-\frac{2x-2}{x}\)
LA
1
6 tháng 8 2020
\(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)+3\left(x^2-4x+4\right)=x^2+2x+1-\left(x^2-4^2\right)+3x^2\)\(\Leftrightarrow x^2-25-x^2-6x-9+3x^2-12x+12=x^2+2x+1-x^2+16+3x^2\)
\(\Leftrightarrow-20x=39\)
\(\Leftrightarrow x=\frac{-39}{20}\)
Vậy \(x=\frac{-39}{20}\)
phá ngoặc rồi giải?
(x3 + x2) + (x2 + x) = 0
<=> x3 + x2 + x2 + x = 0
<=> x3 + 2x2 + x = 0
<=> x(x + 1)(x + 1) = 0
<=> x = 0 hoặc x + 1 = 0
<=> x = 0 hoặc x = -1