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Ta có :
\(\left(2x^2-3x+1\right)-\left(2x^2-3x+4\right)=0\)
\(\Leftrightarrow2x^2-3x+1-2x^2+3x-4=0\)
\(\Leftrightarrow-3=0\left(ktm\right)\)
\(\Leftrightarrow x\in\varnothing\)
M=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{17}{8^2.9^2}+\dfrac{19}{9^2.10^2}\)
=\(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{17}{64.81}+\dfrac{19}{81.100}\)
=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{81}+\dfrac{1}{81}-\dfrac{1}{100}\)
=1-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)<\(\dfrac{100}{100}=1\)
a) | \(\frac{1}{2}\)x| = 3 - 2x
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-\left(3-2x\right)\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x+2x=3\\\frac{1}{2}x=-3+2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=3\\\frac{1}{2}x-2x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3:\frac{5}{2}\\-\frac{3}{2}x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-3:\left(-\frac{3}{2}\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=2\end{cases}}\)
b) |x - 1| = 3x + 2
\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-\left(3x+2\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x-1=-3x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=3\\x+3x=-2+1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{-2}\\4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)
c) | 5x | = x - 12
\(\Rightarrow\orbr{\begin{cases}5x=x-12\\5x=-\left(x-12\right)\end{cases}}\Rightarrow\orbr{\begin{cases}5x-x=-12\\5x=-x+12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=-12\\5x+x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\6x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
d) |7 - x| = 5x + 1
\(\Rightarrow\orbr{\begin{cases}7-x=5x+1\\7-x=-\left(5x+1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}7-1=5x+x\\7-x=-5x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}6=6x\\7+1=-5x+x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\8=-4x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
e) |9 + x| = 2x
\(\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\Rightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}}\Rightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)
Ủng hộ mk nha !!! ^_^
Bài 1 mình đã làm ở olm rồi: https://olm.vn/hoi-dap/question/1325771.html
Bài 1 :
B=|3x-5|+|2-3x|
\(\Rightarrow B=\left|5-3x\right|+\left|3x-2\right|\\ \Rightarrow B\ge\left|5-3x+3x-2\right|=3\\ \Rightarrow B\ge3\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(5-3x\right).\left(3x-2\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5-3x\ge0\\3x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}5-3x\le0\\3x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le5\\3x>2\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge5\\3x< 2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x>\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{3}\\x< \dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Mà \(\left\{{}\begin{matrix}x\ge\dfrac{5}{3}\\x< \dfrac{2}{3}\end{matrix}\right.\) ( vô lí)
\(\Rightarrow\dfrac{2}{3}< x\le\dfrac{5}{3}\)
Vậy \(B_{min}=3\Leftrightarrow\dfrac{2}{3}< x\le\dfrac{5}{3}\)
mặc kệ biến chú tâm vào hệ trong ngoặc rồi mũ nó lên
a)1
b)1
| 3x - 2 | - 2x = -1
| 3x -2 | = -1 + 2x
TH1 : 3x - 2 = -1 + 2x
3x - 2x = -1 + 2
x = 1
TH2 : 3x - 2 = 1 - 2x
3x + 2x = 1 + 2
5x = 3
x = 3/5
Vậy x=1 hoặc x = 3/5
Hok tốt!!!!!