rút gọn biểu thức:
2|x - 3| - |4x - 1|.
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17 tháng 4 2021
\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)
Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)
\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)
15 tháng 8 2021
a: Ta có: \(P=\left(x-1\right)^2-4x\left(x+1\right)\left(x-1\right)+3\)
\(=x^2-2x+1-4x\left(x^2-1\right)+3\)
\(=x^2-2x+4-4x^3+4x\)
\(=-4x^3+x^2+2x+4\)
b: Thay x=-2 vào P, ta được:
\(P=-4\cdot\left(-8\right)+4-4+4=36\)
11 tháng 12 2020
Bài 1 :
\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)
\(=x^2-4x+4-x+9=x^2-5x+13\)
Bài 2 :
a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)
\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)
b, Thay x = -4 ta được :
\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 12 2022
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 12 2022
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
22 tháng 8 2021
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
Xét \(x< \frac{1}{4}\Rightarrow\hept{\begin{cases}4x-1< 0\\x-3< 0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=1-4x\\\left|x-3\right|=3-x\end{cases}}\)
Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(3-x\right)-\left(1-4x\right)=2x+5\)
Xét \(\frac{1}{4}\le x< 3\Rightarrow\hept{\begin{cases}4x-1\ge0\\x-3< 0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=4x-1\\\left|x-3\right|=3-x\end{cases}}\)
Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(3-x\right)-\left(4x-1\right)=-6x+7\)
\(x\ge3\Rightarrow\hept{\begin{cases}4x-1>0\\x-3\ge0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=4x-1\\\left|x-3\right|=x-3\end{cases}}\)
Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(x-3\right)-\left(4x-1\right)=-2x-5\)
TH1: Nếu \(x< \frac{1}{4}\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=-\left(x-3\right)\\\left|4x-1\right|=-\left(4x-1\right)\end{cases}}\)
\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=-2\left(x-3\right)+\left(4x-1\right)\)
\(=-2x+6+4x-1=2x+5\)
TH2: Nếu \(\frac{1}{4}\le x\le3\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=-\left(x-3\right)\\\left|4x-1\right|=4x-1\end{cases}}\)
\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=-2\left(x-3\right)-\left(4x-1\right)\)
\(=-2x+6-4x+1=-6x+7\)
TH3: Nếu \(x>3\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=x-3\\\left|4x-1\right|=4x-1\end{cases}}\)
\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=2\left(x-3\right)-\left(4x-1\right)\)
\(=2x-6-4x+1=-2x-5\)