Tim x:
2x-/x+1/=-1/2
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23 tháng 7 2021
Đặt C(x)=0
\(\Leftrightarrow-2x\left(2x-3\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow-4x^2+6x-2x+2=0\)
\(\Leftrightarrow-4x^2+4x+2=0\)
\(\Leftrightarrow4x^2-4x-2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}\\2x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3}+1\\2x=-\sqrt{3}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{2}\\x=\dfrac{-\sqrt{3}+1}{2}\end{matrix}\right.\)
Đặt Q(x)=0
\(\Leftrightarrow2\left(x-3\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x-6-x+1=0\)
\(\Leftrightarrow x=5\)
23 tháng 1
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
23 tháng 10 2017
\(\left|x-1\right|+2C=\left|x-1,5\right|+\left|1-x\right|\\ \Leftrightarrow\left|x-1\right|+2C=\left|x-1,5\right|+\left|x-1\right|\\ \Rightarrow2C=\left|x-1,5\right|\ge0\\ \Rightarrow C\ge0\)
Để C=0 thì
\(\left|x-1,5\right|=0\\ \Leftrightarrow x-1,5=0\\ \Leftrightarrow x=1,5\)
Vậy...
23 tháng 10 2017
cái này sai r mk xóa nhé
Đề full ko phải vệ,có lẽ bạn đó viết quá gần
TD
0
15 tháng 8 2017
a,\(2x^2-8x=0\)
\(2x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)
\(=2x^2-8x-3x-2x^2\)
=\(-11x\)
c,\(-11x=0\)
\(\Rightarrow x=0\)
15 tháng 8 2017
\(A\left(x\right)=2x^2-8x\)
\(\Rightarrow2x^2-8x=0\)
\(\Rightarrow x\left(2x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)
\(B\left(x\right)=-3x+2x^2\)
\(B\left(x\right)=2x^2-3x\)
\(2x^2-3x=0\)
\(\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
\(2x-\left|x+1\right|=-\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\left(-\frac{1}{2}\right)\)
=> \(\left|x+1\right|=2x+\frac{1}{2}\)
=> \(\left[{}\begin{matrix}x+1=2x+\frac{1}{2}\\x+1=-\left(2x+\frac{1}{2}\right)\end{matrix}\right.\) => \(\left[{}\begin{matrix}x-2x=\frac{1}{2}-1\\x+1=-2x-\frac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}-1x=-\frac{1}{2}\\x+2x=\left(-\frac{1}{2}\right)-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\left(-\frac{1}{2}\right):\left(-1\right)\\3x=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=\left(-\frac{3}{2}\right):3\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{1}{2}\right\}.\)
Chúc bạn học tốt!
\(2x-\left|x+1\right|=-\frac{1}{2}\)
\(\Leftrightarrow\left|x+1\right|=2x+\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2x+\frac{1}{2}\\x+1=-\left(2x+\frac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x+1=-2x-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\3x=-\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},-\frac{1}{2}\right\}\)