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

\(\Leftrightarrow2\left(x-3\right)-\left(x-1\right)=0\)

\(\Leftrightarrow2x-6-x+1=0\)

\(\Leftrightarrow x=5\)

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)

\(\left|x-1\right|+2C=\left|x-1,5\right|+\left|1-x\right|\\ \Leftrightarrow\left|x-1\right|+2C=\left|x-1,5\right|+\left|x-1\right|\\ \Rightarrow2C=\left|x-1,5\right|\ge0\\ \Rightarrow C\ge0\)

Để C=0 thì

\(\left|x-1,5\right|=0\\ \Leftrightarrow x-1,5=0\\ \Leftrightarrow x=1,5\)

Vậy...

ĐKXĐ: x>0; x<>1/2; x<>-1/2; x<>1/4

a: \(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\)

\(=\frac{\left(2\sqrt{x}-1\right)\left(4x+2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}+1\right)\left(4x-2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}+1\right)}\)

\(=\frac{4x+2\sqrt{x}+1}{\sqrt{x}}-\frac{4x-2\sqrt{x}+1}{\sqrt{x}}=\frac{4\sqrt{x}}{\sqrt{x}}=4\)

\(B=\left(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\right):\frac{2x+1}{2x-1}\)

\(=\frac{4\left(2x-1\right)}{2x+1}=\frac{8x-4}{2x+1}\)

b: A/B=1

=>A=B

=>\(\frac{x-2}{\sqrt{x}+2}=\frac{8x-4}{2x+1}\)

=>\(\left(8x-4\right)\left(\sqrt{x}+2\right)=\left(x-2\right)\left(2x+1\right)\)

=>\(8x\sqrt{x}+16x-4\sqrt{x}-8=2x^2+x-4x-2\)

=>\(2x^2-3x-2-8x\sqrt{x}-16x+4\sqrt{x}+8=0\)

=>\(2x^2-8x\cdot\sqrt{x}-19x+4\sqrt{x}+6=0\) (1)

Đặt \(t = \sqrt{x}\) (t>0; \(t<>\frac{1}{\sqrt2}\) )

(1) sẽ trở thành: \(2t^4-8t^3-19t^2+4t+6=0\)

=>t≃0,65 hoặc t≃5,37

t≃0,65

=>\(x=t^2\) ≃0,42(nhận)

t≃5,37

=>\(x=t^2\) ≃28,84(nhận)

x=1hoặc =-1.nếu đúng thì k mình nhé

may ngu nhu bo y sai bet

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,\(-11x=0\)

\(\Rightarrow x=0\)

\(A\left(x\right)=2x^2-8x\)

\(\Rightarrow2x^2-8x=0\)

\(\Rightarrow x\left(2x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)

\(B\left(x\right)=-3x+2x^2\)

\(B\left(x\right)=2x^2-3x\)

\(2x^2-3x=0\)

\(\Rightarrow x\left(2x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)