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a: \(P=\dfrac{2x^2-1-x^2+1+3x}{x\left(x+1\right)}=\dfrac{x^2+3x}{x\left(x+1\right)}=\dfrac{x+3}{x+1}\)
a) P = 2x(-3x + 2) - (x + 2)² + 8x² - 1
= -6x² + 4x - x² - 4x - 4 + 8x² - 1
= (-6x² - x² + 8x²) + (4x - 4x) + (-4 - 1)
= x² - 5
b) Thay x = 3 vào P, ta được:
P = 3² - 5
= 4
c) Để P = -1 thì x² - 5 = -1
x² = -1 + 5
x² = 4
x = 2 hoặc x = -2
Vậy x = 2; x = -2 thì P = -1
\(a,P=2x\left(-3x+2\right)-\left(x+2\right)^2+8x^2-1\)
\(=-6x^2+4x-\left(x^2+4x+4\right)+8x^2-1\)
\(=-6x^2+4x-x^2-4x-4+8x^2-1\)
\(=\left(-6x^2-x^2+8x^2\right) +\left(4x-4x\right)+\left(-4-1\right)\)
\(=x^2-5\)
Vậy \(P=x^2-5\).
\(b,\) Ta có: \(P=x^2-5\)
Thay \(x=3\) vào \(P\), ta được:
\(P=3^2-5=9-5=4\)
Vậy \(P=4\) khi \(x=3\).
\(c,\) Có: \(P=-1\)
\(\Leftrightarrow x^2-5=-1\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(P=-1\) khi \(x\in\left\{2;-2\right\}\).
#\(Toru\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\cdot x+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{\left(x^2+1\right)\left(x+1\right)+x-1}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x^3+x^2+x+1+x-1}{\left(x-1\right)}\cdot\dfrac{x+1}{2x+1}\)
\(=\dfrac{x^3+x^2+2x}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x\left(x^2+x+2\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
b: Khi x=1/2 thì \(A=\dfrac{\dfrac{1}{2}\left(\dfrac{1}{4}+\dfrac{1}{2}+2\right)\left(\dfrac{1}{2}+1\right)}{\left(\dfrac{1}{2}-1\right)\left(2\cdot\dfrac{1}{2}+1\right)}=-\dfrac{33}{16}\)
cứuuuuuuuuuuu