Bài 1: Phân thức đại số.

1)x^2-2x-1=0

<=> (x^2-2x+1)-2=0

<=>(x-1)² =2

=>x-1 = \(\pm\sqrt{2}\)

=> x= \(\pm\sqrt{2}\) +1

2) x^2-x-1=0

<=> (x^2-x+1/4) -5/4=0

<=>(x+1/2)²= 5/4

=> x+1/2 = \(\pm\sqrt{\dfrac{5}{4}}\)

=>x=\(\pm\sqrt{\dfrac{5}{4}}\) - 1/2

3)x^2+x-3=0

<=> (x^2 + x + 1/4) -13/4=0

<=>(x+1/2)² = 13/4

=> x+1/2 = \(\sqrt{\dfrac{13}{4}}\)

=> x= \(\sqrt{\dfrac{13}{4}}\) -1/2

4) 4x^2-4x-1=0

<=> (4x^2-4x+1)-2=0

<=>(2x-1)² -2=0

<=> (2x-1)² - \(\left(\sqrt{2}\right)^2\) =0

<=> (2x-1 - \(\sqrt{2}\) ) . (2x-1 +\(\sqrt{2}\) )=0

=> 2x-1-\(\sqrt{2}\) =0 hoặc 2x-1+\(\sqrt{2}\) =0

=> 2x= 1+\(\sqrt{2}\) hoặc 2x= 1 - \(\sqrt{2}\)

=> x=\(\dfrac{1+\sqrt{2}}{2}\) hoặc x=\(\dfrac{1-\sqrt{2}}{2}\)

tra lời nhanh giúp mình

a.b. \(A=\dfrac{2}{x-1}+\dfrac{2\left(x+1\right)}{x^2+x+1}+\dfrac{x^2-10x+3}{x^3-1}\) ( x ≠ 1 )

\(A=\dfrac{2\left(x^2+x+1\right)+2\left(x+1\right)\left(x-1\right)+x^2-10x+3}{x^3-1}\)

\(A=\dfrac{2x^2+2x+2+2x^2-2+x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\dfrac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{5x^2-5x-3x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{5x\left(x-1\right)-3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{\left(x-1\right)\left(5x-3\right)}{x^2+x+1}=\dfrac{5x-3}{x^2+x+1}\)

c.

\(A=\dfrac{5x-3}{x^2+x+1}\)

\(\Leftrightarrow A\left(x^2+x+1\right)=5x-3\)

\(\Leftrightarrow Ax^2+Ax+A-5x+3=0\)

\(\Leftrightarrow Ax^2+\left(A-5\right)x+A+3=0\)

( \(a=A,b=A-5,c=A+3\) )

* A = 0 \(\Rightarrow x=\dfrac{3}{5}\)

* \(A\ge0\)

\(\Rightarrow\Delta=b^2-4ac\ge0\)

\(\Rightarrow\left(A-5\right)^2-4.A\left(A-3\right)\ge0\)

\(\Rightarrow A^2-10A+25-4A^2-12A\ge0\)

\(\Rightarrow-3A^2-22A+25\ge0\)

\(\Rightarrow-\dfrac{25}{4}\le A\le1\)

\(\Rightarrow Min_A=-\dfrac{25}{3}\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{\dfrac{25}{3}+5}{2.\left(\dfrac{-25}{3}\right)}=-\dfrac{4}{5}\)

giúp với

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(x² - 2x + 4)(x + 2) - x(x² + 2) = 17
⇔x³+8-x³-2x=17
⇔8-2x=17
⇔-2x=9
⇔2x=-9
⇔x=\(\dfrac{-9}{2}\)

(x - 2)³ + 6(x + 1)² - (x² + 3x + 9)(x - 3) = 15
⇔x³-6x²+12x-8+6(x²+2x+1)-(x³-27)=15
⇔x³-6x²+12x-8+6x²+12x+6-x³+27=15
⇔24x+25=15
⇔24x=-10
⇔x=\(\dfrac{-5}{12}\)

(x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 15
⇔(x³-6x²+9x²-27)-(x³-27)+9(x²+2x+1)=15
⇔x³+3x²-27-x³+27+9x²+18x+9=15
⇔12x²+18x-6=0
⇔12x²+12x+6x-6=0
⇔(12x²+12x)-(6x+6)=0
⇔12x(x+1)-6(x+1)=0
⇔(x+1)(12x-6)=0
⇔\(\left[{}\begin{matrix}x+1=0\\12x-6=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=-1\\12x=6\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

(x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15

(x^3-3•x^2•3+3•x•3^2-3^3)-(x^3+3x^2+9x-27)+9(x^2+2•x•1+1^2)=15

(x^3-9x^2+27x-27)-(x^3-27)+9(x^2+2x+1)=15

x^3-9x^2+27x -27 -x^3+27+9x^2+18x+9=15

x^3-9x^2+27x -x^3+9x^2+18x=15-27+27-9

45x=6

x=6/45

x=2/15

(x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 15

\(\Leftrightarrow\) (x³ - 9x² + 27x - 27) - (x³ - 27) + 9(x² + 2x + 1) = 15

\(\Leftrightarrow\) x³ - 9x² + 27x - 27 - x³ + 27 + 9x² + 18x + 9 = 15

\(\Leftrightarrow\) 45x + 9 = 15

\(\Leftrightarrow\) x = \(\frac{6}{45}\)

\(\Leftrightarrow\) x = \(\frac{2}{15}\)

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