Bài 1: Phân thức đại số.

a: \(=\dfrac{12x+3-6x-4}{6}=\dfrac{6x-1}{6}\)

b: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

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

d: \(=\dfrac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{x+2}{\left(3x-2\right)\left(3x+2\right)}\)

e: \(=\dfrac{6x\left(x-y\right)-x\left(x+y\right)}{10\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{6x^2-6xy-x^2-xy}{10\left(x+y\right)\left(x-y\right)}=\dfrac{5x^2-7xy}{10\left(x+y\right)\left(x-y\right)}\)

f: \(=\dfrac{3}{2x\left(x+1\right)}+\dfrac{2x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{2}{x}\)

\(=\dfrac{3x-3+4x^2-2x-4x^2+4}{4x\left(x+1\right)\left(x-1\right)}=\dfrac{x+1}{4x\left(x+1\right)\left(x-1\right)}=\dfrac{1}{4x\left(x-1\right)}\)

a ) \(x^2-13x+36\)

\(=x^2-4x-9x+36\)

\(=x\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-9\right)\left(x-4\right)\)

b ) \(x^2+5x-14\)

\(=x^2-2x+7x-14\)

\(=x\left(x-2\right)+7\left(x-2\right)\)

\(=\left(x+7\right)\left(x-2\right)\)

c ) \(2x^2-5x-12\)

\(=2x^2-8x+3x-12\)

\(=2x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(2x+3\right)\left(x-4\right)\)

d ) \(x^2-4xy-12y^2\)

\(=x^2-4xy+4y^2-16y^2\)

\(=\left(x-2y\right)^2-\left(4y\right)^2\)

\(=\left(x-2y-4y\right)\left(x-2y+4y\right)\)

\(=\left(x-6y\right)\left(x+2y\right)\)

a) x²-13x+36

=x²-4x-9x+36

=(x²-4x)-(9x-36)

=x(x-4)-9(x-4)

=(x-4)(x-9)

b) x²+5x-14

=x²+7x-2x-14

=(x²+7x)-(2x+14)

=x(x+7)-2(x+7)

=(x+7)(x-2)

c) 2x²-5x-12

=2x²+3x-8x-12

=(2x²+3x)-(8x+12)

=x(2x+3)-4(2x+3)

=(2x+3)(x-4)

d) x²-4xy-12y²

=x²-4xy+4y²-16y²

=(x²-4xy+4y²)-16y²

=(x-2y)²-16y²

=(x-2y-4y)(x-2y+4y)

=(x-6y)(x+2y)

a)\(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)b)\(x^2+5x-14=x^2-2x+7x-14=x\left(x-2\right)+7.\left(x-2\right)=\left(7+x\right)\left(x-2\right)\)c)\(2x^2-5x-12=2x^2+3x-8x-12=x\left(2x+3\right)-4\left(2x+3\right)=\left(x-4\right)\left(2x+3\right)\)d)\(x^2-4xy-12y^2=x^2+2xy-6xy-12y^2=x\left(x+2y\right)-6y\left(x+2y\right)=\left(x-6y\right)\left(x+2y\right)\)

Ta có : \(\left\{{}\begin{matrix}\left|5x+1\right|\ge0\forall x\\\left|2y-1\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow3+\left|5x+1\right|+\left|2y-1\right|\ge3\forall x;y\)

\(\Rightarrow\dfrac{12}{3+\left|5x+1\right|+\left|2y-1\right|}\le\dfrac{12}{3}=4\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|5x+1\right|=0\\\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+1=0\\2y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=-1\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy Max của b/t trên là : \(4\Leftrightarrow x=-\dfrac{1}{5};y=\dfrac{1}{2}\)

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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\(=\dfrac{y\left(x^2+2xy+y^2\right)}{2x^2+2xy-xy-y^2}\)

\(=\dfrac{y\left(x+y\right)^2}{\left(x+y\right)\left(2x-y\right)}=\dfrac{y\left(x+y\right)}{2x-y}\)

\(=\dfrac{xy+y^2}{2x-y}\)