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A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x)
= - x+6/x+2
Bài 1:
1.Đặt \(A=x^2+y^2-3x+2y+3\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+y^2+2y+1+2\)
\(=\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{9}{4}+2\)
\(=\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{1}{4}\)
Vì \(\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0;\forall x\\\left(y+1\right)^2\ge0;\forall y\end{cases}}\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2\ge0;\forall x,y\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{1}{4}\ge0-\frac{1}{4};\forall x,y\)
Hay \(A\ge\frac{-1}{4};\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-1\end{cases}}\)
VẬY MIN A=\(\frac{-1}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-1\end{cases}}\)
1. \(x\left(y-4\right)=35-5\left(y-4\right)\) với y= 4 không phải nghiệm y khác 4
\(x=\frac{35}{y-4}-1\)
y=4+35/n
x=n-1
\(\hept{\begin{cases}n=\left\{-7,-5,-1,1,5,7\right\}\\y=\left\{-1,-3,-31,39,11,9\right\}\\x=n-1=\left\{-8,-6,-2,0,4,6\right\}\end{cases}}\)
2.x^2+x+6=y^2
4x^2+4x+1=4y^2-23
(2x+1)^2=4y^2-23
=>4y^2-23=t^2
(2y)^2-t^2=23
=>\(\hept{\begin{cases}y=+-6\\t=+-11\end{cases}\Rightarrow\hept{\begin{cases}2x+1=11\\2x+1=-11\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\x=-6\end{cases}}}\)
\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x=a\)
=> \(\left(a-6\right)\left(a+6\right)=a^2-36\ge-36\)
\(x\left(x+5\right)=0\) thì biểu thức nhỏ nhất
<=> x = 0 hoặc x = -5
1.
<=> 7 - 2x - 4 = -x - 4
<=> -2x + x = -4 -7 + 4
<=> -x = -7
<=> x = 7
Vậy S = { 7 }
2.
<=> \(\frac{2\left(3x-1\right)}{6}\)= \(\frac{3\left(2-x\right)}{6}\)
<=> 2( 3x - 1 ) = 3( 2 - x )
<=> 6x -2 = 6 - 3x
<=> 6x + 3x = 6 + 2
<=> 9x = 8
<=> x = \(\frac{8}{9}\)
Vậy S = \(\left\{\frac{8}{9}\right\}\)
3.
<=> \(\frac{6x+10}{3}-\frac{x}{2}=5-\frac{3x+3}{4}\)
<=> \(\frac{4\left(6x+10\right)}{12}-\frac{6x}{12}=\frac{60}{12}-\frac{3\left(3x+3\right)}{12}\)
<=> 4( 6x + 10 ) - 6x = 60 - 3( 3x + 3 )
<=> 24x + 40 - 6x = 60 - 9x -9
<=> 18x + 40 = 51 - 9x
<=> 18x + 9x = 51 - 40
<=> 27x = 11
<=> x = \(\frac{11}{27}\)
Vậy S = \(\left\{\frac{11}{27}\right\}\)
<=>