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a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(A=\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}\cdot\left(\dfrac{1}{x^2+4x+4}+\dfrac{1}{4-x^2}\right)\)
\(=\dfrac{1}{x+2}-\dfrac{x\left(x+2\right)\left(x-2\right)}{x^2+4}\cdot\dfrac{x-2-x-2}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\dfrac{1}{x+2}-\dfrac{-4x}{\left(x+2\right)\left(x^2+4\right)}\)
\(=\dfrac{x^2+4+4x}{\left(x+2\right)\left(x^2+4\right)}\)
\(=\dfrac{x+2}{x^2+4}\)
b) Để A>0 thì x+2>0
hay x>-2 và \(x\ne2\)
Để A<0 thì x+2<0
hay x<-2
Để A=0 thì x+2=0
hay x=-2(loại)
tìm x biết :
a,(2x-3)^2 =(x+ 5)^2
b,x^2(x-1) -4x^2 +8x -4 =0
c, (x-4)^2 -36 =0
giúp mik nha mik đang gấp
a, (2x-3)^2=(x+5)^2
2x-3=x+5
2x-3-x-5=0
x-8=0
x=8
b, x^2(x-1)-4x^2+8x-4=0
x^2(x-1)-(4x^2-8x+4)=0
x^2(x-1)-4(x^2-2x+1)=0
x^2(x-1)-4(x-1)^2=0
(x-1)(x^2-4)(x-1)=0
(x-1)(x-2)(x+2)(x-1)=0
=>x-1=0=>x=1
=>x-2=0=>x=2
=>x+2=0=>x=-2
=>x-1=0=>x=1
Vậy : x=1 ;x=2 và x=-2
c, (x-4)^2-36=0
(x-4)^2-6^2=0
(x-4-6)(x-4+6)=0
(x-10)(x+2)=0
=>x-10=0=>x=10
=>x+2=0=>x=-2
Vậy : x=10 và x=-2
k đúng cho mình nhé bạn !
1. -4x( x + 3 )( x - 4 ) - 3x( x2 - x + 1 )
= -4x( x2 - x - 12 ) - 3x( x2 - x + 1 )
= -4x3 + 4x2 + 48x - 3x3 + 3x2 - 3x
= -7x3 + 7x2 + 45x
2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5
<=> 4x2 - 20x - ( 4x2 - 7x + 3 ) = 5
<=> 4x2 - 20x - 4x2 + 7x - 3 = 5
<=> -13x - 3 = 5
<=> -13x = 8
<=> x = -8/13
b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4
<=> 6( x2 - 7x + 12 ) - 6x2 + 12x = 4
<=> 6x2 - 42x + 72 - 6x2 + 12x = 4
<=> -30x + 72 = 4
<=> -30x = -68
<=> x = 34/15
Bài 1 :
\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)
\(=-7x^3+7x^2+45x\)
Bài 2 :
a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)
\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)
\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)
b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)
\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)
\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)
a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)
\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)
\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)
\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)
b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)
c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)
\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)