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ĐKXĐ: \(x\ne\pm2\)
Ta có : \(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+3}{4-x^2}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2-3x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+3x+2+x^2-3x+2-x^2-3}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+1}{x^2-4}\)
Vì \(x^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(\Rightarrow\)Để A không âm thì \(x^2-4>0\)(do \(x\ne\pm2\)nên \(x^2-4\ne0\))
\(\Leftrightarrow x^2>4\)
\(\Leftrightarrow\orbr{\begin{cases}x>2\\x< -2\end{cases}}\)
Vậy để A không âm thì \(x>2\)hoặc \(x< -2\)
(3x+2)(x-1)-3(x+1)(x-2)=4
<=>3x2-3x+2x-2+(-3x-3)(x-2)=4
<=>3x2-x-2-3x2+6x-3x+6=4
<=>2x+6=4
<=>2x=-2
<=>x=-1
1) 4x(x-5)-(x-1)(4x-3)=5
<=>4x2-20x-4x2+3x+4x-3=5
<=>-13x=8
<=>x=-8/13
Thôi mỏi tay quá tìm x luôn nha
2) x=1.875
3) x=17/7
Tuyet Anh Nguyen
1.a)(3x-2)(4x+5)=0
12x^2+7x-10=0>>x1=2/3,x2=-5/4
b)4x^3+2x^2+4x+2=0>>x=-1
c)0,23x^2-4,21x-13,8=0>>x1=21,14,x2=-2,8...
d)10x^3-13x^2-178x-35=0>>x1=5,x2=-1/5
b2/a)2x^3+5x^2-3x=0>>x1=1/2,x2=-3
b)(3x-1)(x^2-7x+12)=0>>x1=1/3,x2=4,x3=...
b3/
a)x^2+x-2=0>>x1=1,x2=-2
b)x1=-1,x2=-6
b4/a)0,5x^2-1,5x-1,5x^2+x+4,5x-3=0>>-x...
b)3x/7-1=3x/7-x>>x=1
c)2x^2-13x+15=0>>x1=5,x2=3/2
P/s: Tham khảo nha