Cho các biểu thức: A=x+2/x-2 - x-2/x+2 và B= -16/x^2-4
a) Tìm x để A=B
b) Tìm x để A:B < 0
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29 tháng 8 2023
a) \(A=B\) khi
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=-16\)
\(\Leftrightarrow8x=-16\)
\(\Leftrightarrow x=\dfrac{-16}{8}\)
\(\Leftrightarrow x=-2\left(ktmdk\right)\)
b) \(A:B< 0\) khi:
\(\left(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}\right):\left(\dfrac{-16}{x^2-4}\right)< 0\)
\(\Leftrightarrow\left[\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\right]:\left[\dfrac{-16}{\left(x+2\right)\left(x-2\right)}\right]< 0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-16}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{x}{-2}< 0\)
Mà: -2 < 0
\(\Leftrightarrow x>0\)
So với đk:
Vậy: \(A:B< 0\) khi
\(x>0;x\ne2\)
29 tháng 8 2023
a: A=B
=>A-B=0
=>\(\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
=>x^2+4x+4-x^2+4x-4=-16
=>8x=-16
=>x=-2(loại)
b: A:B<0
=>\(\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
=>\(\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
=>\(\dfrac{-8x}{16}< 0\)
=>x>0
Kết hợp ĐKXĐ, ta được: x>0 và x<>2
S
9 tháng 1 2019
a, ĐKXĐ: \(x\ne0;x\ne\pm1\)
\(P=\left(\frac{2x}{x^2-1}+\frac{x-1}{2x+2}\right):\frac{x+1}{2x}=\left(\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{2\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(=\left(\frac{2x.2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(=\frac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}:\frac{x+1}{2x}=\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{2x}{x+1}=\frac{x}{x-1}\)
b,Để \(P=2\Leftrightarrow\frac{x}{x-1}=2\Leftrightarrow2\left(x-1\right)=x\Leftrightarrow2x-2-x=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)
Vậy để P=2 <=> x=2
a.
\(A=B\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)
\(\Leftrightarrow8x+16=0\)
\(\Leftrightarrow8\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn A=B
b.
\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{16}>0\)
\(\Leftrightarrow\dfrac{x}{2}>0\)
\(\Leftrightarrow x>0\)