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Giải:
a) ĐKXĐ: \(x\ne0\)
\(H=1-\dfrac{1}{x^2+4}\left(\left(1-\dfrac{x^2+4}{4x}\right):\left(\dfrac{1}{x}-\dfrac{1}{2}\right)\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\left(\dfrac{4x}{4x}-\dfrac{x^2+4}{4x}\right):\left(\dfrac{4}{4x}-\dfrac{2x}{4x}\right)\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\dfrac{4x-x^2+4}{4x}:\dfrac{4-2x}{4x}\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\dfrac{4x-x^2+4}{4x}.\dfrac{4x}{4-2x}\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{4x-x^2+4}{4-2x}\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{\left(2-x\right)^2}{2\left(2-x\right)}\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{2-x}{2}\)
\(\Leftrightarrow H=1-\dfrac{2-x}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2\left(x^2+4\right)}{2\left(x^2+4\right)}-\dfrac{2-x}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2\left(x^2+4\right)-\left(2-x\right)}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2x^2+8-2+x}{2x^2+8}\)
\(\Leftrightarrow H=\dfrac{2x^2+6+x}{2x^2+8}\)
b) Để \(H=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2x^2+6+x}{2x^2+8}=\dfrac{1}{4}\)
\(\Rightarrow4\left(2x^2+6+x\right)=2x^2+8\)
\(\Leftrightarrow8x^2+24+4x=2x^2+8\)
\(\Leftrightarrow8x^2+24+4x-2x^2-8=0\)
\(\Leftrightarrow6x^2+4x+16=0\)
Đến đây phân tích đa thức thành nhân tử rồi tìm nghiệm.
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a: \(M=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^2+1}{2}\)
a) \(M=\left(\dfrac{x^2-1}{x^4-x^2+1}-\dfrac{1}{x^2+1}\right).\left(x^4+\dfrac{1-x^4}{1+x^2}\right)\)
\(M=\left(\dfrac{(x^2-1)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{(x^4-x^2+1)\left(x^2+1\right)}\right).\left(\dfrac{x^6+x^4+1-x^4}{1+x^2}\right)\)
\(M=\left(\dfrac{x^4-1-\left(x^4-x^2+1\right)}{x^6+1}\right).\left(\dfrac{x^6+1}{1+x^2}\right)\)
\(M=\left(\dfrac{x^2-2}{x^6+1}\right).\left(\dfrac{x^6+1}{1+x^2}\right)\)
\(M=\dfrac{x^2-2}{x^2+1}\)
b) Ta có:
\(x^2\ge0\left(\forall x\right)\)
\(x^2-2\ge-2\)
\(\dfrac{x^2-2}{x^2+1}\ge-2\)
do mẫu \(x^2+1\) lớn hơn 0 nên chia ko cần đổi dấu
\(\Rightarrow M\ge-2\)