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a)\(\dfrac{5}{2x-1}>0\)
ĐKXĐ: \(x\ne\dfrac{1}{2}\)
Để phân thức nhận giá trị lớn hơn 0 thì:\(2x-1>0\)\(\Leftrightarrow x>\dfrac{1}{2}\)
b) \(\dfrac{x-1}{2x^2+3}>0\)
Dễ dàng nhận thấy:
\(2x^2+3\ge3>0\) với \(\forall x\)
Để phân thức nhận giá trị lớn hơn 0 thì:
\(x-1>0\Leftrightarrow x>1\)
c)\(\dfrac{x-2}{x+3}>0\). ĐKXĐ: \(x\ne-3\)
Lập bảng xét dấu:
| \(x\) | \(-3\) \(2\) |
| \(x-2\) | \(-\) \(-\) \(0\) \(+\) |
| \(x+3\) | \(-\) \(0\) \(+\) \(+\) |
| \(\dfrac{x-2}{x+3}\) | \(+\) \(-\) \(+\) |
Vì \(\dfrac{x-2}{x+3}>0\) nên từ bảng xét dấu ta có:
\(x< -3\) hoặc \(x>2\)
d)\(\dfrac{5x^2+1}{x-3}< 0\) ĐKXĐ: \(x\ne3\)
Dễ dàng nhận thấy:
\(5x^2+1\ge1>0\) với \(\forall x\)
Để biểu thức nhận giá trị nhỏ hơn 0 thì:
\(x-3< 0\Leftrightarrow x< 3\)
\(a,a^3+3a^2+3a+1-27b^3\\ =\left(a+1\right)^3-\left(3b\right)^3\\ =\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)\left(3b\right)+\left(3b\right)^2\right]\\ =\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
\(c,x^6-x^4+2x^3+2x^2\\ =x^4\left(x^2-1\right)+2x^2\left(x+1\right)\\ =x^4\left(x+1\right)\left(x-1\right)+2x^2\left(x+1\right)\\ =x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\\ =x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
ĐK: \(x\ne1;x\ne-1\)
\(Q=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}-\dfrac{1}{\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\right)\left(x-1\right)\left(x+1\right)\)
\(Q=\left(\dfrac{x-1}{x+1}-\dfrac{1}{x+1}+\dfrac{x+1}{x-1}\right)\left(x-1\right)\left(x+1\right)\)
\(Q=\left(x-1\right)^2-\left(x-1\right)+\left(x+1\right)^2\)
\(Q=x^2-2x+1-x+1+x^2+2x+1=2x^2-x+3\)
c/ \(Q=2\left(x^2-\dfrac{1}{2}x\right)+3=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)-\dfrac{1}{8}+3\)
\(Q=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{23}{8}\ge\dfrac{23}{8}\)
\(\Rightarrow Q_{min}=\dfrac{23}{8}\) khi \(x=\dfrac{1}{4}\)
\(1.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{|\sqrt{7}+1|-|\sqrt{7}-1|}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(3a.x+1-\dfrac{x-1}{3}< x-\dfrac{2x+3}{2}+\dfrac{x}{3}+5\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)-2\left(x-1\right)}{6}< \dfrac{6x-3\left(2x+3\right)+2x+30}{6}\)
\(\Leftrightarrow6x+6-2x+2< 6x-6x-9+2x+30\)
\(\Leftrightarrow6x-2x-2x+6+2+9-30< 0\)
\(\Leftrightarrow2x-13< 0\)
\(\Leftrightarrow x< \dfrac{13}{2}\)
KL...............
\(b.5+\dfrac{x+4}{5}< x-\dfrac{x-2}{2}+\dfrac{x+3}{3}\)
\(\Leftrightarrow\dfrac{150+6\left(x+4\right)}{30}< \dfrac{30x-15\left(x-2\right)+10\left(x+3\right)}{30}\)
\(\Leftrightarrow150+6x+24< 30x-15x+30+10x+30\)
\(\Leftrightarrow6x-30x+15x-10x+150+24-30-30< 0\)
\(\Leftrightarrow-19x+114< 0\)
\(\Leftrightarrow x>6\)
KL..................
Câu 4 :
Ta có :
\(A=\dfrac{3}{1-x}+\dfrac{4}{x}\)
\(=\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\)
Theo BĐT Bu - nhi a - cốp xki ta có :
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\Leftrightarrow\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\ge\left(\sqrt{\dfrac{3\left(1-x\right)}{1-x}}+\sqrt{\dfrac{4x}{x}}\right)^2=\left(\sqrt{3}+2\right)^2=7+4\sqrt{3}\)
Dấu \("="\) xảy ra khi \(\dfrac{3}{\left(1-x\right)^2}=\dfrac{4}{x^2}\)
\(\Leftrightarrow3x^2=4x^2-8x+4\)
\(\Leftrightarrow x^2-8x+4=0\)
\(\Delta=64-16=48>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)
Vậy GTNN của\(A=7+4\sqrt{3}\) khi \(\left[{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)
1) \(\dfrac{x^2-4}{x^2+2x+1}:\dfrac{4-2x}{2x+2}=\dfrac{\left(x-2\right)\left(x+2\right)2\left(x+1\right)}{\left(x+1\right)^22\left(2-x\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x+1\right)}{-2\left(x-2\right)\left(x+1\right)\left(x+1\right)}=\dfrac{-\left(x+2\right)}{x+1}=\dfrac{-x-2}{x+1}\)
2) \(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)=\dfrac{x+1}{x+2}:\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}=\dfrac{x^2+6x+9}{x^2+4x+4}\)
Ta có:
\(x^2+2x=x\left(x+2\right)\)
\(x^2+6x+8=x^2+2x+4x+8=x\left(x+2\right)+4\left(x+2\right)=\left(x+2\right)\left(x+4\right)\)
\(x^2+10x+24=x^2+4x+6x+24=x\left(x+4\right)+6\left(x+4\right)=\left(x+4\right)\left(x+6\right)\)
\(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)
Phương trình trở thành:
\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=3\)
\(\Leftrightarrow2\left(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+...+\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=3\)
\(\Leftrightarrow2\left(\dfrac{1}{x}-\dfrac{1}{x+8}\right)=3\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{3}{2}\Leftrightarrow3x\left(x+8\right)=16\Leftrightarrow x^2+8x=\dfrac{16}{3}\Leftrightarrow x=0,6188021535\)
a) ĐKXĐ của A là \(x\ne1\)
\(A=\dfrac{x^2-1}{x^2-2x+1}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\)
ĐKXĐ của B là \(x\ne2;x\ne-2\)
\(B=\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right):\dfrac{6}{x-2}=\left(\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\right).\dfrac{x-2}{6}=\left(\dfrac{x^2-3x+2-x^2-3x-2}{\left(x+2\right)\left(x-2\right)}\right).\dfrac{x-2}{6}=\dfrac{-6x}{\left(x+2\right)\left(x-2\right)}.\dfrac{x-2}{6}=\dfrac{-x}{x+2}\)b)
Với \(x\ne1\)
\(A>1\Leftrightarrow A-1>0\Leftrightarrow\dfrac{x+1}{x-1}>0\)
TH1 \(\left\{{}\begin{matrix}x+1>0\\x-1>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x>1\end{matrix}\right.\)\(\Leftrightarrow x>1\)
TH2 \(\left\{{}\begin{matrix}x+1< 0\\x-1< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x< 1\end{matrix}\right.\)\(\Leftrightarrow x< -1\)
c) Với \(x\ne1;x\ne2;x\ne-2\)
\(A=B\Leftrightarrow\dfrac{x+1}{x-1}=\dfrac{-x}{x+2}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}+\dfrac{x}{x+2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2+3x+2+x^2-x=0\)
\(\Leftrightarrow2x^2-2x+2=0\)
\(\Leftrightarrow2\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x^2-x+1=0\) \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
Với mọi x ta luôn có \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=> ko có giá trị nào của x để A=B
c, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
\(=x^2\left(x+1\right)[x^2\left(x+1\right)-2x\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
d,
\(2x^3-x^2-1\)
\(=2x^3-2x^2+x^2-x+x-1\)
\(=2x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(2x^2+x+1\right)\)