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\(1.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{|\sqrt{7}+1|-|\sqrt{7}-1|}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(3a.x+1-\dfrac{x-1}{3}< x-\dfrac{2x+3}{2}+\dfrac{x}{3}+5\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)-2\left(x-1\right)}{6}< \dfrac{6x-3\left(2x+3\right)+2x+30}{6}\)
\(\Leftrightarrow6x+6-2x+2< 6x-6x-9+2x+30\)
\(\Leftrightarrow6x-2x-2x+6+2+9-30< 0\)
\(\Leftrightarrow2x-13< 0\)
\(\Leftrightarrow x< \dfrac{13}{2}\)
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 \(\left[{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)
\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0
\(\Leftrightarrow x=2013\)
Vậy ........
\(x^2-5x+6=\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
1, <=>x^2-x-2 = x^2-4
<=>x^2-4-x^2+x+2 = 0
<=> x-2 = 0
<=> x=2
2, <=> (x-2).(x-3)=0
<=> x-2 = 0 hoặc x-3 = 0
<=> x=2 hoặc x=3
Từ \(x\ge2\) cộng cả hai vế với \(\dfrac{1}{2}\) ta được
\(x+\dfrac{1}{2}\ge2+\dfrac{1}{2}=\dfrac{5}{2}\)
\(VT=x+\dfrac{1}{2}=x-2+2+\dfrac{1}{2}=\left(x-2\right)+\dfrac{5}{2}\)
\(\left\{{}\begin{matrix}x\ge2\Rightarrow x-2\ge0\\VT=\left(x-2\right)+\dfrac{5}{2}\ge\dfrac{5}{2}=VP\rightarrow dpcm\end{matrix}\right.\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)
Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)
Lâu lắm không làm nhể
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)
Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
Khi đó \(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)
\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)
Suy ra \(x=3\)hoặc \(x=-3\)
\(\left(x+1\right)^2-\left(x-1\right)^2=6\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x+1+x-1\right)\left(x+1-x+1\right)=6\left(x^2+x+1\right)\)
\(\Leftrightarrow2x.2=6x^2+6x+6\)
\(\Leftrightarrow4x=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2x+6=0\)
Ta có \(\Delta=2^2-4.6.6< 0\)
Vậy pt vô nghiệm
\(\left(x+1\right)^2-\left(x-1\right)^2=6\left(x^2+x+1\right)\)
\(\Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)+\left(x-1\right)\right]=6\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=6x^2+6x+6\)
\(\Leftrightarrow2.2x=6x^2+6x+6\)\(\Leftrightarrow4x=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2x+6=0\)\(\Leftrightarrow3x^2+x+3=0\)( vô nghiệm vì \(1^2< 4.3.3\)hay \(1< 36\))
Vậy tập nghiệm của phương trình là \(S=\varnothing\)
Điều kiện x > 0
Ta có:
\(x=\sqrt{x-\dfrac{1}{x}}\sqrt{1-\dfrac{1}{x}}\)
\(\Leftrightarrow1=\dfrac{1}{\sqrt{x}}\left(1-\dfrac{1}{x^2}\right)+\dfrac{1}{x}\left(1-\dfrac{1}{x}\right)\)
Áp dụng bunhia ta có:
\(\dfrac{1}{\sqrt{x}}\left(1-\dfrac{1}{x^2}\right)+\dfrac{1}{x}\left(1-\dfrac{1}{x}\right)\le\sqrt{\left(\dfrac{1}{x}+1-\dfrac{1}{x}\right)\left(\dfrac{1}{x^2}+1-\dfrac{1}{x^2}\right)}=1\)
Dấu = xảy ra khi
\(\dfrac{1}{\sqrt{x}}.\dfrac{1}{x}=\sqrt{1-\dfrac{1}{x}}.\sqrt{1-\dfrac{1}{x^2}}\)
\(\Leftrightarrow x^3-x^2-x=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)