Tìm Min B=\(\frac{x}{\sqrt{x-4}}\)
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cho 3 số dương a,b,c thỏa mãn abc = 1 và a+b+c > 1/a + 1/b + 1/. chứng minh rằng (a-1)(b-1)(c-1) > 0
PT \(\Leftrightarrow x^3+3x^2+x-2+\left(x+1\right)-\sqrt[3]{2x+3}=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-1\right)+\frac{\left(x+2\right)\left(x^2+x-1\right)}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{2x+3}+\left(\sqrt[3]{2x+3}\right)^2}=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-1\right)\left[1+\frac{1}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{2x+3}+\left(\sqrt[3]{2x+3}\right)^2}\right]=0\)
Cái ngoặc to yên tâm là vô nghiệm từ đó...
P/s: em chi có mỗi cách này thôi, ko biết có đúng không nữa..
Đặt \(\sqrt{x^2+1}=t>\left|x\right|\ge x\) (dễ dàng chứng minh.
PT \(\Leftrightarrow\) \(t^2-\left(x+3\right)t+3x=0\)
\(\Leftrightarrow\left(t-3\right)\left(t-x\right)=0\)
\(\Leftrightarrow t=3\Leftrightarrow\orbr{\begin{cases}x=2\sqrt{2}\\x=-2\sqrt{2}\end{cases}}\)
True?Em tính liên hợp nhưng thôi:v Mệt lắm
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=\sqrt{x+1}\left(4-3+2+1\right)=4\sqrt{x+1}\)
\(DK:x>4\)
\(\Rightarrow B>0\)
Ta co:
\(B^2=\frac{x^2}{x-4}\)
\(\Leftrightarrow x^2-B^2x+4B^2=0\)
Ta lai co:\(\Delta\ge0\)
\(\Leftrightarrow B^4-16B^2\ge0\)
\(\Leftrightarrow B^2\left(B+4\right)\left(B-4\right)\ge0\)
TH1:
\(\hept{\begin{cases}B+4\ge0\\B-4\ge0\end{cases}\Leftrightarrow B\ge4\left(n\right)}\)
TH2:
\(\hept{\begin{cases}B+4\le0\\B-4\le0\end{cases}\Leftrightarrow B\le-4\left(l\right)}\)
Dau '=' xay ra khi \(x=8\)
Vay \(B_{min}=4\)khi \(x=8\)