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Ta có: \(x=\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}+\sqrt[3]{\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}\)
=> \(x^3=\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+3\cdot\sqrt[3]{\left(\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)\left(\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)}\cdot x.\)
= \(-b+\sqrt[3]{\frac{b^2}{4}-\left(\frac{b^2}{4}+\frac{a^3}{27}\right)}\cdot x\)
=\(-b+\sqrt[3]{\frac{a^3}{27}}\cdot x=-b+\frac{a}{27}\cdot x\)
=> \(x^3+b=\frac{a}{27}\cdot x\)
Vậy \(x^3+ax+b=\frac{a}{27}\cdot x+ax=\frac{28a}{27}\cdot x\)
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)
Bài 1:
a) Ta có: \(\sqrt{243}-\frac{1}{2}\sqrt{12}-2\sqrt{75}+\sqrt{27}\)
\(=\sqrt{3}\cdot9-\frac{1}{2}\cdot\sqrt{3}\cdot2-2\cdot\sqrt{3}\cdot5+\sqrt{3}\cdot3\)
\(=\sqrt{3}\left(9-1-10+3\right)\)
\(=\sqrt{3}\cdot1=\sqrt{3}\)
b) Ta có: \(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{5}{1+\sqrt{6}}-6\sqrt{\frac{1}{6}}\)
\(=\frac{\left(2\sqrt{3}-3\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)}+\frac{5\cdot\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\sqrt{36\cdot\frac{1}{6}}\)
\(=-\sqrt{6}+\frac{5\left(\sqrt{6}-1\right)}{5}-\sqrt{6}\)
\(=-2\sqrt{6}+\sqrt{6}-1\)
\(=-\sqrt{6}-1\)
Bài 2: Rút gọn
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Bất đẳng thức cần chứng minh tương đương \(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}}{\sqrt[3]{\frac{1}{\left(a+b\right)^3}+\frac{1}{\left(b+c\right)^3}+\frac{1}{\left(c+a\right)^3}}}\le2.\sqrt{2}.\sqrt[3]{9}\)
Ta quy bài toán về chứng minh hai bất đẳng thức sau
\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\le3\sqrt{2}\)và \(\sqrt[3]{\frac{1}{\left(a+b\right)^3}+\frac{1}{\left(b+c\right)^3}+\frac{1}{\left(c+a\right)^3}}\ge\frac{\sqrt[3]{3}}{2}\)
Áp dụng bất đẳng thức Bunyakovsky ta được \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\le\sqrt{6\left(a^2+b^2+c^2\right)}\)\(\le\sqrt{6\sqrt{3\left(a^4+b^4+c^4\right)}}\le3\sqrt{2}\)
Mặt khác ta lại có \(\left[\left(x^3+y^3+z^3\right)\left(x+y+z\right)\right]^2\ge\left(x^2+y^2+z^2\right)^4\); \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
Do đó ta được \(\left(x^3+y^3+z^3\right)^2\ge\frac{\left(x^2+y^2+z^2\right)^3}{3}\)
Áp dụng kết quả trên ta thu được \(\left[\frac{1}{\left(a+b\right)^3}+\frac{1}{\left(b+c\right)^3}+\frac{1}{\left(c+a\right)^3}\right]^2\ge\frac{1}{3}\left[\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\right]^3\)
Mà theo bất đẳng thức Cauchy-Schwarz ta có\(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\ge\frac{1}{2\left(a^2+b^2\right)}+\frac{1}{2\left(b^2+c^2\right)}+\frac{1}{2\left(c^2+a^2\right)}\) \(\ge\frac{9}{4\left(a^2+b^2+c^2\right)}\ge\frac{9}{4\sqrt{3\left(a^4+b^4+c^4\right)}}\ge\frac{9}{4\sqrt{9}}=\frac{3}{4}\)
Do đó ta có \(\left[\frac{1}{\left(a+b\right)^3}+\frac{1}{\left(b+c\right)^3}+\frac{1}{\left(c+a\right)^3}\right]^2\ge\frac{1}{3}\left[\frac{3}{4}\right]^3=\frac{9}{64}\)
Suy ra \(\sqrt[3]{\frac{1}{\left(a+b\right)^3}+\frac{1}{\left(b+c\right)^3}+\frac{1}{\left(c+a\right)^3}}\ge\frac{\sqrt[3]{3}}{2}\)
Từ các kết quả trên ta được \(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}}{\sqrt[3]{\frac{1}{\left(a+b\right)^3}+\frac{1}{\left(b+c\right)^3}+\frac{1}{\left(c+a\right)^3}}}\le\frac{3\sqrt{2}}{\frac{\sqrt[3]{3}}{2}}=2.\sqrt{2}.\sqrt[3]{9}\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi a = b = c = 1
<=> x3 = -b/2 + -b/2 + 3x.\(\sqrt[3]{\left(\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right).\left(\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)}\)
<=> x³ = -b + 3x.\(\sqrt[3]{\left(\frac{-b}{2}\right)^2-\frac{b^2}{4}-\frac{a^3}{27}}\)
<=> x³ = -b + 3x.\(\sqrt[3]{\frac{-a^3}{27}}\)
<=> x³ = -b + 3x\(\frac{-a}{3}\)
<=> x³ = -b - ax
=> Q = -b - ax + ax + b = 0