Bài 9: Căn bậc ba

Ta có:

\(x=\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}\)

\(\Leftrightarrow x^3=2+\sqrt{3}+2-\sqrt{3}+3\left(\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}\right).\sqrt[3]{2+\sqrt{3}}.\sqrt[3]{2-\sqrt{3}}\)

\(\Leftrightarrow x^3=4+3x\)

\(\Leftrightarrow x^3-3x-4=0\)

Thế vào ta có:

\(M=2015\left(x^3-3x-5\right)^{2014}=2015\left[\left(x^3-3x-4\right)-1\right]^{2014}\)

\(=2015.\left(-1\right)^{2014}=2015\)

\(x=\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}\)

\(x^3=2+\sqrt{3}+2-\sqrt{3}+3.\sqrt[3]{2+\sqrt{3}}.\sqrt[3]{2-\sqrt{3}}\left(\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}\right)\)

\(x^3=4+3x\)

\(x^3-3x=4\)

\(M=2015\left(x^3-3x-5\right)^{2014}=2015\left(4-5\right)^{2014}=2015\)

Câu 1:

\(\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{\left(3x-1\right)^2}+\sqrt[3]{9x^2-1}=1\)

\(\Leftrightarrow\left(\sqrt[3]{3x+1}\right)^2+\left(\sqrt[3]{3x-1}\right)^2+\sqrt[3]{\left(3x-1\right)\left(3x+1\right)}=1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{3x-1}=m\end{matrix}\right.\), ta có hpt:

\(\left\{{}\begin{matrix}a^2+m^2+am=1\\a^3-m^3=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2+am+m^2=1\\\left(a-m\right)\left(a^2+am+m^2\right)=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2+am+m^2=1\left(1\right)\\a-m=2\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow a=m+2\). Thay vào (1)

\(\Rightarrow\left(m+2\right)^2+\left(m+2\right)m+m^2=1\)

\(\Leftrightarrow3m^2+6m+3=0\)

\(\Leftrightarrow3\left(m+1\right)^2=0\)

\(\Leftrightarrow m=-1\)

\(\Rightarrow\sqrt[3]{3x-1}=-1\)

\(\Leftrightarrow3x-1=-1\)

\(\Leftrightarrow x=0\)

Câu 2: Đặt ẩn phụ và giải hpt như câu 1 >v<"

A=\(\sqrt[3]{4+\sqrt{5}}\)-\(\sqrt[3]{4-\sqrt{5}}\)

\(^{A^3}\)= (\(\sqrt[3]{4+\sqrt{5}}-\sqrt[3]{4-\sqrt{5}}\))³

=(\(\sqrt[3]{4+\sqrt{5}}\))³ -\(\left(\sqrt[3]{4-\sqrt{5}}\right)^3\)\(-3\sqrt[3]{\left(4+\sqrt{5}\right)^2}.\sqrt[3]{4-\sqrt{5}}+3\sqrt[3]{\left(4-\sqrt{5}\right)^2}.\sqrt[3]{4+\sqrt{5}}\)

=4+\(\sqrt{5}\) -4+\(\sqrt{5}\)\(-3\sqrt[3]{4+\sqrt{5}}.\sqrt[3]{4-\sqrt{5}}\left(\sqrt[3]{4+\sqrt{5}}-\sqrt[3]{4-\sqrt{5}}\right)=2\sqrt{5}-3\sqrt[3]{\left(4+\sqrt{5}\right).\left(4-\sqrt{5}\right)}.A=2\sqrt{5}-3\sqrt[3]{16-5}.A=2\sqrt{5}-3\sqrt[3]{11}.A\Rightarrow A^3+3\sqrt[3]{11}.A-2\sqrt{5}=0\)

\(\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

=\(\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\dfrac{\sqrt[3]{4-\sqrt{15}}}{1}\)

=\(\dfrac{\left(\sqrt[3]{4-\sqrt{15}}\right)^2}{\sqrt[3]{4-\sqrt{15}}}\)

=\(\sqrt[3]{4-\sqrt{15}}\).

mình nhầm từ bước 3 thực xin lỗi:

=\(\dfrac{1+\sqrt[3]{\left(4-\sqrt{15}\right)^2}}{\sqrt[3]{4-\sqrt{15}}}\)

=\(=\dfrac{1+\sqrt[3]{31-8\sqrt{15}}}{\sqrt[3]{4-\sqrt{15}}}\)

đặt \(\sqrt{x+1}=a\),\(\sqrt[3]{x-2}=b\)

rồi suy ra hệ:

\(\left\{{}\begin{matrix}a+b=3\\a^2+b^3=2x-1\end{matrix}\right.\)

a) \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

\(P=\left[\dfrac{2\sqrt{x}}{x\left(\sqrt{x}-1\right)+\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)

\(P=\left[\dfrac{2\sqrt{x}-\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)

\(P=\left[\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)

\(P=\left[\dfrac{-\left(\sqrt{x}-1\right)}{\left(x+1\right)}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)

\(P=\left[\dfrac{-\left(\sqrt{x}-1\right)}{x+1+\sqrt{x}}\right]\)

b) điều kiện \(x+\sqrt{x}+1\ne0\)

Giả sử P\(\le0\)

\(\Rightarrow\left[\dfrac{-\left(\sqrt{x}-1\right)}{x+1+\sqrt{x}}\right]\le0\)

\(-(\sqrt{x}-1)\le0\)

\(\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)

Vậy khi x\(\le1\) thì P \(\le0\)

\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+1+\sqrt{x}}{x+1}=P\)

\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right).\dfrac{x+1}{x+1+\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x+1+\sqrt{x}}\)

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}.\dfrac{1}{x+1+\sqrt{x}}\)

\(=\dfrac{1-\sqrt{x}}{x+\sqrt{x}+1}\)

b/DKXD:x≥0,x≠1,x≠-1

\(P=\dfrac{1-\sqrt{x}}{x+\sqrt{x}+1}< 0\)

Vi x+can(x)+1>0 , P<0

\(=>1-\sqrt{x}< 0\)

=>x>1

Vay để P<0 thì x>1