Giải các phương trình sau:
a \(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
b \(\sqrt[3]{\left(65+x\right)^2}+4\sqrt[3]{\left(65-x\right)^2}=5\sqrt[3]{65^2-x^2}\)
c \(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
d \(\sqrt[3]{x-2}+\sqrt[3]{x+3}=\sqrt[3]{2x+1}\)
e \(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}\)
a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x+1}=b\end{matrix}\right.\)
\(\Rightarrow a+b=1+ab\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=1\\b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=1\\x+1=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-2}=a\\\sqrt[3]{x+3}=b\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt[3]{a^3+b^3}\)
\(\Leftrightarrow\left(a+b\right)^3=a^3+b^3\)
\(\Leftrightarrow ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a=-b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=0\\b^3=0\\a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\\x-2=-\left(x+3\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
e.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2x-1}=a\\\sqrt[3]{x-1}=b\end{matrix}\right.\)
Ta có: \(3x+1=4\left(2x-1\right)-5\left(x-1\right)\) nên pt trở thành:
\(a+b=\sqrt[3]{4a^3-5b^3}\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=4a^3-5b^3\)
\(\Leftrightarrow a^3-ab\left(a+b\right)-2b^3=0\)
\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+b^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\a=b=0\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow a^3=8b^3\)
\(\Leftrightarrow2x-1=8\left(x-1\right)\)
\(\Leftrightarrow...\)
cảm ơn tất cả moi người,đó là bài cuối rồi,chúc mọi người cuối tuần vui vẻ
