b.
Ta có:
\(\dfrac{x}{1+\sqrt{1+x}}=\dfrac{x\left(\sqrt{1+x}-1\right)}{\left(1+\sqrt{1+x}\right)\left(\sqrt{1+x}-1\right)}=\dfrac{x\left(\sqrt{1+x}-1\right)}{x}=\sqrt{1+x}-1\)
\(\Rightarrow2+\dfrac{x}{1+\sqrt{1+x}}=1+\sqrt{1+x}\)
Do đó:
\(\dfrac{x}{2+\dfrac{x}{2+...}}=\dfrac{x}{1+\sqrt{1+x}}=\sqrt{1+x}-1\)
\(\Rightarrow\sqrt{1+x}-1=8\)
\(\Rightarrow\sqrt{1+x}=9\)
\(\Rightarrow x=80\)
177a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+1}=a\\\sqrt[3]{7-x}=b\end{matrix}\right.\) \(\Rightarrow a^3+b^3=8\)
Pt trở thành:
\(\left\{{}\begin{matrix}a+b=2\\a^3+b^3=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=2-a\\a^3+b^3=8\end{matrix}\right.\)
\(\Rightarrow a^3+\left(2-a\right)^3=8\)
\(\Leftrightarrow6a^2-12a=0\Rightarrow\left[{}\begin{matrix}a=0\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x+1}=0\\\sqrt[3]{x+1}=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)
177b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{25+x}=a\\\sqrt[3]{3-x}=b\end{matrix}\right.\) \(\Rightarrow a^3+b^3=28\)
Ta được hệ:
\(\left\{{}\begin{matrix}a+b=4\\a^3+b^3=28\end{matrix}\right.\)
\(\Rightarrow a^3+\left(4-a\right)^3=28\)
\(\Leftrightarrow12a^2-48a+36=0\Rightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{25+x}=1\\\sqrt[3]{25+x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}25+x=1\\25+x=27\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-24\\x=2\end{matrix}\right.\)
177c.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+3}=a\\\sqrt[3]{x-16}=b\end{matrix}\right.\) \(\Rightarrow a^3-b^3=19\)
Ta được hệ:
\(\left\{{}\begin{matrix}1+b=a\\a^3-b^3=19\end{matrix}\right.\)
\(\Rightarrow\left(1+b\right)^3-b^3=19\)
\(\Rightarrow3b^2+3b-18=0\)
\(\Rightarrow\left[{}\begin{matrix}b=2\\b=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x-16}=2\\\sqrt[3]{x-16}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-8\\x=-11\end{matrix}\right.\)
177d.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+3}=a\\\sqrt[3]{6-x}=b\end{matrix}\right.\) \(\Rightarrow a^3+b^3=9\)
Ta được hệ:
\(\left\{{}\begin{matrix}a-b=1\\a^3+b^3=9\end{matrix}\right.\)
\(\Rightarrow a^3+\left(a-1\right)^3=9\)
\(\Leftrightarrow2a^3-3a^2+2a-10=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a^2+a+5\right)=0\)
\(\Leftrightarrow a=2\)
\(\Rightarrow\sqrt[3]{x+3}=2\)
\(\Rightarrow x=5\)
178a.
ĐKXĐ: \(x\le12\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt[]{12-x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^3+b^2=36\)
ta được hệ:
\(\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)
\(\Rightarrow a^3+\left(6-a\right)^2=36\)
\(\Leftrightarrow a^3+a^2-12a=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-24\\x=-16\\x=-88\end{matrix}\right.\)
178b.
ĐKXĐ: \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-x}=a\\\sqrt[]{x-1}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^3+b^2=1\)
Ta được hệ:
\(\left\{{}\begin{matrix}a+b=1\\a^3+b^2=1\end{matrix}\right.\)
\(\Rightarrow a^3+\left(1-a\right)^2=1\)
\(\Leftrightarrow a^3+a^2-2a=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{2-x}=0\\\sqrt[3]{2-x}=1\\\sqrt[3]{2-x}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=10\end{matrix}\right.\)
178c.
ĐKXĐ: \(x\ge-1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-2}=a\\\sqrt[]{x+1}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^3-b^2=-3\)
Ta được hệ:
\(\left\{{}\begin{matrix}a+b=3\\a^3-b^2=-3\end{matrix}\right.\)
\(\Rightarrow a^3-\left(3-a\right)^2=-3\)
\(\Leftrightarrow a^3-a^2+6a-6=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+6\right)=0\)
\(\Leftrightarrow a=1\)
\(\Rightarrow\sqrt[3]{x-2}=1\)
\(\Rightarrow x=3\)
179a.
\(\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}\)
\(\Leftrightarrow2x+3\sqrt[3]{\left(x-1\right)\left(x+1\right)}.\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)=5x\)
\(\Leftrightarrow\sqrt[3]{x^2-1}.\sqrt[3]{5x}=x\)
\(\Leftrightarrow\left(x^2-1\right).5x=x^3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x^2-5=x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)
179b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+1}=a\\\sqrt[3]{x-1}=b\end{matrix}\right.\) \(\Rightarrow a^3-b^3=2\)
Ta được hệ:
\(\left\{{}\begin{matrix}a^2+b^2+ab=1\\a^3-b^3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+ab=1\\\left(a-b\right)\left(a^2+b^2+ab\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+ab=1\\a-b=2\end{matrix}\right.\)
\(\Rightarrow a^2+\left(a-2\right)^2+a\left(a-2\right)=1\)
\(\Leftrightarrow3a^2-6a+3=0\)
\(\Rightarrow a=1\)
\(\Rightarrow\sqrt[3]{x+1}=1\)
\(\Rightarrow x=0\)
179c.
Đặt \(\sqrt[3]{x+2}=a\)
\(\Rightarrow\sqrt[3]{a^3-1}+a+\sqrt[3]{a^3+1}=0\)
\(\Leftrightarrow\sqrt[3]{a^3-1}+\sqrt[3]{a^3+1}=-a\)
\(\Leftrightarrow2a^3+3\sqrt[3]{\left(a^3-1\right)\left(a^3+1\right)}\left(\sqrt[3]{a^3-1}+\sqrt[3]{a^3+1}\right)=-a^3\)
\(\Leftrightarrow2a^3+3\sqrt[3]{a^6-1}.\left(-a\right)=-a^3\)
\(\Leftrightarrow a\sqrt[3]{a^6-1}=a^3\)
\(\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow x=-2\\\sqrt[3]{a^6-1}=a^2\Rightarrow a^6-1=a^6\left(vn\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=-2\)
180.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{7-x}=a\\\sqrt[3]{x-5}=b\end{matrix}\right.\) \(\Rightarrow a^3-b^3=2\left(6-x\right)\)
Đồng thời ta cũng có:
\(a^3+b^3=2\) \(\Rightarrow a^3-b^3=\left(a^3+b^3\right)\left(6-x\right)\)
\(\Rightarrow6-x=\dfrac{a^3-b^3}{a^3+b^3}\)
Pt trở thành:
\(\dfrac{a-b}{a+b}=\dfrac{a^3-b^3}{a^3+b^3}\)
\(\Leftrightarrow\dfrac{a-b}{a+b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)\left(a^2-ab+b^2\right)}\)
\(\Leftrightarrow a-b=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2-ab+b^2}\)
TH1: \(a-b=0\Rightarrow7-x=x-5\)
\(\Rightarrow x=6\)
TH2: \(a^3+ab+b^2=a^3-ab+b^2\)
\(\Rightarrow ab=0\Rightarrow\left[{}\begin{matrix}\sqrt[3]{7-x}=0\\\sqrt[3]{x-5}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
181a.
ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{1-x}=a\ge0\\\sqrt[4]{1+x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^4+b^4=2\)
ta được:
\(\left\{{}\begin{matrix}a+b+ab=3\\a^4+b^4=2\end{matrix}\right.\)
Mà: \(2=a^4+b^4=a^4+1+b^4+1-2\ge2a^2+2b^2-2\)
\(\Rightarrow a^2+b^2\le2\) (1)
\(3=a+b+ab\le\dfrac{a^2+1}{2}+\dfrac{b^2+1}{2}+\dfrac{a^2+b^2}{2}=\dfrac{2\left(a^2+b^2\right)+2}{2}\)
\(\Rightarrow a^2+b^2\ge2\) (2)
Từ (1);(2) \(\Rightarrow a^2+b^2=2\)
Đẳng thức xảy ra khi và chỉ khi \(a=b=1\)
\(\Rightarrow\sqrt[4]{1-x}=\sqrt[4]{1+x}=1\Rightarrow x=0\)
181b.
ĐKXĐ: \(x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{1-x}=a\ge0\\\sqrt[4]{2-x}=b\ge1\end{matrix}\right.\)
\(\Rightarrow3-2x=a^4+b^4\)
Pt trở thành: \(a+b=\sqrt[4]{a^4+b^4}\)
\(\Leftrightarrow\left(a+b\right)^4=a^4+b^4\)
\(\Leftrightarrow2ab\left(2a^2+3ab+2b^2\right)=0\)
\(\Leftrightarrow ab=0\) (do \(b\ge1\Rightarrow2a^2+3ab+2b^2>0\))
\(\Leftrightarrow a=0\) (do \(b\ge1>0\))
\(\Leftrightarrow x=1\)
