Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Đặt $\sqrt[3]{x+1}=a;\sqrt[3]{x-1}=b$ thì pt trở thành:
\(\left\{\begin{matrix} a^2+b^2+ab=1\\ a^3-b^3=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^2+ab+b^2=1\\ (a-b)(a^2+ab+b^2)=2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a^2+ab+b^2=1\\ a-b=2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} (a-b)^2+3ab=1\\ a-b=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a(-b)=1\\ a+(-b)=2\end{matrix}\right.\)
Theo đl Viet đảo thì $a,-b$ là nghiệm của pt $X^2-2X+1=0$
$\Rightarrow a=-b=1$
$\Leftrightarrow \sqrt[3]{x+1}=1; \sqrt[3]{x-1}=-1$
$\Rightarrow x=0$
Vậy.........
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))
\(\Leftrightarrow x=4y\)
Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)
\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)
a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}+1\)
=3
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
\(\dfrac{2\left(\sqrt{2}-\sqrt{6}\right)}{3\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{2\sqrt{2}\left(1-\sqrt{3}\right)}{3\cdot\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{4\left(1-\sqrt{3}\right)}{3\cdot\sqrt{4-2\sqrt{3}}}\)
\(=\dfrac{-4\left(\sqrt{3}-1\right)}{3\cdot\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{-4\left(\sqrt{3}-1\right)}{3\cdot\left(\sqrt{3}-1\right)}=-\dfrac{4}{3}\)
1) đkxđ \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\y\ge0\end{matrix}\right.\)
Xét biểu thức \(P=x^3+y^3+7xy\left(x+y\right)\)
\(P=\left(x+y\right)^3+4xy\left(x+y\right)\)
\(P\ge4\sqrt{xy}\left(x+y\right)^2\)
Ta sẽ chứng minh \(4\sqrt{xy}\left(x+y\right)^2\ge8xy\sqrt{2\left(x^2+y^2\right)}\) (*)
Thật vậy, (*)
\(\Leftrightarrow\left(x+y\right)^2\ge2\sqrt{2xy\left(x^2+y^2\right)}\)
\(\Leftrightarrow\left(x+y\right)^4\ge8xy\left(x^2+y^2\right)\)
\(\Leftrightarrow x^4+y^4+6x^2y^2\ge4xy\left(x^2+y^2\right)\) (**)
Áp dụng BĐT Cô-si, ta được:
VT(**) \(=\left(x^2+y^2\right)^2+4x^2y^2\ge4xy\left(x^2+y^2\right)\)\(=\) VP(**)
Vậy (**) đúng \(\Rightarrowđpcm\). Do đó, để đẳng thức xảy ra thì \(x=y\).
Thế vào pt đầu tiên, ta được \(\sqrt{2x-3}-\sqrt{x}=2x-6\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{2x-3}+\sqrt{x}}=2\left(x-3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}=2\end{matrix}\right.\)
Rõ ràng với \(x\ge\dfrac{3}{2}\) thì \(\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}\le\dfrac{1}{\sqrt{\dfrac{2.3}{2}-3}+\sqrt{\dfrac{3}{2}}}< 2\) nên ta chỉ xét TH \(x=3\Rightarrow y=3\) (nhận)
Vậy hệ pt đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(3;3\right)\)
\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)
\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)
\(x^3=\)
\(4-3\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)
\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)
\(x^3=4-3.2x\)
\(x^3=4-6x\)
thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)