Đặt \(P\left(x\right)=\frac{3(x-\sqrt{3})(x-\sqrt{5})}{(1-\sqrt{3})(1-\sqrt{5})}+\frac{4(x-1)(x-\sqrt{5})}{(\sqrt{3}-1)(\sqrt{3}-\sqrt{5})}+\frac{5(x-1)(x-\sqrt{3})}{(\sqrt{5}-1)(\sqrt{5}-\sqrt{3})}\)

Khi x=1 thì \(P\left(1\right)=\frac{3(1-\sqrt{3})(1-\sqrt{5})}{(1-\sqrt{3})(1-\sqrt{5})}+\frac{4(1-1)(1-\sqrt{5})}{(\sqrt{3}-1)(\sqrt{3}-\sqrt{5})}+\frac{5(1-1)(1-\sqrt{3})}{(\sqrt{5}-1)(\sqrt{5}-\sqrt{3})}\) =3

Khi x=\(\sqrt3\) thì \(P\left(\sqrt3\right)=\frac{3(\sqrt3-\sqrt{3})(\sqrt3-\sqrt{5})}{(1-\sqrt{3})(1-\sqrt{5})}+\frac{4(\sqrt3-1)(\sqrt3-\sqrt{5})}{(\sqrt{3}-1)(\sqrt{3}-\sqrt{5})}+\frac{5(\sqrt3-1)(\sqrt3-\sqrt{3})}{(\sqrt{5}-1)(\sqrt{5}-\sqrt{3})}=4\)

Khi x=\(\sqrt5\) thì \(P\left(\sqrt5\right)=\frac{3(\sqrt5-\sqrt{3})(\sqrt5-\sqrt{5})}{(1-\sqrt{3})(1-\sqrt{5})}+\frac{4(\sqrt5-1)(\sqrt5-\sqrt{5})}{(\sqrt{3}-1)(\sqrt{3}-\sqrt{5})}+\frac{5(\sqrt5-1)(\sqrt5-\sqrt{3})}{(\sqrt{5}-1)(\sqrt{5}-\sqrt{3})}\) =5

Đặt \(P\left(x\right)=a\cdot x^2+bx+c\)

Vì P(x) đi qua các điểm A(1;3); B\(\left(\sqrt3;4\right)\) ; \(C\left(\sqrt5;5\right)\) nên ta sẽ có:

\(\begin{cases}a\cdot1^2+b\cdot1+c=3\\ a(\sqrt{3})^2+b\sqrt{3}+c=4\\ a(\sqrt{5})^2+b\sqrt{5}+c=5\end{cases}\iff\begin{cases}a+b+c=3\quad(1)\\ 3a+\sqrt{3}b+c=4\quad(2)\\ 5a+\sqrt{5}b+c=5\quad(3)\end{cases}\)

=>a=1/2; b=0; c=5/2

=>\(P(x) = \frac{1}{2}x^2 + \frac{5}{2}\)

P(x)=3x

=>\(\frac12x^2+\frac52=3x\)

=>\(0,5x^2-3x+2,5=0\)

=>\(x^2-6x+5=0\)

=>(x-5)(x-1)=0

=>x=5 hoặc x=1

Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{4-x}=b\end{cases}}\)

PT <=> a + b + ab = 5 và a² + b² = 5

Tới đây thì đơn giản rồi

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Đặt \(x+y=a\)   ;  \(\sqrt{x+1}=b\)

Ta được hpt sau:

\(\left\{{}\begin{matrix}2a+b=4\\a-3b=-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+1}=2\)

\(\Leftrightarrow x=3\)

\(\Rightarrow y=-2\)

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 \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)

\(\Rightarrow a^2+4b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)