ĐKXĐ: \(-1\le x\le1\)

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

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

Khi đó phương trình đề trở thành:

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

Vì \(2+\sqrt{1-x^2}>0\)nên ta có thể chia 2 vế cho \(2+\sqrt{1-x^2}\):

\(\Rightarrow\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{1+x}-\sqrt{1-x}\right)=\frac{1}{\sqrt{3}}\),Bình phương 2 vế:

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

\(\Leftrightarrow\left(1+\sqrt{1-x^2}\right)\left(2-2\sqrt{1-x^2}\right)=\frac{1}{3}\Leftrightarrow2\left(1+\sqrt{1-x^2}\right)\left(1-\sqrt{1-x^2}\right)=\frac{1}{3}\)\(\Leftrightarrow1-\left(1-x^2\right)=\frac{1}{3}\Leftrightarrow x^2=\frac{1}{6}\Leftrightarrow x=\pm\frac{1}{\sqrt{6}}\)

Ta xét phương trình đề: vế phải luôn không âm vì vậy vế trái phải không âm 

Khi đó \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\ge0\Leftrightarrow1+x\ge1-x\Leftrightarrow x\ge0\)

Vậy ta chỉ nhận nghiệm duy nhất là \(x=\frac{1}{\sqrt{6}}\)

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

545rfdff

dsd

bai nao cung kho zay bn co bai nao de de thi minh lam duoc chu bai nay thi minh chiu thoi!

chuc bn hoc gioi nha!

tth