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

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

.

a: ĐKXĐ: x>=1

Đặt \(a=\sqrt{5x-1};b=\sqrt{x-1}\)

\(a^2-b^2=\left(5x-1\right)-\left(x-1\right)=4x\)

\(\frac{a^2+b^2}{2}=\frac{5x-1+x-1}{2}=\frac{6x-2}{2}=3x-1\)

\(ab=\sqrt{\left(5x-1\right)\left(x-1\right)}=\sqrt{5x^2-6x+1}\)

Thay vào phương trình ban đầu, ta được: \(\left(a+b\right)\left(\frac{a^2+b^2}{2}-ab\right)=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a+b\right)\left(a-b\right)^2}{2}=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a-b\right)^2}{2}=\left(a-b\right)\)

=>\(\left(a-b\right)^2=2\left(a-b\right)\)

=>(a-b)(a-b-2)=0

TH1: a-b=0

=>a=b

=>5x-1=x-1

=>x=0(loại)

TH2: a-b-2=0

=>\(\sqrt{5x-1}-\sqrt{x-1}-2=0\)

=>\(\sqrt{5x-1}=\sqrt{x-1}+2\)

=>\(5x-1=x-1+4+2\cdot2\cdot\sqrt{x-1}\)

=>\(4\sqrt{x-1}+x+3=5x-1\)

=>\(4\sqrt{x-1}=5x-1-x-3=4x-4\)

=>\(\sqrt{x-1}=x-1\)

=>\(\sqrt{x-1}\left(\sqrt{x-1}-1\right)=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=2\left(nhận\right)\end{array}\right.\)

bài này mình chưa giải dc triệt để ở cái cuối

\(2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\) \(\left(y\le\dfrac{3}{2}\right)\)

\(\Leftrightarrow4x^3-8x^2+6x-2=2x^3\left(4-2y\right)\sqrt{3-2y}\left(1\right)\)

\(đặt:\sqrt{3-2y}=a\ge0\Rightarrow a^2+1=4-2y\)

\(\left(1\right)\Leftrightarrow4x^3-8x^2+6x-2=2x^3.\left(a^2+1\right)a\)

\(\Leftrightarrow4x^3-8x^2+6x-2-2x^3\left(a^2+1\right)a\)

\(\Leftrightarrow-2\left(xa-x+1\right)\left[\left(xa\right)^2+x^2a+2x^2-xa-2x+1\right]=0\)

\(\Leftrightarrow x.a-x+1=0\Leftrightarrow x\left(a-1\right)=-1\Leftrightarrow x=\dfrac{-1}{a-1}\)

\(\left(\sqrt{x\sqrt{3-2y}-\sqrt{x}}\right) ^2=x\sqrt{3-2y}-\sqrt{x}\)

\(=\dfrac{-a}{a-1}-\sqrt{\dfrac{-1}{a-1}}\)

\(\left(\sqrt{x\sqrt{3-2y}+2}+\sqrt{x+1}\right)=\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\)

\(\Rightarrow\left(\dfrac{-a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\left(\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\right)-4=0\)

\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right).2\sqrt{\dfrac{a-2}{a-1}}=4\)

\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{\dfrac{a-2}{a-1}}=2\)

\(\Leftrightarrow\left(-1+\dfrac{-1}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{1-\dfrac{1}{a-1}}=2\)(3)

\(đặt:1-\dfrac{1}{a-1}=u\Rightarrow\sqrt{-\dfrac{1}{a-1}}=\sqrt{u-1}\)

\(\left(3\right)\Leftrightarrow\left(u-2-\sqrt{u-1}\right)\sqrt{u}=2\)

bình phương lên tính được u

\(\Rightarrow u=.....\Rightarrow a\Rightarrow y=...\Rightarrow x=....\)

Với \(x=0\) không phải nghiệm

Với \(x>0\) chia 2 vế cho pt đầu cho \(x^3\)

\(\Rightarrow2-\dfrac{4}{x}+\dfrac{3}{x^2}-\dfrac{1}{x^3}=2\left(2-y\right)\sqrt{3-2y}\)

\(\Leftrightarrow1-\dfrac{1}{x}+\left(1-\dfrac{1}{x}\right)^3=\sqrt{3-2y}+\sqrt{\left(3-2y\right)^3}\)

Xét hàm \(f\left(t\right)=t+t^3\Rightarrow f'\left(t\right)=1+3t^2>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow1-\dfrac{1}{x}=\sqrt{3-2y}\)

Thế vào pt dưới:

\(\left(\sqrt{x\left(1-\dfrac{1}{x}\right)-\sqrt{x}}\right)^2\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}+\sqrt{x+1}\right)=4\)

\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\left(\sqrt{x+1}+\sqrt{x+1}\right)=4\)

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

Phương trình này ko có nghiệm đẹp, chắc bạn ghi nhầm đề bài của pt dưới