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

\(\Leftrightarrow x^2+4+2\sqrt{x-1}-5x=0\)

\(\Leftrightarrow x^2-5x+2\sqrt{x-1}+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

Đk: \(x\ge1\)

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

\(\Leftrightarrow\left(x^2-4x+4\right)-\left[\left(x-1\right)-2\sqrt{x-1}+1\right]=0\)

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

\(\Leftrightarrow\left(x+\sqrt{x-1}-3\right)\left(x-\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x-1}-3=0\left(2\right)\\x-\sqrt{x-1}-1=0\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(x-1\right)+\sqrt{x-1}-2=0\)

\(\Leftrightarrow\left(x-1\right)-\sqrt{x-1}+2\sqrt{x-1}-2=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-1\right)+2\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}+2\right)=0\)

\(\Leftrightarrow\sqrt{x-1}-1=0\) (vì \(\sqrt{x-1}+2>0\))

\(\Leftrightarrow x=2\left(nhận\right)\)

\(\left(3\right)\Leftrightarrow\left(x-1\right)-\sqrt{x-1}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\left(nhận\right)\)

Vậy phương trình (1) có 2 nghiệm là \(x=1\text{v}àx=2\)

a: =>\(x\cdot\left(\sqrt{3}-1\right)=16\)

=>\(x=\dfrac{16}{\sqrt{3}-1}=8\left(\sqrt{3}+1\right)\)

b: =>(x-căn 15)^2=0

=>x-căn 15=0

=>x=căn 15

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1

\(\Leftrightarrow\left(x^2+2\right)\sqrt{x^2+x+1}-2\left(x^2+2\right)+x^3-x^2-5x+6=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(\sqrt{x^2+x+1}-2\right)+\left(x-2\right)\left(x^2+x-3\right)=0\)

\(\Leftrightarrow\dfrac{\left(x^2+2\right)\left(x^2+x-3\right)}{\sqrt{x^2+x+1}+2}+\left(x-2\right)\left(x^2+x-3\right)=0\)

\(\Leftrightarrow\left(x^2+x-3\right)\left(\dfrac{x^2+2}{\sqrt{x^2+x+1}+2}+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-3=0\Rightarrow x=...\\x^2+2=\left(2-x\right)\left(\sqrt{x^2+x+1}+2\right)\left(1\right)\end{matrix}\right.\)

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

Đặt \(\sqrt{x^2+x+1}=t>0\Rightarrow x^2=t^2-x-1\)

\(\Rightarrow t^2+x-3=\left(2-x\right)t\)

\(\Leftrightarrow t^2+\left(x-2\right)t+x-3=0\)

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

\(\Leftrightarrow\left(t+1\right)\left(t+x-3\right)=0\)

\(\Leftrightarrow t=3-x\)

\(\Leftrightarrow\sqrt{x^2+x+1}=3-x\) (\(x\le3\))

\(\Leftrightarrow x^2+x+1=x^2-6x+9\)

\(\Leftrightarrow x=\dfrac{8}{7}\)

Đặt \(\sqrt{x^2-2x+5}=t>0\)

\(\Rightarrow x^2-2x=t^2-5\)

Phương trình trở thành:

\(t=t^2-5-1\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-2x+5}=3\)

\(\Rightarrow x^2-2x+5=9\)

\(\Rightarrow x^2-2x-4=0\)

\(\Rightarrow...\)