Dùng định lí Viète vào pt cho ta:
\(\left\{{}\begin{matrix}S=x_1+x_2=2\\P=x_1x_2=\dfrac{1}{3}\end{matrix}\right.\)

a) \(A=\left(x_1-1\right)\left(x_2-1\right)=x_1x_2-\left(x_1+x_2\right)+1=-\dfrac{2}{3}\)

b)\(B=x_1\left(x_2-1\right)+x_2\left(x_1-1\right)=2x_1x_2-\left(x_1+x_2\right)=-\dfrac{4}{3}\)

c)\(C=\sqrt{x_1}+\sqrt{x_2}=\sqrt{\left(\sqrt{x_1}+\sqrt{x_2}\right)^2}=\sqrt{x_1+x_2+2\sqrt{x_1x_2}}=\sqrt{2+2\sqrt{\dfrac{1}{3}}}\)

Tới đó hết giải được tiếp :)
d)\(D=x_1\sqrt{x_2}+x_2\sqrt{x_1}=\sqrt{x_1x_2}.\left(\sqrt{x_1}+\sqrt{x_2}\right)\) rồi thế kết quả câu C và biểu thức từ trên.

`x^2+3x+1=(x+3)sqrt{x^2+1}`
`<=>x(x+3)+1=(x+3)sqrt{x^2+1}`
`<=>(x+3)(sqrt{x^2+1}-x)=1`
`<=>((x+3)(sqrt{x^2+1-x)(sqrt{x^2+1}+x))/(sqrt{x^2+1}+x)=1`
`<=>(x+3)/(sqrt{x^2+1}+x)=1`
`<=>x+3=sqrt{x^2+1}+x`
`<=>sqrt{x^2+1}=3`
`<=>x^2+1=9`
`<=>x^2=8`
`<=>x=+-2sqrt2`
Vậy `S={2sqrt2,-2sqrt2}`

`x^2+3x+1=(x+3)\sqrt(x^2+1)`

`<=> x^4+6x^3+11x^2+6x+1=(x^2+6x+9)(x^2+1)`

`<=> x^4+6x^3+11x^2+6x+1=x^4+6x^3+10x^2+6x+9`

`<=> 11x^2+1=10x^2+9`

`<=> x^2=8`

`<=> x=\pm 2\sqrt2`.

Sửa lại nha lỗi quá '-'

`x^2+3x+1=(x+3)sqrt{x^2+1}`
`<=>x(x+3)+1=(x+3)sqrt{x^2+1}`
`<=>(x+3)(sqrt{x^2+1}-x)=1`
`<=>((x+3)(sqrt{x^2+1}-x)(sqrt{x^2+1}+x))/(sqrt{x^2+1}+x)=1`
`<=>(x+3)/(sqrt{x^2+1}+x)=1`
`<=>x+3=sqrt{x^2+1}+x`
`<=>sqrt{x^2+1}=3`
`<=>x^2+1=9`
`<=>x^2=8`
`<=>x=+-2sqrt2`
Vậy `S={2sqrt2,-2sqrt2}`

Lời giải:

Áp dụng định lý Viete cho pt bậc 2 ta có:

\(\left\{\begin{matrix} x_1+x_2=3\\ x_1x_2=1\end{matrix}\right.\)

Khi đó:

\(A=x_1\sqrt{x_1}+x_2\sqrt{x_2}=(\sqrt{x_1})^3+(\sqrt{x_2})^3\)

\(=(\sqrt{x_1}+\sqrt{x_2})(x_1-\sqrt{x_1x_2}+x_2)\)

\(=\sqrt{(\sqrt{x_1}+\sqrt{x_2})^2}(x_1+x_2-\sqrt{x_1x_2})\)

\(=\sqrt{x_1+x_2+2\sqrt{x_1x_2}}(x_1+x_2-\sqrt{x_1x_2})\)

\(=\sqrt{3+2}(3-1)=2\sqrt{5}\)

∆=9-4=5

x1=(3+√5)/2; x2=(3-√5)/2

4x1=(√5+1)^2; 4x2=(√5-1)^2

4.A=(3+√5)(√5+1)+(3-√5)(√5-1)

=(4√5+3+5)+(4√5-3-5)=8√5

A=2√5

≠ gi robot

CHÚC BẠN HỌC TỐT NHÁ

ĐK:x\(\ge\)0

Đặt t=x²+3x(t\(\ge\) 0)ta được:

\(\sqrt{t+12}=t\Leftrightarrow t^2=t+12\)

<=>t²-t-12=0

\(\Delta=49\Rightarrow\sqrt{\Delta}=7\)

\(\Delta>0,\text{phương trình có 2 nghiệm phân biệt}\)

\(t_1=4\left(thỏa\right);t_2=-3\left(loại\right)\)

t=4=>x²+3x=4

<=>x²+3x-4=0

\(\Delta=25\Rightarrow\sqrt{\Delta}=5;\Delta>0,pt\text{ có 2 nghiệm phân biệt:}\)

\(x_1=1\left(thỏa\right);x_2=-4\left(loại\right)\)

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

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

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

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

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

\(\Leftrightarrow t=3-x\)

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

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

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

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

Đặt \(\sqrt{x^2+1}=t>0\)

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

\(\Delta=\left(x+3\right)^2-12x=\left(x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3+x-3}{2}=x\\t=\dfrac{x+3-x+3}{2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\left(x\ge0\right)\\\sqrt{x^2+1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\left(vô-nghiệm\right)\\x=\pm2\sqrt{2}\end{matrix}\right.\)

ĐK: Với mọi x thuộc R.

Ta có: \(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

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

\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=\left(x+3\right)^2\left(x^2+1\right)\)

\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=x^4+6x^3+10x^2+6x+9\)

\(\Leftrightarrow x^2-8=0\)

\(\Leftrightarrow x^2=8\)

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

Vậy....

a.

\(\Leftrightarrow2x^2\ge3\Leftrightarrow x^2\ge\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\le-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)

b.

\(\Leftrightarrow\left(1-x\right)\left(x-3\right)\ge0\Rightarrow1\le x\le3\)

c.

\(\Leftrightarrow\sqrt{1-3x}\le2-x\Leftrightarrow\left\{{}\begin{matrix}1-3x\ge0\\2-x\ge0\\1-3x\le x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\le2\\x^2-x+3\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{1}{3}\)