\(x^2+6x+6m-m^2=0\left(1\right)\)

Áp dụng định lý Viet ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-6\\P=x_1.x_2=6m-m^2\end{matrix}\right.\)

\(\Delta'=9-6m+m^2=\left(m-3\right)^2\ge0,\forall m\in R\)

\(\Rightarrow\sqrt[]{\Delta'}=\left|m-3\right|\)

Phương trình \(\left(1\right)\) có 2 nhiệm phân biệt

\(\left[{}\begin{matrix}x_1=-3+\left|m-3\right|\\x_2=-3-\left|m-3\right|\end{matrix}\right.\)

\(\Rightarrow x_1-x_2=2\left|m-3\right|\)

Theo đề bài ta có :

\(x^3_1-x^3_2+2x^2_1+12x_1+72=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x^2_1+x^2_2+x_1.x_2\right)+2x^2_1+12x_1+72=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1.x_2\right]+2x^2_1+12x_1+72=0\)

\(\Leftrightarrow2\left|m-3\right|\left(36-6m+m^2\right)+2\left[-3+\left|m-3\right|\right]^2+12\left[-3+\left|m-3\right|\right]+72=0\)

\(\Leftrightarrow2\left|m-3\right|\left(9-6m+m^2+27\right)+2\left[-3+\left|m-3\right|\right]^2+12\left[-3+\left|m-3\right|\right]+72=0\)

\(\Leftrightarrow2\left|m-3\right|\left[\left(m-3\right)^2+27\right]+2\left[-3+\left|m-3\right|\right]^2+12\left[-3+\left|m-3\right|\right]+72=0\left(a\right)\)

- Với \(m>3\)

\(\left(a\right)\Leftrightarrow2\left(m-3\right)\left[\left(m-3\right)^2+27\right]+2\left[-3+m-3\right]^2+12\left[-3+m-3\right]+72=0\)

\(\Leftrightarrow2\left(m-3\right)\left[\left(m-3\right)^2+27\right]+2\left(m-6\right)^2+12\left(m-6\right)+72=0\)

Đặt \(t=m-3>0\)

\(pt\Leftrightarrow2t\left(t^2+27\right)+2\left(t-3\right)^2+12\left(t-3\right)+72=0\)

\(\Leftrightarrow2t^3+54t+2t^2-12t+18+12t-36+72=0\)

\(\Leftrightarrow2t^3+2t^2+54t+54=0\)

\(\Leftrightarrow2t^2\left(t+1\right)+54\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(2t^2+54\right)=0\)

\(\Leftrightarrow t+1=0\left(2t^2+54>0,\forall t\in R\right)\)

\(\Leftrightarrow t=-1\left(ktm\right)\)

- Với \(m< 3\)

\(\left(a\right)\Leftrightarrow2\left(3-m\right)\left[\left(3-m\right)^2+27\right]+2\left[-3-m+3\right]^2+12\left[-3-m+3\right]+72=0\)

\(\Leftrightarrow2\left(3-m\right)\left[\left(3-m\right)^2+27\right]+2m^2-12m+72=0\)

\(\Leftrightarrow2\left(3-m\right)\left[\left(3-m\right)^2+27\right]-2m\left(6-m\right)+72=0\)

Đặt \(t=3-m< 0\)

\(pt\Leftrightarrow2t\left(t^2+27\right)-2\left(3-t\right)\left(3+t\right)+72=0\)

\(\Leftrightarrow2t^3+54t-18+2t^2+72=0\)

\(\Leftrightarrow2t^3+2t^2+54t+54=0\)

\(\Leftrightarrow2t^2\left(t+1\right)+54\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(2t^2+54\right)=0\)

\(\Leftrightarrow t+1=0\left(2t^2+54>0,\forall t\in R\right)\)

\(\Leftrightarrow t=-1\)

\(\Leftrightarrow3-m=-1\)

\(\Leftrightarrow m=4\left(ktm\right)\)

- Với \(m=3\)

\(\left(a\right)\Leftrightarrow0+2.9-36+72=54=0\left(vô.lý\right)\)

\(\Rightarrow m=3\left(loại\right)\)

Vậy không có m nào để thỏa yêu cầu đề bài.

Cảm ơn cậu nhiều .

sai đenta bạn eiii, thế thì làm thế nào đc hả bạn :) muốn làm đc thì phân tích A thành (x₁-x₂)²-x₁x₂ xong thay theo Vi-ét là ok bạn nhé :)

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

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)

cái pt thứ 2 bạn nhân 2 vế vs x

Sau đó chuyển hết sang 1 vế,,,dùng máy băm nghiệm

x⁴+x³-6x³-6x²+6x²+6x+4x+4=0

học lớp 6 thôi,sai thôi nhé

Ta có (1-2x)y=-3²+5x-2

Do x nguyên nên 1-2x khác 0

=>y=\(\frac{3x^2-5x+2}{2x-1}\)<=>4y=\(\frac{12x^2-20x+8}{2x-1}\)=6x-7+\(\frac{1}{2x-1}\)

Do x,y là số nguyên =>\(\frac{1}{2x-1}\)là số nguyên,nên 2x-1 thuộc (1;-1).Từ đó tìm đc (x;y) là (1;0),(0;-2)

trả lời đúng có tích ko?

Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)

phương trình \(\Leftrightarrow2x^2\left(x+1003\right)^2+\left(\sqrt{2x+2007}-1\right)^2=0\)

x^4+2006^x^3+1006009x^2=-x+\(\sqrt{2x+2007}\)-1004

x^2(x+1003)^2=-x+2\(\sqrt{2x+2007}\)-1004

2x^2(x+1003)^2=-2x-2007+2\(\sqrt{2x+2007}\)-1 rồi tách hđt 1 vế âm 1 vế dương

`\sqrt{4-x^2}+x^2 > 4`         `ĐK: -2 <= x <= 2`

`<=>\sqrt{4-x^2}-(4-x^2) > 0`

`<=>\sqrt{4-x^2}(1-\sqrt{4-x^2}) > 0`

  Mà `\sqrt{4-x^2} > 0 AA -2 < x < 2`

  `=>1-\sqrt{4-x^2} > 0`

`<=>\sqrt{4-x^2} < 1`

`<=>4-x^2 < 1`

`<=>x^2 > 3`

`<=>[(x > \sqrt{3}),(x < -\sqrt{3}):}`

   Kết hợp `-2 < x < 2`

 ` =>[(-2 < x < -\sqrt{3}),(\sqrt{3} < x < 2):}`