\(\sqrt{6x-x^2}+2x^2-12+15=0\left(x;\left[0;6\right]\right)\)

<=> \(2\left(6x-x^2\right)-\sqrt{6x-x^2}-15=0\)

\(\Delta_{\left(\sqrt{6x-x^2}\right)}=1+4.2.15=121=11^2\)

\(\sqrt{6x-x^2}=\dfrac{1-11}{4}=\dfrac{-5}{2}\left(l\right)\)

\(\sqrt{6x-x^2}=\dfrac{1+11}{4}=3\Leftrightarrow6x-x^2=9\)

\(\Leftrightarrow\left(x-3\right)^2=0;x=3\left(n\right)\)

mik vẫn k hiểu ở dòng thứ 2

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

a) - 3 =0

<=> (√x-3)(√x+3)-3√x-3=0

<=> (√x-3)(√x+3-3)=0

<=> (√x-3)√x=0

<=> √x-3=0

<=>x=9

b)=x - 3

<=> √(2x -3)² =x-3

<=> 2x-3=x-3

<=>2x-x=-3+3

<=>x=0

c)=3x-1

<=> √(x+3)² =3x-1

<=> x+3=3x-1

<=> -2x=-4

<=>  x=2

Nhớ cho mình 1 tim nha bạn

Sau em nên gõ các kí hiệu toán học ở phần Σ để mọi người dễ dàng đọc hơn nhé.

ĐK: \(\hept{\begin{cases}x\ge2\\y\ge-\frac{1}{3}\end{cases}}\)

\(\sqrt{x-2}+x^3-6x^2+12x=\sqrt{3y+1}+27y^3+27y^2+9y+9\)

<=> \(\sqrt{x-2}+x^3-6x^2+12x-8=\sqrt{3y+1}+27y^3+27y^2+9y+1\)

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

<=> \(\left(\sqrt{x-2}-\sqrt{3y+1}\right)+\left[\left(x-2\right)^3-\left(3y+1\right)^3\right]=0\)

<=> \(\frac{x-3y-3}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-3y-3\right)\left[\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right]=0\)

<=> \(\left(x-3y-3\right)\left(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right)=0\)

<=> \(x-3y-3=0\)

vì \(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2>0\)

<=> x = 3y + 3

Thế vào phương trình trên ta có: 

\(2+2\left(3y+3\right)^2-2y^2+3\left(3y+3\right)y-4\left(3y+3\right)-3y=0\)

<=> \(25y^2+30y+8=0\Leftrightarrow\orbr{\begin{cases}y=-\frac{2}{5}\\y=-\frac{4}{5}\end{cases}}\)không thỏa mãn đk 

Vậy hệ vô nghiệm.

gái xinh

\(ĐKXĐ:0\le x\le6\)

\(\Leftrightarrow\sqrt{6x-x^2}-2\left(6x-x^2\right)+15=0\)

Đặt \(\sqrt{6x-x^2}=t\left(t\ge0\right)\)

PT trở thành:

\(2t^2-t-15=0\)

\(\Leftrightarrow\left(t-3\right)\left(2t+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=\frac{-5}{2}\end{cases}}\)

\(TH1:t=3\Rightarrow\sqrt{6x-x^2}=3\Rightarrow6x-x^2=9\)

\(\Leftrightarrow x^2-6x+9=0\)

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

\(\Leftrightarrow x=3\)

\(TH2:t=\frac{-5}{2}\)không TMĐK \(t\ge0\)

Vậy PT có nghiệm là \(S=\left\{3\right\}\)

\(x^3-2\sqrt{2}x^2+6x-4\sqrt{2}=0\)

\(\Leftrightarrow\left(x^3-\sqrt{2}x^2+4x\right)-\left(\sqrt{2}x^2+2x-4\sqrt{2}\right)=0\)

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

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

\(\Leftrightarrow x=\sqrt{2}\)