Điều kiện xác định của pt : \(6x^2-12x+7\ge0\) => Với mọi số thực thì pt xác định

Ta có : \(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\Leftrightarrow-\left(6x^2-12x+7\right)+6\sqrt{6x^2-12x+7}+7=0\)

Đặt \(t=\sqrt{6x^2-12x+7},t\ge0\) . pt trở thành : \(-t^2+6t+7=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}t=7\left(\text{nhận}\right)\\t=-1\left(\text{loại}\right)\end{array}\right.\)

Với \(t=7\) ta có pt : \(6x^2-12x+7=49\)

\(\Leftrightarrow6x^2-12x-42=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=1-2\sqrt{2}\\x=1+2\sqrt{2}\end{array}\right.\)

\(pt\Leftrightarrow\sqrt{6\left(x^2-2x\right)+7}=x^2-2x\)

Đặt \(t=2x-x^2\left(t\ge0\right)\) pt trở thành

\(\sqrt{6t+7}=t\).Ta có 2 vế dương bình phương đc:

\(6t+7=t^2\)

\(\Leftrightarrow t^2-6t-7=0\)

\(\Leftrightarrow t^2-7t+t-7=0\)

\(\Leftrightarrow t\left(t-7\right)+\left(t-7\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-1\left(loai\right)\\t=7\left(tm\right)\end{array}\right.\).

Từ t=7 ta tìm được các giá trị của \(\left[\begin{array}{nghiempt}x=1-\sqrt{8}\\x=\sqrt{8}+1\end{array}\right.\)

ĐKXĐ: \(0\le x\le6\)

Đặt \(\sqrt{6x-x^2}=t\ge0\)

\(t-2t^2+15=0\Leftrightarrow-2t^2+t+15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-\frac{5}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{6x-x^2}=3\Leftrightarrow x^2-6x+9=0\Rightarrow x=3\)

https://olm.vn/hoi-dap/question/1102059.html

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

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

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

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

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

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

a/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+5x+2}=2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)\)

\(\Leftrightarrow6x^2+15x-26=0\)

b/ ĐKXĐ: ...

Đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)

\(a+\frac{1}{a}=\frac{5}{2}\Leftrightarrow a^2-\frac{5}{2}a+1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[5]{\frac{16x}{x-1}}=2\\\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}16x=32\left(x-1\right)\\16x=\frac{1}{32}\left(x-1\right)\end{matrix}\right.\)

c/ĐKXĐ: ...

\(\Leftrightarrow x^2-2x-\sqrt{6x^2-12x+7}=0\)

Đặt \(\sqrt{6x^2-12x+7}=a\ge0\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

\(\frac{a^2-7}{6}-a=0\Leftrightarrow a^2-6a-7=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=7\end{matrix}\right.\) \(\Rightarrow\sqrt{6x^2-12x+7}=7\)

\(\Leftrightarrow6x^2-12x-42=0\)

d/ \(\Leftrightarrow x^2+x+4-\sqrt{x^2+x+4}-2=0\)

Đặt \(\sqrt{x^2+x+4}=a>0\)

\(a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+x+4}=2\Rightarrow x^2+x=0\)

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

Đặt \(\sqrt{3x^2+6x+4}=a>0\Rightarrow x^2+2x=\frac{a^2-4}{3}\)

\(\frac{a^2-4}{3}+a-2=0\)

\(\Leftrightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2+6x+4}=2\Rightarrow3x^2+6x=0\)

ĐKXĐ:...

a/ \(\sqrt{2x^2+5x+2}=1+2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)+1+4\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow3\left(2x^2+6x-6\right)+4\sqrt{2x^2+5x-6}-7=0\)

Đặt \(\sqrt{2x^2+5x-6}=a\ge0\)

\(3a^2+4a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+5x-6}=1\)

\(\Leftrightarrow2x^2+5x-7=0\)

Đặt \(\sqrt{6x^2-12x+7}=t\left(t\ge0\right)\)

<=>\(t^2-7=6x^2-12x\)

\(\Leftrightarrow\dfrac{t^2-7}{6}=x^2-2x\)

Ta có pt mới:

\(\dfrac{7-t^2}{6}+t=0\)

\(\Leftrightarrow t^2-6t-7=0\)

\(\Leftrightarrow t^2-2\cdot t\cdot3+9-9-7=0\)

\(\Leftrightarrow\left(t-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}t=7\\t=-1\end{matrix}\right.\)(loại t=-1)

Với t=7

=>\(\sqrt{6x^2-12x+7}=7\)

<=>6x²-12x+7=49

<=>6x²-12x-42=0

<=>x²-2x-7=0

<=>(x-1)²=8

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