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

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a) \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=1+\sqrt{3}\)

\(pt\Leftrightarrow\sqrt{-x^2+2x+1+1}+\sqrt{-x^2-6x-9+1}=1+\sqrt{3}\)

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

Dễ thấy: \(VT\le2< 1+\sqrt{3}=VP\) (vô nghiệm)

b)\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)

\(pt\Leftrightarrow\sqrt{9x^2-6x+1+1}+\sqrt{45x^2-30x+5+4}=\sqrt{-9x^2+6x-1+9}\)

\(\Leftrightarrow\sqrt{\left(3x-1\right)^2+1}+\sqrt{5\left(3x-1\right)^2+4}=\sqrt{-\left(3x-1\right)^2+9}\)

Dễ thấy: \(VT\ge1+\sqrt{4}=3=VP\)

Đẳng thức xảy ra khi \(x=\dfrac{1}{3}\)

Đặt \(\sqrt{6x-9}=a\ge0\Rightarrow x=\frac{a^2+9}{6}\) pt trở thành:

\(\sqrt{\frac{a^2+9}{6}+a}+\sqrt{\frac{a^2+9}{6}-4a}=\sqrt{6}\)

\(\Leftrightarrow\sqrt{a^2+6a+9}+\sqrt{a^2-24a+9}=6\)

\(\Leftrightarrow a+3+\sqrt{a^2-24a+9}=6\)

\(\Leftrightarrow\sqrt{a^2-24a+9}=3-a\) (\(a\le3\))

\(\Leftrightarrow a^2-24a+9=a^2-6a+9\)

\(\Rightarrow a=0\Rightarrow\sqrt{6x-9}=0\Rightarrow x=\frac{3}{2}\)

Do ban đầu ko đặt ĐKXĐ nên phải thay nghiệm vào để thử, thấy đúng, vậy pt có nghiệm duy nhất \(x=\frac{3}{2}\)

a: ĐKXĐ: \(x^2-6x+6\ge0\)

=>\(x^2-6x+9-3\ge0\)

=>\(\left(x-3\right)^2-3\ge0\)

=>\(\left(x-3\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

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

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

TH1: \(\sqrt{x^2-6x+6}-3=0\)

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

=>\(x^2-6x+6=9\)

=>\(x^2-6x-3=0\)

=>\(x^2-6x+9-12=0\)

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

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)

TH2: \(\sqrt{x^2-6x+6}-1=0\)

=>\(x^2-6x+6=1\)

=>\(x^2-6x+5=0\)

=>(x-1)(x-5)=0

=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

\(x^2-x+8-4\sqrt{x^2-x+4}=0\)

=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)

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

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

=>\(\sqrt{x^2-x+4}=2\)

=>\(x^2-x+4=4\)

=>\(x^2-x=0\)

=>x(x-1)=0

=>x=0 hoặc x=1

c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)

=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)

=>\(\sqrt{x^2-3x+11}-3=0\)

=>\(\sqrt{x^2-3x+11}=3\)

=>\(x^2-3x+11=9\)

=>\(x^2-3x+2=0\)

=>(x-1)(x-2)=0

=>x=1(nhận) hoặc x=2(nhận)

\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)

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

\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)

Lần sau bạn chú ý viết đầy đủ đề.

1.

\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)

2.

\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)

Có thấy ẩn gì đâu bạn

ko có ẩn ak anh lớp 8