a.

\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:

\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)

\(\Leftrightarrow3a^2=b^2\)

\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)

\(\Leftrightarrow...\)

b.

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

Lặp lại cách làm câu a

hết cứu đi mà làm

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)

\(x^4-16x^2+32x-16=0\)

\(\Leftrightarrow x^4-2x^3+2x^3-4x^2-12x^2+24x+8x-16=0\)

\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)-12x\left(x-2\right)+8\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x^2-4x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)

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

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

Vậy.............

\(x^4-16x^2+32x-16=0\)

\(\Leftrightarrow x^4-16\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow x^4-16\left(x-1\right)^2=0\)

\(\Leftrightarrow x^4-\left(4\left(x-1\right)\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\left(x-1\right)\right).\left(x^2+4\left(x-1\right)\right)=0\)

\(\Leftrightarrow\left(x^2-4x+4\right).\left(x^2+4x-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2.\left(x^2+4x-4\right)=0\)

\(\Leftrightarrow\)\(\left(x-2\right)^2=0\) hoặc \(x^2+4x-4=0\)

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

\(2\)) \(x^2+4x-4=0\Leftrightarrow x^2+4x+4-8=0\)

\(\Leftrightarrow\left(x+2\right)^2=8\)

\(\Leftrightarrow x+2=\sqrt{8}\) hoặc \(x+2=-\sqrt{8}\)

\(\Leftrightarrow x=\sqrt{8}-2\) \(x=-\sqrt{8}-2\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;\sqrt{8}-2;-\sqrt{8}-2\right\}\)

\(x^4-16x^2+32x-16=0\)

\(\Leftrightarrow\left(x^4-16\right)-\left(16x^2-32x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2+4\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x^2-4x+8\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x-4=0\end{matrix}\right.\)

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

Vậy x có tập \(n_o\) \(S=\left\{2;-2+2\sqrt{2};-2-2\sqrt{2}\right\}\)

3x⁴-13x³+16x²-13x+3=0

⇔(x-3)(3x³-4x²+4x-1)=0

⇔(x-3)(x-\(\dfrac {1}{3}\))(3x²-3x+3)=0

⇔3(x-3)(x-\(\dfrac{1}{3}\))(x²-x+1)=0

⇔x-3=0 hoặc x-1/3=0

⇔x=3 hoặc x=1/3

a/ ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=a>0\)

\(\Rightarrow a^2=2x+2\sqrt{x^2-16}\)

Phương trình trở thành:

\(a=a^2-12\Leftrightarrow a^2-a-12=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+4}+\sqrt{x-4}=4\)

\(\Leftrightarrow2x+2\sqrt{x^2-16}=16\)

\(\Leftrightarrow\sqrt{x^2-16}=8-x\left(x\le8\right)\)

\(\Leftrightarrow x^2-16=x^2-16x+64\)

\(\Rightarrow x=5\)

b/ \(x\ge-\frac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+1}=a\\\sqrt{4x^2-2x+1}=b\end{matrix}\right.\) ta được:

\(a+3b=3+ab\)

\(\Leftrightarrow ab-a-\left(3b-3\right)=0\)

\(\Leftrightarrow a\left(b-1\right)-3\left(b-1\right)=0\)

\(\Leftrightarrow\left(a-3\right)\left(b-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=3\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{4x^2-2x+1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=9\\4x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\\x=\frac{1}{2}\end{matrix}\right.\)

Bài 2:

a/ \(\left\{{}\begin{matrix}\left(x+2y\right)^2-4xy-5=0\\4xy\left(x+2y\right)+5\left(x+2y\right)-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2y\right)^2-\left(4xy+5\right)=0\\\left(4xy+5\right)\left(x+2y\right)-1=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+2y=a\\4xy+5=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2-b=0\\ab=1\end{matrix}\right.\) \(\Rightarrow a^2-\frac{1}{a}=0\Rightarrow a^3-1=0\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+2y=1\\4xy+5=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1-2y\\4y\left(1-2y\right)+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-2y\\-8y^2+4y+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-1\\y=-\frac{1}{2}\Rightarrow x=2\end{matrix}\right.\)

b/Cộng vế với vế:

\(17x^2-2\left(4y^2+1\right)x+y^4+1=0\)

\(\Delta'=\left(4y^2+1\right)^2-17\left(y^4+1\right)=-y^4+8y^2-16\)

\(\Delta'=-\left(y^2-4\right)^2\ge0\Rightarrow y^2-4=0\Rightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)

- Với \(y=2\) \(\Rightarrow x^2-2x+1=0\Rightarrow x=1\)

\(\)- Với \(y=-2\Rightarrow x^2-2x-7=0\Rightarrow x=1\pm2\sqrt{2}\)