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

a) \(\left\{{}\begin{matrix}3x-4y=-2\\2x+y=6\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}3x-4y=-2\\8x+4y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\3x-4y=-2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a: =>3x-4y=-2 và 8x+4y=24

=>11x=22 và 2x+y=6

=>x=2 và y=6-2x=6-2*2=2

b: 2x-y=0 và 3x+y=4

=>5x=4 và y=2x

=>x=4/5 và y=8/5

c: x+3y=-2 và x-y=-1

=>4y=-1 và x=y-1

=>y=-1/4 và x=-1/4-1=-5/4

d: x+y=3 và 4x-3y=-2

=>4x+4y=12 và 4x-3y=-2

=>7y=14 và x+y=3

=>y=2 và x=1

- Trừ hai pt ta được :\(x^3-y^3-x^2+y^2+x-y+1-1=2y-2x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+xy+y^2-x-y+3=0\end{matrix}\right.\)

TH1 : x = y

PT ( I ) TT : \(x^3-x^2+x+1-2x=x^3-x^2-x+1=0\)

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

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

\(\Leftrightarrow x=y=\pm1\)

TH₂ : \(x^2+xy+y^2-x-y+3=0\)

\(\Leftrightarrow x^2+\dfrac{y^2}{4}+\dfrac{1}{4}+xy-x-\dfrac{1}{2}y+\dfrac{3}{4}y^2-\dfrac{1}{2}y+\dfrac{11}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2+\dfrac{8}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2=-\dfrac{8}{3}\left(VL\right)\)

Vậy ....

\(ĐK:x\ge0;y\ge2;5x-y\ge0\\ PT\left(1\right)\Leftrightarrow\sqrt{y+3x}-\sqrt{5x-y}+\sqrt{2x+7y}-3\sqrt{x}=0\\ \Leftrightarrow\dfrac{2y-2x}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7y-7x}{\sqrt{2x+7y}+3\sqrt{x}}=0\\ \Leftrightarrow\left(y-x\right)\left(\dfrac{2}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7}{\sqrt{2x+7y}+3\sqrt{x}}\right)=0\\ \Leftrightarrow x=y\left(\dfrac{2}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7}{\sqrt{2x+7y}+3\sqrt{x}}>0\right)\)

Thay vào \(PT\left(2\right)\Leftrightarrow x-4+\sqrt{x-2}=\sqrt{x^3-10x^2+33x-34}-\sqrt{x^3-9x^2+24x-16}\)

\(\Leftrightarrow\dfrac{x^2-9x+18}{x-4+\sqrt{x-2}}=\dfrac{-x^2+9x-18}{\sqrt{x^3-10x^2+33x-34}+\sqrt{x^3-9x^2+24x-16}}\\ \Leftrightarrow\left(x^2-9x+18\right)\left(\dfrac{1}{x-4+\sqrt{x-2}}+\dfrac{1}{\sqrt{x^3-10x^2+33x-34}+\sqrt{x^3-9x^2+24x-16}}\right)=0\\ \Leftrightarrow x^2-9x+18=0\left(\text{ngoặc lớn luôn }>0,\forall x\ge2\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=6\end{matrix}\right.\)

Vậy ...

Do \(2y^2-x^2=1\)

\(\Rightarrow2x^3-y^3=\left(2y-x\right)\left(2y^2-x^2\right)\)

\(\Leftrightarrow2x^3-y^3=4y^3-2yx^2-2y^2x+x^3\)

\(\Leftrightarrow x^3+2yx^2+2y^2x-5y^3=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+3xy+5y^2\right)=0\)

\(\Rightarrow x=y\)

easy from here

\(\left\{{}\begin{matrix}5x^2+5y^2-6xy=2\\2x^2+3x-2y^2-y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x^2+5y^2-6xy=2\\4x^2+6x-4y^2-2y=6\end{matrix}\right.\)

\(\Rightarrow9x^2+y^2-6xy+6x-2y+1=9\)

\(\Leftrightarrow\left(3x-y+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-y+1=3\\3x-y+1=-3\end{matrix}\right.\)

Đến đây chia 2 trường hợp và thế vào 1 trong 2 pt để giải