a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)

a: =>2x-4+3y+3=-2 và 3x-6+2y+2=-3

=>2x+3y=-2-3+4=-1 và 3x+2y=-3+6-2=1

=>x=1;y=-1

b: =>1/2x=4/3 và 5x-8y=3

=>x=4/3:1/2=4/3*2=8/3 và 8y=5x-3=5*8/3-3=40/3-3=31/3

=>y=31/24; x=8/3

\(\begin{cases} x - 4y + 3\sqrt{y} = \sqrt{2x + y} & (1) \\ \sqrt{8y - 1} + x^2 - 12y + 1 = 0 & (2) \end{cases}\)

ĐKXĐ: y>=1/8; x>=-y/2

Đặt \(a=\sqrt{2x+y}\ge0;b=\sqrt{y}\ge\frac{1}{2\sqrt2}>0\)

=>\(a^2=2x+y\)

=>\(x=\frac{a^2 - b^2}{2}\)

(1): \(x - 4y + 3\sqrt{y} = \sqrt{2x + y}\)

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

=>\(a^2-b^2-8b^2+6b-2a=0\)

=>\(a^2-2a+1-9b^2+6b-1=0\)

=>\(\left(a-1\right)^2-\left(3b-1\right)^2=0\)

=>(a-1-3b+1)(a-1+3b-1)=0

=>(a-3b)(a+3b-2)=0

TH1: a-3b=0

=>a=3b

=>2x+y=9y

=>2x=8y

=>x=4y

Thay x=4y vào (2), ta được:

\(\sqrt{8y - 1} + (4y)^2 - 12y + 1 = 0\)

=>\(\sqrt{8y - 1}+16y^2-12y+1=0\)

Đặt \(t=\sqrt{8y - 1}\ge0\)

=>\(8y=t^2+1\)

=>\(y=\frac{t^2 + 1}{8}\)

\(\sqrt{8y - 1} + 16y^2 - 12y + 1 = 0\)

=>\(t+16\cdot\left(\frac{t^2 + 1}{8}\right)^2-12\cdot\left(\frac{t^2 + 1}{8}\right)+1=0\)

=>\(t+\frac{(t^2 + 1)^2}{4}-\frac{3(t^2 + 1)}{2}+1=0\)

=>\(4t+(t^4+2t^2+1)-6(t^2+1)+4=0\)

=>\(t^4-4t^2+4t-1=0\)

=>\((t^4-1)-4t(t-1)=0\)

=>\((t-1)(t+1)(t^2+1)-4t(t-1)=0\)

=>\((t-1)\left[(t+1)(t^2+1)-4t\right]=0\)

=>\((t-1)(t^3+t^2-3t+1)=0\)

=>\((t-1)^2(t^2+2t-1)=0\)

TH1: t-1=0

=>t=1

=>8y-1=1

=>8y=2

=>y=1/4(nhận)

=>x=4y=1(nhận)

TH2: \(t^2+2t-1=0\)

=>\(t^2+2t+1=2\)

=>\(\left(t+1\right)^2=2\)

=>\(t+1=\sqrt2\) (Do t>0)

=>\(t=\sqrt2-1\)

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

=>\(8y=4-2\sqrt{2}\)

=>\(y=\frac{2 - \sqrt{2}}{4}\) (nhận)

=>\(x=4y=2-\sqrt2\) (nhận)

TH2:a+3b-2=0

=>\(\sqrt{2x+y}+3\sqrt{y}=2\) (3)

y>=1/8

=>\(3\sqrt{y}\ge3\cdot\sqrt{\frac18}=\frac{3}{2\sqrt2}\) ≃1,0606

(2)=>\(x^2 = 12y - 1 - \sqrt{8y - 1}\)

Với y>=1/8, ta xét hàm số \(f\left(y\right)=12y-1-\sqrt{8y - 1}\)

=>f(y)>=0

Khi kết hợp cùng (3), ta sẽ thấy các nghiệm giống hệt với TH1

Vậy: \((x; y) \in \left\{ \left(1; \frac{1}{4}\right), \left(2 - \sqrt{2}; \frac{2 - \sqrt{2}}{4}\right) \right\}\)

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:

\(\dfrac{a^2-b^2}{2}-4b^2+3b=a\Leftrightarrow a^2-9b^2+6b=2a\)

\(\Leftrightarrow\left(a-3b\right)\left(a+3b\right)-2\left(a-3b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a+3b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=2-3b\end{matrix}\right.\) \(\Rightarrow...\)

1: ĐKXĐ: x>=-1

\(\begin{cases}\sqrt{x+1}=\sqrt{2}(8y^2+8y+1)\left(1\right)\\ 4(x^3-8y^3)-6(x^2+4y^2)+3(x+2y)-1=0\left(2\right)\end{cases}\)

(2): \(4(x^3-8y^3)-6(x^2+4y^2)+3(x+2y)-1=0\)

=>\(8x^3 - 64y^3 - 12x^2 - 48y^2 + 6x + 12y - 2 = 0\)

=>\((8x^3-12x^2+6x-1)-(64y^3+48y^2+12y+1)=0\)

=>\((2x-1)^3-(4y+1)^3=0\)

=>\(\left(2x-1\right)^3=\left(4y+1\right)^3\)

=>2x-1=4y+1

=>2x-4y=2

=>x-2y=1

=>x=2y+1

(1) sẽ tương đương: \(\sqrt{2y + 1 + 1} = \sqrt{2}(8y^2 + 8y + 1)\)

=>\(\sqrt{2(y + 1)}=\sqrt{2}(8y^2+8y+1)\)

=>\(\sqrt{y + 1}=8y^2+8y+1\)

Đặt \(t = 8y^2 + 8y + 1 \ge 0\)

Ta có: \(\sqrt{y + 1}=8y^2+8y+1\)

=>\(\sqrt{y+1}=t\)

=>\(y+1=t^2\)

Phương trình sẽ trở thành:

\(8(t^2-1)^2+8(t^2-1)+1=t\)

=>\(8t^4-8t^2-t+1=0\)

=>\((t-1)(8t^3+8t^2-1)=0\)

=>t-1=0

=>t=1

=>8y(y+1)=0

=>y=0 hoặc y=-1

Khi y=0 thì x=1(nhận)

Khi y=-1 thì x=-1(loại)