a: \(=\left(4xy^2+2xy^2\right)+\left(3x^2y-3x^2y\right)=6xy^2\)

b: \(=xy\left(\dfrac{1}{5}+\dfrac{1}{3}\right)+xy^2\left(\dfrac{4}{3}-\dfrac{2}{5}\right)=\dfrac{8}{15}xy+\dfrac{14}{15}xy^2\)

d: \(=\dfrac{-4}{9}\cdot\dfrac{3}{2}\cdot xy^2\cdot xy^3=-\dfrac{2}{3}x^2y^5\)

ĐKXĐ: x>=0; y>=1; 2y-3x-4>=0

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

(1): \(x^3 + x^2 + y^2 - x^2y - xy - y = 0\)

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

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

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

\(\Delta = (x^2 + x + 1)^2 - 4 \cdot 1 \cdot (x^3 + x^2)\)

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

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

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

=>(1) có hai nghiệm là:

\(\left[\begin{array}{l}y=\frac{(x^2+x+1)+(x^2-x-1)}{2}=\frac{x^2+x+1+x^2-x-1}{2}=\frac{2x^2}{2}=x^2\\ y=\frac{(x^2+x+1)-(x^2-x-1)}{2}=\frac{x^2+x+1-x^2+x+1}{2}=\frac{2x+2}{2}=x+1\end{array}\right.\)

TH1: \(y = x^2\)

(2) sẽ trở thành: \(\sqrt{x} + \sqrt{x^2 - 1} = \sqrt{2x^2 - 3x - 4}\)

=>\(x + x^2 - 1 + 2\sqrt{x(x^2 - 1)} = 2x^2 - 3x - 4\)

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

=>\(\begin{cases}4(x^3-x)=(x^2-4x-3)^2\\ x^2-4x-3\ge0\end{cases}\)

=>\(\begin{cases}4x^3-4x=x^4+16x^2+9-8x^3-6x^2+24x\\ x^2-4x+4-7\ge0\end{cases}\)

=>\(\begin{cases}x^4-12x^3+10x^2+28x+9=0\\ \left(x-2\right)^2\ge7\end{cases}\Rightarrow\begin{cases}(x^2-10x-9)(x^2-2x-1)=0\\ \left(x-2\right)^2\ge7\end{cases}\)

=>\(x=5+\sqrt{34}\)

=>\(y=x^2=(5+\sqrt{34})^2=59+10\sqrt{34}\)

TH2: y=x+1

Thay y=x+1 vào (2), ta được:

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

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

=>\(2\sqrt{x}=\sqrt{-x-2}\)

=>-x-2>=0 và 4x=-x-2

=>-x>=2 và 5x=-2

=>x<=-2 và x=-2/5

=>x∈∅

=>Loại

Đặt \(\left\{{}\begin{matrix}x-2y=a\\\dfrac{1}{2x+3y}=b\end{matrix}\right.\) 

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=2\\2a+3b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\7y=-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2.-1\\y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Vậy nghiệm hpt \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

4:

x+3y=4m+4 và 2x+y=3m+3

=>2x+6y=8m+8 và 2x+y=3m+3

=>5y=5m+5 và x+3y=4m+4

=>y=m+1 và x=4m+4-3m-3=m+1

x+y=4

=>m+1+m+1=4

=>2m+2=4

=>2m=2

=>m=1

3:

x+2y=3m+2 và 2x+y=3m+2

=>2x+4y=6m+4 và 2x+y=3m+2

=>3y=3m+2 và x+2y=3m+2

=>y=m+2/3 và x=3m+2-2m-4/3=m+2/3

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)