1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

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

=> Hệ có vô số nghiệm.

3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)

=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75

=>x=7; y=5

b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)

=>4x+9y=8 và -8x+3y=5

=>x=-1/4; y=1

c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)

=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5

=>2x-3y=-5,5 và 3x-2y=-4,5

=>x=-1/2; y=3/2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

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

b)\(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(5-2x\right)=4\\y=5-2x\end{matrix}\right.\)\(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-10+4x=4\\y=5-2x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=14\\y=5-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy nghiệm duy nhất của hpt là: (2;1)

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

    \(\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy nghiệm duy nhất của hpt là: (0;1)

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

Từ (1): \(x=2-2y\) (3)

Thế (3) vào (2), ta được: \(-2\left(2-2y\right)+y=1< =>-4+4y+y=1\)

                                          \(\Leftrightarrow y=1\)\(\Rightarrow\)\(x=2-2.1=0\)

Vậy nghiệm duy nhất của hpt là:  (0;1)

1)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

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

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

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

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\\x=\dfrac{y^2-1}{4y}\end{matrix}\right.\)

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

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