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)

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

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

=>\(\left\{{}\begin{matrix}7x=14\\2x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=5-2x=5-2\cdot2=1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}y=1\\x=-2+2y=-2+2\cdot1=0\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}2x-y=13\\y-5=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-y=13\\y=-7+5=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y+13=-2+13=11\\y=-2\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}9x+3y=24\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=25\\3x+y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{25}{11}\\y=8-3x=8-3\cdot\dfrac{25}{11}=8-\dfrac{75}{11}=\dfrac{13}{11}\end{matrix}\right.\)

Lời giải:

Tất cả những bài này đều có hướng giải y chang nhau, nên mình hướng dẫn mẫu 1 bài, các bài khác bạn triển khai tương tự

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

Từ PT(1) ta có: $y=2x-5$ (biểu diễn $y$ theo $x$). Thay vào PT(2):

$-x+(2x-5)=-2$

$\Leftrightarrow x-5=-2$

$\Leftrightarrow x=3$

Khi đó: $y=2x-5=2.3-5=1$

Vậy $(x,y)=(3,1)$

a, b và c có thể dùng phương pháp thế hoặc cộng trừ đại số

\(a,\left\{{}\begin{matrix}x=1-y\\1-y-y=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\1-2y=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\2y=6\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\y=3\end{matrix}\right.=>\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

Kết luận hpt có 1 nghiệm duy nhất (x;y)=(-2;3)

b và c làm tương tự

a.\(\Leftrightarrow\left\{{}\begin{matrix}2x=-4\\x-y=-5\end{matrix}\right.\) ( cộng đại số bạn nhé )

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

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

c.\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\9x-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}13x=13\\9x-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\9.1-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

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

\(\Leftrightarrow x+y+x-y=-4\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-2\)

Thay \(x=-2\) vào \(x+y=1\)\(\Leftrightarrow-2+y=1\)\(\Leftrightarrow y=3\)

Vậy \(x=-2;y=3\)

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

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)

a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}12y-9+2y=-2\\x=4y-3\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}y=\frac{1}{2}\\x=\frac{4.1}{2}-3=-1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-1;\frac{1}{2}\right)\)

b, Ta có : \(\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}x=11-2y\\55-10y-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\-13y=-52\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2.4=3\\y=4\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

c, Ta có : \(\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}30x-27y=3\\30x+42y=72\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9y=1\\-69y=-69\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9=1\\y=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(1;1\right)\)

d, Ta có : \(\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.\)

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

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

=> \(\left\{{}\begin{matrix}y=3-2.0=3\\x=0\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(0;3\right)\)

e, Ta có : \(\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\2\left(2-y\right)-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\4-2y-3y=9\end{matrix}\right.\)

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

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

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-1\right)\)

f, Ta có : \(\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}x=11+2y\\55+10y+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\13y=-52\end{matrix}\right.\)​

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-4\right)\)

g, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=9-5=4\\x=3\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

h, Ta có : \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+3\left(3x+8\right)=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+9x+24=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-\frac{31}{14}\\y=3.\left(-\frac{31}{14}\right)+8=\frac{19}{14}\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)

...????

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

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

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

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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