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

2) 2 pt 3 ẩn không giải được.

3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

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

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

6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

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

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

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

Vì \(\dfrac{1}{2}\ne\dfrac{-2}{3}\)

nên hệ luôn có nghiệm duy nhất

a: \(\left\{{}\begin{matrix}x-2y=-3m-4\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y=-6m-8\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y-2x-3y=-6m-8-8m+1\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=-14m-7\\2x=8m-1-3y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2m+1\\2x=8m-1-6m-3=2m-4\end{matrix}\right.\)

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

Đặt \(A=y^2+3x-1\)

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

\(=4m^2+4m+1+3m-6-1\)

\(=4m^2+7m-6\)

\(=4\left(m^2+\dfrac{7}{4}m-\dfrac{3}{2}\right)\)

\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{8}+\dfrac{49}{64}-\dfrac{145}{64}\right)\)

\(=4\left(m+\dfrac{7}{8}\right)^2-\dfrac{145}{16}>=-\dfrac{145}{16}\)
Dấu '=' xảy ra khi m=-7/8

b: Đặt B=x^2-y^2

\(=\left(m-2\right)^2-\left(2m+1\right)^2\)

\(=m^2-4m+4-4m^2-4m-1\)

\(=-3m^2-8m+3\)

\(=-3\left(m^2+\dfrac{8}{3}m-1\right)\)

\(=-3\left(m^2+2\cdot m\cdot\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{25}{9}\right)\)

\(=-3\left(m+\dfrac{4}{3}\right)^2+\dfrac{25}{3}< =\dfrac{25}{3}\)

Dấu '=' xảy ra khi m=-4/3

Đề bài yêu cầu gì?

Bài 1.

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)

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

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

\(x_0^2+y_0^2=9m\)

\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)

\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)

\(\Leftrightarrow2m^2-7m+5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )

a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)

\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

các câu khác tương tự

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 \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)

...????

có ai giúp ko nè

Câu 1:

\(\left\{{}\begin{matrix}m^2x+y=3m\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2x-4x=3m+6\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(m^2-4\right)=3m+6\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+6}{m^2-4}=\dfrac{3}{m-2}\\y=6-\dfrac{3}{m-2}=\dfrac{6m-15}{m-2}\end{matrix}\right.\)Câu 2:

\(\left\{{}\begin{matrix}5x-y=13\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15x-3y=39\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16x=32\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

Anh vẫn thấy bình thường mà em, vẫn thấy cách làm

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

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