\(\Leftrightarrow\left\{{}\begin{matrix}8x+2y=-10\\3x-2y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=-22\\4x+y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-5-4x=-5-4\cdot\left(-2\right)=-5+8=3\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}-8x-2y=10\\3x-2y=-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-11x=22\\3x-2y=-12\end{matrix}\right.\)

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

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

Vậy hệ phương trình có nghiệm duy nhất là \(\left(-2;3\right)\)

\(\left\{{}\begin{matrix}4x+y=-5\\3x-2y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+2y=-10\\3x-2y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=-22\\4x+y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

ĐKXĐ : \(\left\{{}\begin{matrix}4x^2+2y+2\ge0\\3x+y\ge0\end{matrix}\right.\)

Ta có : \(\left(\sqrt{4x^2+3}-2x\right)\left(\sqrt{y^2-2y+4}-y+1\right)=3\)

\(\Leftrightarrow\dfrac{3}{\sqrt{4x^2+3}+2x}.\dfrac{3}{\sqrt{y^2-2y+4}+y-1}=3\)

\(\Leftrightarrow\left(\sqrt{4x^2+3}+2x\right)\left(\sqrt{y^2-2y+4}+y-1\right)=3\)

\(\Rightarrow\left(\sqrt{4x^2+3}+2x\right)\left(\sqrt{y^2-2y+4}+y-1\right)=\left(\sqrt{4x^2+3}-2x\right)\left(\sqrt{y^2-2y+4}-y+1\right)\)

\(\Leftrightarrow2x\sqrt{y^2-2y+4}+\left(y-1\right).\sqrt{4x^2+3}=0\)

\(\Leftrightarrow2x\sqrt{y^2-2y+4}=\left(1-y\right).\sqrt{4x^2+3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2.\left(y^2-2y+4\right)=\left(y^2-2y+1\right).\left(4x^2+3\right)\\2x.\left(1-y\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2=y^2-2y+1\\2x\left(1-y\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=y-1\\2x=1-y\end{matrix}\right.\\2x\left(1-y\right)\ge0\end{matrix}\right.\)

Với 2x = 1 - y

Khi đó ta có \(\sqrt{4x^2+2y+2}-\sqrt{3x+y}=2x+1\)

\(\Leftrightarrow\sqrt{4x^2-4x+4}-\sqrt{x+1}=2x+1\)      (ĐK : \(x\ge-1\))

\(\Leftrightarrow2\sqrt{x^2-x+1}-\sqrt{x+1}=2x+1\)

\(\Leftrightarrow2\left(\sqrt{x^2-x+1}-1\right)=2x+\sqrt{x+1}-1\)

\(\Leftrightarrow\dfrac{2x\left(x-1\right)}{\sqrt{x^2-x+1}+1}=2x+\dfrac{x}{\sqrt{x+1}+1}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{2x-2}{\sqrt{x^2-x+1}}=2+\dfrac{1}{\sqrt{x+1}+1}\left(1\right)\end{matrix}\right.\)

Phương trình (1) 

<=> \(\dfrac{2\left(x+1\right)}{\sqrt{x^2-x+1}}=2+\dfrac{1}{\sqrt{x+1}+1}+\dfrac{4}{\sqrt{x^2-x+1}}\)

Xét vế trái : \(\dfrac{2\left(x+1\right)}{\sqrt{x^2-x+1}}=\sqrt{\dfrac{4x^2+4x+1}{x^2-x+1}}=\sqrt{\dfrac{5x^2-5x+5-x^2+9x-4}{x^2-x+1}}\)

\(=\sqrt{5-\dfrac{x^2-9x+4}{x^2-x+1}}< \sqrt{5}\) (2) 

Lại có \(2+\dfrac{1}{\sqrt{x+1}+1}+\dfrac{4}{\sqrt{x^2-x+1}}\)

\(=2+\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{x^2-x+1}}+\dfrac{1}{\sqrt{x^2-x+1}}+\dfrac{1}{\sqrt{x^2-x+1}}+\dfrac{1}{\sqrt{x^2-x+1}}\)

\(\ge2+\dfrac{\left(1+1+1+1+1\right)^2}{\sqrt{x+1}+1+4\sqrt{x^2-x+1}}=2+\dfrac{25}{\sqrt{x+1}+1+4\sqrt{x^2-x+1}}\)

Dấu "=" khi \(\dfrac{1}{\sqrt{x+1}+1}=\dfrac{1}{\sqrt{x^2-x+1}}\Leftrightarrow\left[{}\begin{matrix}x\approx3,498374325\\x\approx-0,7385661113\end{matrix}\right.\)

Khi đó \(VP\ge3,6\) (3) 

Từ (3) và (2) => (1) vô nghiệm 

Vậy x = 0 => y = 1

Với 2x = y - 1 kết hợp điều kiện 2x(1 - y) \(\ge0\)

ta được x = 0 ; y = 1 

Vậy (x ; y) = (0;1) 

a: \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-4y+2z=6\\8x+4y-8z=-3\\3x-4y-z=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-6z=3\\11x-9z=1\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\z=\dfrac{1}{2}\\4y=3x-z-4=\dfrac{3}{2}-\dfrac{1}{2}-4=1-4=-3\end{matrix}\right.\)

=>x=1/2;z=1/2;y=-3/4

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

Mấy bài này đơn giản , bạn chỉ cần rút x hoặc y ra là đc

mk làm ví dụ một câu ha

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

Thay (1) vào bt (2) ta có -3(1-2y)-y=2

Bạn giải ra y rồi giải ra x là xong

a)\(\left\{{}\begin{matrix}x=7-2y\\3\left(7-2y\right)-4y=1\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7-2y\\21-6y-4y=1\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7-2y\\20=10y\end{matrix}\right.< =>\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Vậy ...

b)\(\left\{{}\begin{matrix}y=7-2x\\4x-3\left(7-2x\right)=-1\end{matrix}\right.< =>\left\{{}\begin{matrix}y=7-2x\\4x-21+6x=-1\end{matrix}\right.< =>\left\{{}\begin{matrix}y=7-2x\\10x=20\end{matrix}\right.< =>\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}x+2y=7\left(1\right)\\3x-4y=1\left(2\right)\end{matrix}\right.\)
Nhân cả 2 vế pt (1) với 3 ta được hệ phương trình
\(\left\{{}\begin{matrix}3x+6y=21\left(3\right)\\3x-4y=1\left(4\right)\end{matrix}\right.\)
Trừ 2 vế pt (3) cho pt (4)
=>10y=20
\(\Leftrightarrow y=2\) thay vào (1) ta có: x+4=7\(\Leftrightarrow x=3\)
Vậy nghiệm của hpt (x;y)=(3;2)
b,\(\left\{{}\begin{matrix}2x+y=7\left(1\right)\\4x-3y=-1\left(2\right)\end{matrix}\right.\)
Nhân 2 vế pt (1) vs 2 ta được
4x+2y=14(3)
Trừ 2 vế pt(3) cho pt(2)ta có
5y=15
\(\Leftrightarrow\)y=3 thay vào (1)
=>2x+3=7\(\Leftrightarrow x=2\)
Vậy nghiệm của hpt (x;y)=(2;3)

e.

\(\left\{{}\begin{matrix}2x-3y+5=0\\3x+5y-21=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y=-25\\9x+15y=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=38\\3x+5y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{21-3x}{5}\end{matrix}\right.\)

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

f.

\(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x\sqrt{2}+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y\sqrt{2}=0\\4x+y\sqrt{2}=5\sqrt{2}\end{matrix}\right.\)

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

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

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

a.

\(\Leftrightarrow\left\{{}\begin{matrix}5x=-25\\3x-5y=-30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{3x+30}{5}\end{matrix}\right.\)

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

b.

\(\Leftrightarrow\left\{{}\begin{matrix}8x-6y=-10\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}17x=-34\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{-24-9x}{6}\end{matrix}\right.\)

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

c.

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

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

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

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

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

d.

\(\left\{{}\begin{matrix}5x-4y=32\\6x+2y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=32\\12x+4y=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=32\\17x=68\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3x-32}{4}\end{matrix}\right.\)

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

a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

vậy  hệ pt có ndn \(\left\{2;0\right\}\)

b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)

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

vậy hệ pt có ndn \(\left\{2;1\right\}\)

d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)

vậy hệ pt có ndn\(\left\{28;21\right\}\)

9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)

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

12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)

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

14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)

15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))

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

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