a,Ta có hệ phương trình\(\left\{{}\begin{matrix}7x-2y=1\left(1\right)\\2x+3y=11\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}21x-6y=3\\4x+6y=22\end{matrix}\right.\)

=> \(21x-6y+4x+6y=25\)

=> \(25x=25\)

=> \(x=1\)

- Thay x = 1 vào phương trình 1 ta được :

\(7-2y=1\)

=> \(y=3\)

Vậy hệ phương trình có duy nhất 1 nghiệm là ( x, y ) = ( 1, 3 )

b, Ta có hệ phương trình\(\left\{{}\begin{matrix}3x+2y=16\\2x-y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+2y=16\\y=2x+1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}3x+2\left(2x+1\right)=16\\y=2x+1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}3x+4x+2=16\\y=2x+1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2\\y=2x+1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x=2\\y=2.2+1=5\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là ( x, y ) = ( 2, 5 )

c, Ta có hệ phương trình \(\left\{{}\begin{matrix}x+2y=5\\3x-2y=-1\end{matrix}\right.\)

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

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

=> \(\left\{{}\begin{matrix}x=5-2y\\15-6y-2y=-1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=5-2y\\y=2\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x=5-2.2=1\\y=2\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là ( x, y ) = ( 1, 2 )

nhâ vế 1 vs 2

nhân vế 2 vs 3 là ra thôi bn

trừ 2 vế cho nhau nữa

\(\left\{{}\begin{matrix}3x+2y=4\\2x-3y=7\end{matrix}\right.< =>\left\{{}\begin{matrix}6x+4y=8\\6x-9y=21\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}13y=-13\\3x+2y=4\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-1\\3x=4+2=6\end{matrix}\right.\)

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

giúp mình nhé

\(a,\left\{{}\begin{matrix}2x-y=1\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4x-2y=2\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}7x=7\\2x-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\2.1-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

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

\(b,\left\{{}\begin{matrix}4x+3y=-1\\3x-2y=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4.2x+3.2y=-1.2\\3.3x-2.3y=2.3\end{matrix}\right.\\ =>\left\{{}\begin{matrix}8x+6y=-2\\9x-6y=6\end{matrix}\right.\\ =>\left\{{}\begin{matrix}17x=4\\3x-2y=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=\dfrac{4}{17}\\y=-\dfrac{11}{17}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{4}{17};-\dfrac{11}{17}\right)\)

khó + lười + nhiều = không làm

Hello

ko thích làm

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

⇒ \(3x=-5\)

⇒ \(x=-\dfrac{5}{3}\)

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}3.2+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6+5y=11\\x=2\end{matrix}\right.\)

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

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

A)\(5xyz.4x^2y^2.\left(-2x^3y\right)=\left(5.4.\left(-2\right)\right).\left(xx^2x^3\right).\left(yy^2y\right)=\left(-40\right)x^6y^4\)

- BẬC : 10

- HỆ SỐ: -40 

B) \(-xy.\left(\frac{1}{2}x^3y^4\right).\left(\frac{-4}{7}x^2y^5\right)=\left(\frac{1}{2}.\frac{-4}{7}.\left(-1\right)\right).\left(xx^3x^2\right).\left(y^4y^5y\right)=\frac{2}{7}x^6y^{10}\)

- BẬC : 16

- HỆ SỐ: 2/7

C) \(\frac{5}{3}x^2y^4.\left(\frac{-6}{5}xy^3\right).\left(-xy\right)=\left(\frac{5}{3}.\frac{-6}{5}.\left(-1\right)\right).\left(x^2xx\right).\left(y^4y^3y\right)=2x^4y^8\)

- BẬC : 12

- HỆ SỐ : 2

D) \(\left(\frac{-1}{3}x^2y^5\right).\left(\frac{3}{4}xy\right).5x=\left(\frac{-1}{3}.\frac{3}{4}.5\right).\left(x^2xx\right).\left(y^5y\right)=\frac{-5}{4}x^4y^6\)

- BẬC : 10

- HỆ SỐ : -5 /4

CHÚC BN HỌC TỐT!!

cảm on

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 \(x=-2;y=3\)

Chọn B