Bài 4: Giải hệ phương trình bằng phương pháp cộng đại số

\(\left\{{}\begin{matrix}2x-3y=5\\4x+7y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-6y=10\\4x+7y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-13y=2\\2x-3y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{2}{13}\\2x=3y+5=3\cdot\dfrac{-2}{13}+5=-\dfrac{6}{13}+5=\dfrac{65-6}{13}=\dfrac{59}{13}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{59}{26}\\y=-\dfrac{2}{13}\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}x-y=14\\2x+y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y+2x+y=14+1\\x-y=14\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=15\\x-y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=x-14=5-14=-9\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}3x+5y=-1\\3x+y=7\end{matrix}\right.\)

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

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

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

c: \(\left\{{}\begin{matrix}4x-3y=-10\\2x+5y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-3y=-10\\4x+10y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-3y-4x-10y=-10-16\\4x+10y=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-13y=-26\\2x+5y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\2x=8-5y=8-5\cdot2=-2\end{matrix}\right.\)

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

d: \(\left\{{}\begin{matrix}\dfrac{1}{4}x-\dfrac{1}{5}y=1\\5x-4y=20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-4y=20\\5x-4y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=0\\5x-4y=20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in R\\4y=5x-20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=\dfrac{5}{4}x-5\end{matrix}\right.\)

\(a,\left\{{}\begin{matrix}x-y=14\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=15\\x-y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{15}{3}=5\\y=x-14=5-14=-9\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(5;-9\right)\)

\(b,\left\{{}\begin{matrix}3x+5y=-1\\3x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=-8\\3x+y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{8}{4}=-2\\3x+\left(-2\right)=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=\dfrac{7+2}{3}=3\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(3;-2\right)\)

a) \(\left\{{}\begin{matrix}x-y=14\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=15\\2x+y=1\end{matrix}\right.\)

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

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

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

Vậy hệ PT có nghiệm duy nhất: \(\left\{{}\begin{matrix}x=5\\y=-9\end{matrix}\right.\)

b)\(\left\{{}\begin{matrix}3x+5y=-1\\\text{3x + y = 7}\end{matrix}\right.\)

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

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

Vậy hệ PT có nghiệm duy nhất: \(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x-3y=-10\\4x+10y=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+10y=16\\y=\dfrac{6}{\text{7}}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+\dfrac{60}{\text{7}}=16\\y=\dfrac{6}{\text{7}}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=\dfrac{52}{\text{7}}\\y=\dfrac{6}{\text{7}}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{52}{28}\\y=\dfrac{6}{\text{7}}\end{matrix}\right.\)'

Vậy hệ PT có nghiệm duy nhất: \(\left\{{}\begin{matrix}x=\dfrac{52}{28}\\y=\dfrac{6}{\text{7}}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}2x+5y-2x+3y=7+1=8\\2x-3y=-1\end{matrix}\right.\)

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

=>x=1 và y=1

\(Đặt:a=\dfrac{1}{x};b=\dfrac{1}{y}\left(x,y\ne0\right)\\ \left\{{}\begin{matrix}\dfrac{108}{x}+\dfrac{63}{y}=7\\\dfrac{81}{x}+\dfrac{84}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}108a+63b=7\\81a+84b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}324a+189b=21\\324a+336b=28\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-147b=-7\\81a+84b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{-7}{-147}=\dfrac{1}{21}\\81a+84.\dfrac{1}{21}=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\81a=7-4=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{y}=\dfrac{1}{21}\left(TM\right)\\a=\dfrac{1}{x}=\dfrac{3}{81}=\dfrac{1}{27}\left(TM\right)\end{matrix}\right.\\ Vậy:\left\{{}\begin{matrix}x=27\\y=21\end{matrix}\right. \)

\(\left\{{}\begin{matrix}-2,1x+1,4y=3,5\\4,5x-2,25y=-2,4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-4,5x+3y=7,5\\4,5x-2,25y=-2,4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,75y=5,1\\4,5x-2,25y=-2,4\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}y=6,8\\4,5x-2,25\cdot6,8=-2,4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=6,8\\4,5x=-2,4+15,3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=6,8\\x=\dfrac{12,9}{4,5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=6,8\\x=\dfrac{43}{15}\end{matrix}\right.\)

Xem lại giúp tớ dấu căn ở câu c và d nhé.  

5:

a: \(\left\{{}\begin{matrix}x-2y=-m-7\\2x+y=3m-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2m-14\\2x+y=3m-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-5y=-5m-10\\x-2y=-m-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=m+2\\x=-m-7+2m+4=m-3\end{matrix}\right.\)

Thay x=m-3 và y=m+2 vào y=3x-1, ta được:

3(m-3)-1=m+2

=>3m-10=m+2

=>2m=12

=>m=6

b: A(x,y) nằm trong góc phần tư thứ II

\(\Leftrightarrow\left\{{}\begin{matrix}m-3< 0\\m+2>0\end{matrix}\right.\Leftrightarrow-2< m< 3\)

mà m nguyên

nên \(m\in\left\{-1;0;1;2\right\}\)

Gọi thời gian hoàn thành của đội 1 là x

=>Thời gian hoàn thành của đội 2 là x+60

Theo đề, ta có: 1/x+1/(x+60)=1/40

=>(x+60+x)/(x^2+60x)=1/40

=>x^2+60x=80x+2400

=>x^2-20x-2400=0

=>x=60

=>Thời gian hoàn thành của đội 2 là 120h

mọi người giúp với

Công suất làm việc mỗi giờ của người thứ nhất, người thứ hai lần lượt là a,b (a,b>0)

Ta lập hpt:

\(\left\{{}\begin{matrix}4a+4b=1\\a+2b=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{6}\\b=\dfrac{1}{12}\end{matrix}\right.\)

Vậy nếu làm một mình người thứ nhất cần 6 giờ để hoàn thành công việc, người thứ hai cần đến 12 giờ để hoàn thành công việc đó.