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a) \(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+5x+3y+15=xy+8x+y+8\\10xy+14x-15y-21=10xy+10x-12y-12\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-3x+2y=-7\\4x-3y=9\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-9x+6y=-21\\8x-6y=18\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x=-3\\8x-6y=18\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=3\\8.3-6y=18\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm (x;y)=(3;1)
b) ĐKXĐ:\(\left\{{}\begin{matrix}2y-5\ne0\\3y-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\ne\dfrac{5}{2}\\y\ne\dfrac{4}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x-3}{2y-5}=\dfrac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(2x-3\right)\left(3y-4\right)=\left(3x+1\right)\left(2y-5\right)\\2x-6-3y-6=-16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}6xy-8x-9y+12=6xy-15x+2y-5\\2x-3y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}7x-11y=-17\\2x-3y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}14x-22y=-34\\14x-21y=-28\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}14x-22y=-34\\-y=-6\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}14x-22.6=-34\\y=6\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=7\left(TM\right)\\y=6\left(TM\right)\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm (x;y)=(7;6)
a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)
Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)
=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)
\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
a,\(\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=18\left(1\right)\\10x-6y=10\left(2\right)\end{matrix}\right.\)
Cộng (1) và (2) => 7x=28
\(\Leftrightarrow\) x=4
thay x vào (1) ta có -4+2y=6
=> 2y=10
=>y=5
Vậy nghiệm của phương trình (x;y)=(4;5)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{4}y=2\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\5y=15\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)
a) x2=14−5x⇔x2+5x−14=0x2=14−5x⇔x2+5x−14=0
Δ=52−4.1.(−14)=25+56=81>0√Δ=√81=9x1=−5+92.1=42=2x2=−5−92.1=−142=−7Δ=52−4.1.(−14)=25+56=81>0Δ=81=9x1=−5+92.1=42=2x2=−5−92.1=−142=−7
b)
3x2+5x=x2+7x−2=0⇔2x2−2x+2=0⇔x2−x+1=0Δ=(−1)2−4.1.1=1−4=−3<03x2+5x=x2+7x−2=0⇔2x2−2x+2=0⇔x2−x+1=0Δ=(−1)2−4.1.1=1−4=−3<0
Phương trình vô nghiệm
c)
(x+2)2=3131−2x⇔x2+4x+4+2x−3131=0⇔x2+6x−3127=0Δ=62−4.1.(−3127)=36+12508=12544>0√Δ=√12544=112x1=−6+1122.1=1062=53x2=−6−1122.1=−59(x+2)2=3131−2x⇔x2+4x+4+2x−3131=0⇔x2+6x−3127=0Δ=62−4.1.(−3127)=36+12508=12544>0Δ=12544=112x1=−6+1122.1=1062=53x2=−6−1122.1=−59
d)
(x+3)25+1=(3x−1)25+x(2x−3)2⇔2(x+3)2+10=2(3x−1)2+5x(2x−3)⇔2x2+12x+18+10=18x2−12x+2+10x2−15x⇔26x2−39x−26=0⇔2x2−3x−2=0Δ=(−3)2−4.2.(−2)=9+16=25>0√Δ=√25=5x1=3+52.2=84=2x2=3−52.2=−12
A. Cách B sai vì 5 : 2/5 thì ko thể nào = 25 đc.