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3(2x+y)-2(3x-2y)=3.19-11.2
6x+3y-6x+4y=57-22
7y=35
y=5
thay vào :
2x+y=19
2x+5=19
2x=14
x=7
2/ x2+21x-1x-21=0
x(x+21)-1(x+21)=0
(x+21)(x-1)=0
TH1 x+21=0
x=-21
TH2 x-1=0
x=1
vậy x = {-21} ; {1}
3/ x4-16x2-4x2+64=0
x2(x2-16)-4(x2-16)=0
(x2-16)-(x2-4)=0
TH1 x2-16=0
x2=16
<=>x=4;-4
TH2 x2-4=0
x2=4
x=2;-2
Bài 1 :
\(\hept{\begin{cases}2x+y=19\\3x-2y=11\end{cases}\Leftrightarrow\hept{\begin{cases}4x+2y=38\\3x-2y=11\end{cases}\Leftrightarrow\hept{\begin{cases}7x=49\\2x+y=19\end{cases}}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7\\2x+y=19\end{cases}}\)Thay vào x = 7 vào pt 2 ta được :
\(14+y=19\Leftrightarrow y=5\)Vậy hệ pt có một nghiệm ( x ; y ) = ( 7 ; 5 )
Bài 2 :
\(x^2+20x-21=0\)
\(\Delta=400-4\left(-21\right)=400+84=484\)
\(x_1=\frac{-20-22}{2}=-24;x_2=\frac{-20+22}{2}=1\)
Bài 3 : Đặt \(x^2=t\left(t\ge0\right)\)
\(t^2-20t+64=0\)
\(\Delta=400+4.64=656\)
\(t_1=\frac{20+4\sqrt{41}}{2}\left(tm\right);t_2=\frac{20-4\sqrt{41}}{2}\left(ktm\right)\)
Theo cách đặt : \(x^2=\frac{20+4\sqrt{41}}{2}\Rightarrow x=\sqrt{\frac{20+4\sqrt{41}}{2}}=\frac{\sqrt{20\sqrt{2}+4\sqrt{82}}}{2}\)
\(Đặt:\dfrac{1}{y+2}=a\left(y\ne-2\right)\\ Hpt\Leftrightarrow\left\{{}\begin{matrix}2x+12a=5\\3x-4a=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+12a=5\\9x-12a=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}11x=11\\3x-4a=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\3.1-4a=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\a=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(1;2\right)\)
\(\hept{\begin{cases}3x+y=14\\2x-y=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x+y=14\\5x=15\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=3\\y=5\end{cases}}\)
Vậy hệ pt có nghiệm (x,y) =( 3,5)
\(\hept{\begin{cases}3x+y=14\\2x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=15\\3x+y=14\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\3x+y=14\end{cases}}}\)
Thay x = 3 vào pt 2 ta được
\(\left(2\right)\Rightarrow9+y=14\Leftrightarrow y=5\)
Vậy hệ pt có một nghiệm là ( x ; y ) = ( 3 ; 5 )
\(\left\{{}\begin{matrix}2x+y=3\\3x+2y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}2.(2x+y)=3.2\\3x+2y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}4x+2y=6\\3x+2y=4\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}4x+2y-3x-2y=6-4\\3x+2y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\3x+2y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\3.2+2y=4\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=2\\2y=-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy (\(x;y\)) =( 2; -1)
1. \(2x^2-3x-5=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2,5\\x=-1\end{cases}}\)
Vậy tập ngiệm của phương trình là \(S=\left\{2,5;-1\right\}\)
2x2-3x-5=0
2x2+2x-5x-5=0
2x(x+1)+5(x+1)=0
(x+1)(2x+5)=0
TH1 x+1=0 <=>x=-1
TH2 2x+5=0<=>2x=-5<=>x=-5/2
2. ta có:
2(x-2y)-(2x+y)=-1.2-8
2x-4y-2x-y=-2-8
-5y=-10
y=2
thay vào
x-2y=-1 ( với y=2)
<=> x-2.2=-1
x-4=-1
x=3
đk: \(y\ge1\)
Ta có: \(\hept{\begin{cases}2\left(x+2\right)-\sqrt{y-1}=6\\5\left(x+2\right)-2\sqrt{y-1}=16\end{cases}}\Leftrightarrow\hept{\begin{cases}4\left(x+2\right)-2\sqrt{y-1}=12\\5\left(x+2\right)-2\sqrt{y-1}=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2=4\\2\left(x+2\right)-\sqrt{y-1}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\\sqrt{y-1}=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\y-1=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=5\end{cases}}\)
Vậy \(\hept{\begin{cases}x=2\\y=5\end{cases}}\)
\(\left\{{}\begin{matrix}\dfrac{3}{x-1}+\dfrac{4}{y}=13\\\dfrac{2}{x-1}-\dfrac{5}{y}=1\end{matrix}\right.\)(1)
ĐK: \(\left\{{}\begin{matrix}x-1\ne0\\y\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\y\ne0\end{matrix}\right.\)
Đặt \(u=\dfrac{1}{x-1};v=\dfrac{1}{y}\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}3u+4v=13\\2u-5v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6u+8v=26\\6u-15v=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}23v=23\\2u-5v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}v=1\\2u=1-5v=1+5.1=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=1\\u=\dfrac{6}{2}=3\end{matrix}\right.\)
- Khi u= 3, ta có \(\dfrac{1}{x-1}=3\Leftrightarrow1=3\left(x-1\right)\Leftrightarrow1=3x-3\)
\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)(thỏa mãn)
- Khi v= 1, ta có: \(\dfrac{1}{y}=1\Leftrightarrow y=1\)(thỏa mãn)
Vậy nghiệm của hệ phương trình là: \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-3y=5\\2x+3y=1\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x-3y+2x+3y=5+1\\2x+3y=1\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}3x=6\\2x+3y=1\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=6:3\\2x+3y=1\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=2\\2x+3y=1\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=2\\2.2+3y=1\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=2\\3y=1-4\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=2\\3y=-3\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy (\(x\);y) =(2; -1)
<=>\(\left\{{}\begin{matrix}4x+3y=11\\4x-y=7\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}4y=4\\4x-y=7\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}y=1\\4x-1=7\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}y=1\\4x=8\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
vậy hệ đã cho có nghiệm (x;y) = (2;1)