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\(\left\{{}\begin{matrix}x+y=500\\\dfrac{8}{10}x+\dfrac{9}{10}y=420\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=500-y\\\dfrac{8}{10}\left(500-y\right)+\dfrac{9}{10}y=420\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=500-y\\400+\dfrac{y}{10}=420\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=500-y=300\\y=200\end{matrix}\right.\)
Vậy (x,y)=(300,200)
hpt <=> \(\left\{{}\begin{matrix}\dfrac{8}{10}x+\dfrac{8}{10}y=400\\\dfrac{8}{10}x+\dfrac{9}{10}y=420\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x+y=500\\\dfrac{1}{10}y=20\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x+y=500\\y=200\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=300\\y=200\end{matrix}\right.\)
ĐKXĐ: \(x,y\ne0\)\(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=4\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=4\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}\right)-3\left(y+\dfrac{1}{y}\right)=4\end{matrix}\right.\)
Đặt \(x+\dfrac{1}{x}=a;y+\dfrac{1}{y}=b\left(a,b\ne0\right)\)
\(\Rightarrow hpt\) trở thành:
\(\left\{{}\begin{matrix}a+b=4\left(1\right)\\a^3+b^3-3a-3b=4\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow a=4-b\) Thay vào (2) ta được:
\(\left(4-b\right)^3+b^3-3\left(4-b\right)-3b=4\Leftrightarrow64-48b+12b^2-b^3+b^3-12+3b-3b-4=0\Leftrightarrow12b^2-48b+60=0\Leftrightarrow b^2-4b+5=0\Leftrightarrow b^2-4b+4+1=0\Leftrightarrow\left(b-2\right)^2+1=0\) Vô lí \(\Rightarrow\) ko có a,b \(\Rightarrow\) ko có x,y
Vậy hpt vô nghiệm
\(\left\{{}\begin{matrix}x+y=80\\2x+3y=198\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y=240\\2x+3y=198\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=80\\x=240-198=42\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=42\\y=38\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=80\\2x+3y=198\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}2x+2y=160\\2x+3y=198\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=38\\2x+3\cdot38=198\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=38\\2x=84\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=38\\x=42\end{matrix}\right.\)
Vậy (42;38) là nghiệm
\(\Leftrightarrow\left\{{}\begin{matrix}1.18x+1.18y=914.5\\1.18x+1.12y=889\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=425\\x=350\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2y=3\\2x+3y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\2\left(3+2y\right)+3y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\6+4y+3y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\7y=-7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2\left(-1\right)\\y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=6\\2x+3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3+2y=3-2=1\end{matrix}\right.\)
Bài 1:
a) Ta có: \(\Delta=\left(2m-1\right)^2-4\cdot m\cdot\left(m+2\right)\)
\(\Leftrightarrow\Delta=4m^2-4m+1-4m^2-8m\)
\(\Leftrightarrow\Delta=-12m+1\)
Để phương trình có nghiệm kép thì \(\Delta=0\)
\(\Leftrightarrow-12m+1=0\)
\(\Leftrightarrow-12m=-1\)
hay \(m=\dfrac{1}{12}\)
b) Ta có: \(\Delta=\left(4m+3\right)^2-4\cdot2\cdot\left(2m^2-1\right)\)
\(\Leftrightarrow\Delta=16m^2+24m+9-16m^2+8\)
\(\Leftrightarrow\Delta=24m+17\)
Để phương trình có nghiệm kép thì \(\Delta=0\)
\(\Leftrightarrow24m+17=0\)
\(\Leftrightarrow24m=-17\)
hay \(m=-\dfrac{17}{24}\)
\(ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=9\end{matrix}\right.\)
`x-y=2<=>x=y+2` thay vào trên
`=>m(y+2)+2y=m+1`
`<=>y(m+2)=m+1-2m`
`<=>y(m+2)=1-2m`
Để hpt có nghiệm duy nhất
`=>m+2 ne 0<=>m ne -2`
`=>y=(1-2m)/(m+2)`
`=>x=y+2=5/(m+2)`
`xy=x+y+2`
`<=>(5-10m)/(m+2)=(6-2m)/(m+2)+2`
`<=>(5-10m)/(m+2)=10/(m+2)`
`<=>5-10m=10`
`<=>10m=-5`
`<=>m=-1/2(tm)`
Vậy `m=-1/2` thì HPT có nghiệm duy nhât `xy=x+y+2`
`a)m=2`
$\begin{cases}2x+2y=3\\x-y=2\end{cases}$
`<=>` $\begin{cases}2x+2y=3\\2x-2y=4\end{cases}$
`<=>` $\begin{cases}4y=-1\\x=y+2\end{cases}$
`<=>` $\begin{cases}y=-\dfrac14\\y=\dfrac74\end{cases}$
Vậy m=2 thì `(x,y)=(7/4,-1/4)`
Mình năm nay mấy lên lớp 8 , lên ko biết giải phương trình .
Ta có :
x + y = 900
=> 1 , 1 ( x+ y ) = 990 ( 1 )
Lại có :
1 , 1 x + 1 , 12 y = 1000
=> 1, 1 x + 1 , 1 y + 0 ,02 y = 1000
=> 1, 1 ( x + y ) + 0 , 02 y = 1000 ( 2 )
Lấy ( 2 ) trừ ( 1 ) , được :
0 , 02 y = 1000- 990
=> 0 ,02 y = 10
=> y = 500
Mà x + y = 900
=> x = 400
Vậy ...
\(1,1x+1,12y=1000\)
\(1,1\left(x+y\right)+0,02y=1000\)
1,1 x 900 + 0,02y = 1000
990 + 0,02y=1000
0,02y = 10
y = 500
x = 400