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a) Với m = -2
=> hpt trở thành: \(\left\{{}\begin{matrix}x+y=2\\-2x-y=-2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=2-x\\-x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Vậy S = {0; 2}
b) Ta có: \(\left\{{}\begin{matrix}x+y=2\left(1\right)\\mx-y=m\left(2\right)\end{matrix}\right.\)
=> x + mx = 2 + m
<=> x(m + 1) = 2 + m
Để hpt có nghiệm duy nhất <=> \(m\ne-1\)
<=> x = \(\dfrac{m+2}{m+1}\) thay vào pt (1)
=> y = \(2-\dfrac{m+2}{m+1}=\dfrac{2m+2-m-2}{m+1}=\dfrac{m}{m+1}\)
Mà 3x - y = -10
=> \(3\cdot\dfrac{m+2}{m+1}-\dfrac{m}{m+1}=-10\)
<=> \(\dfrac{2m+6}{m+1}=-10\) <=> m + 3 = -5(m + 1)
<=> 6m = -8
<=> m = -4/3
c) Để hpt có nghiệm <=> m \(\ne\)-1
Do x;y \(\in\) Z <=> \(\left\{{}\begin{matrix}\dfrac{m+2}{m+1}\in Z\\\dfrac{m}{m+1}\in Z\end{matrix}\right.\)
Ta có: \(x=\dfrac{m+2}{m+1}=1+\dfrac{1}{m+1}\)
Để x nguyên <=> 1 \(⋮\)m + 1
<=> m +1 \(\in\)Ư(1) = {1; -1}
<=> m \(\in\) {0; -2}
Thay vào y :
với m = 0 => y = \(\dfrac{0}{0+1}=0\)(tm)
m = -2 => y = \(\dfrac{-2}{-2+1}=2\)(tm)
Vậy ....
a, Thay m = 2 ta được \(\left\{{}\begin{matrix}2x+y=1\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
b, \(\Leftrightarrow\left\{{}\begin{matrix}3x=3m-3\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m-1\\y=m-3\end{matrix}\right.\)
Ta có : \(x^2+y^2=m^2-2m+1+m^2-6m+9=2m^2-8m+10\)
\(=2\left(m^2-4m+4-4\right)+10=2\left(m-2\right)^2+2\ge2\forall m\)
Dấu''='' xảy ra khi m =2
Vậy ...
\(\left\{{}\begin{matrix}mx-y=2\\x+my=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\x+m\left(mx-2\right)=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\x+m^2x-2m=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\x\left(m^2+1\right)=3+2m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m.\dfrac{3+2m}{m^2+1}-2\\x=\dfrac{3+2m}{m^2+1}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3m+2m^2-2m^2-2}{m^2+1}\\x=\dfrac{3+2m}{m^2+1}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3m-2}{m^2+1}\\x=\dfrac{3+2m}{m^2+1}\end{matrix}\right.\)
\(x+y=0\\ \Leftrightarrow\dfrac{3m-2}{m^2+1}+\dfrac{3+2m}{m^2+1}=0\\ \Leftrightarrow\dfrac{3m-2+3+2m}{m^2+1}=0\\ \Rightarrow4m+1=0\\ \Leftrightarrow m=-\dfrac{1}{4}\)
x+y=0 \(\Rightarrow\) y=-x.
\(\left\{{}\begin{matrix}mx-y=2\\x+my=3\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}mx+x=2\\x-mx=3\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}x\left(m+1\right)=2\\x\left(1-m\right)=3\end{matrix}\right.\) \(\Rightarrow\) \(\dfrac{2}{m+1}=\dfrac{3}{1-m}\) \(\Rightarrow\) m=-1/5 (nhận).
a) Thay m=-1 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}3x+y=7\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)
Vậy: Khi m=-1 thì (x,y)=(1;4)
b) Ta có: \(\left\{{}\begin{matrix}3x+y=2m+9\\x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+y=2m+9\\x=5-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left(5-y\right)+y=2m+9\\x=5-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}15-3y+y=2m+9\\x=5-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2y=2m-6\\x=5-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-m+3\\x=5-\left(-m+3\right)=5+m-3=m+2\end{matrix}\right.\)
Ta có: \(x^2+2y^2=18\)
\(\Leftrightarrow\left(m+2\right)^2+2\cdot\left(-m+3\right)^2=18\)
\(\Leftrightarrow m^2+4m+4+2\left(m^2-6m+9\right)-18=0\)
\(\Leftrightarrow m^2+4m-14+2m^2-12m+18=0\)
\(\Leftrightarrow3m^2-8m+4=0\)
\(\Leftrightarrow3m^2-2m-6m+4=0\)
\(\Leftrightarrow m\left(3m-2\right)-2\left(3m-2\right)=0\)
\(\Leftrightarrow\left(3m-2\right)\left(m-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3m-2=0\\m-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3m=2\\m=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{2}{3}\\m=2\end{matrix}\right.\)
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{3}\ne-\dfrac{1}{m}\)
=>\(m^2\ne-3\)(luôn đúng)
\(\left\{{}\begin{matrix}mx-y=2\\3x+my=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-2\\3x+m\cdot\left(mx-2\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\x\left(m^2+3\right)=5+2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-2\\x=\dfrac{2m+5}{m^2+3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=\dfrac{2m^2+5m}{m^2+3}-2=\dfrac{2m^2+5m-2m^2-6}{m^2+3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=\dfrac{5m-6}{m^2+3}\end{matrix}\right.\)
\(x+y=\dfrac{3}{m^2+3}\)
=>\(\dfrac{2m+5+5m-6}{m^2+3}=\dfrac{3}{m^2+3}\)
=>\(7m-1=3\)
=>7m=4
=>m=4/7(nhận)
Ta có: D = m − 1 3 m = m 2 + 3 ; D x = 2 − 1 5 m = 2 m + 5 ; D y = m 2 3 5 = 5 m − 6
Vì m 2 + 3 ≠ 0 , ∀ m nên hệ phương trình luôn có nghiệm duy nhất x = D x D = 2 m + 5 m 2 + 3 y = D y D = 5 m − 6 m 2 + 3
Theo giả thiết, ta có:
x + y < 1 ⇔ 2 m + 5 m 2 + 3 + 5 m − 6 m 2 + 3 < 1 ⇔ 7 m − 1 m 2 + 3 < 1
⇔ 7 m − 1 < m 2 + 3 ⇔ m 2 − 7 m + 4 > 0 ⇔ m > 7 + 33 2 m < 7 − 33 2
Đáp án cần chọn là: A