\(\left|x+y\right|+\left|2x-y\right|=0\)
Những câu hỏi liên quan
a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
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b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
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c.
\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
Xét pt:
\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)
TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)
\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
TH2: \(y=6-x\) thế vào...
\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)
\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)
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21 tháng 6 2023 lúc 21:08
Tìm tất cả các hàm số \(f:\left(0;+\infty\right)\rightarrow\left(0;+\infty\right)\) thỏa mãn
\(f\left(x+f\left(y\right)+y\right)=f\left(2x\right)+f\left(y\right),\forall x,y\in\left(0;+\infty\right)\)
Giải hệ pta)\(\left\{{}\begin{matrix}x^3+xy^2+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+2y^2+xy+2x-4=0\end{matrix}\right.\)b) \(\left\{{}\begin{matrix}x^4+x^2y^2-y^2=y^3+x^2y+x^2\\5\left(x\sqrt{y+7}+\left(x+1\right)\sqrt{2x+y+8}\right)=13\left(2x...
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a.
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2+y^2\right)+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+y^2+\left(x+y-2\right)\left(y+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x^2+y^2-4\right)\left(y+2\right)=-x\left(x^2+y^2\right)\\-\left(x^2+y^2\right)=\left(x+y-2\right)\left(y+2\right)\end{matrix}\right.\)
\(\Rightarrow\left(x^2+y^2-4\right)\left(y+2\right)=x\left(x+y-2\right)\left(y+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\left(\text{không thỏa mãn}\right)\\x^2+y^2-4=x\left(x+y-2\right)\end{matrix}\right.\)
\(\Rightarrow x^2+y^2-4=x^2+x\left(y-2\right)\)
\(\Leftrightarrow\left(y+2\right)\left(y-2\right)=x\left(y-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\x=y+2\end{matrix}\right.\)
Thế vào pt dưới:
\(\Rightarrow\left[{}\begin{matrix}x^2+8+2x+2x-4=0\\\left(y+2\right)^2+2y^2+y\left(y+2\right)+2\left(y+2\right)-4=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Câu b chắc chắn đề sai, nhìn 2 vế pt đầu đều có \(x^2\) thì chúng sẽ rút gọn, không ai cho đề như thế hết
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dễ dàng phân tích được sqrt{2x-y}frac{left(x^2-x-xyright)}{left(y+1right)}left(y+1right)frac{left(x^2-x-xyright)}{sqrt{2x-y}}left(y+1right)sqrt{2x-y}frac{left(x^2-x-xyright)^2}{sqrt{2x-y}left(y+1right)}thay vào pt 1 ta đượcleft(x^2-x-xyright)left(frac{x^2-x-xy-1}{sqrt{2x-y}left(y+1right)}right)0x^2-x-xy0Leftrightarrow x^2xleft(1+yright)Leftrightarrow x1+y thay xy+1 vào pt2 ta đượcleft(y+1right)^2+y^2-2yleft(y+1right)-3left(y+1right)+20left(y^2+y^2-2y^2right)+left(2y-2y-3yright)+left(1-3+2right)0...
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dễ dàng phân tích được
\(\sqrt{2x-y}=\frac{\left(x^2-x-xy\right)}{\left(y+1\right)}\)
\(\left(y+1\right)=\frac{\left(x^2-x-xy\right)}{\sqrt{2x-y}}\)
\(\left(y+1\right)\sqrt{2x-y}=\frac{\left(x^2-x-xy\right)^2}{\sqrt{2x-y}\left(y+1\right)}\)
thay vào "pt" 1 ta được
\(\left(x^2-x-xy\right)\left(\frac{x^2-x-xy-1}{\sqrt{2x-y}\left(y+1\right)}\right)=0\)
\(x^2-x-xy=0\Leftrightarrow x^2=x\left(1+y\right)\Leftrightarrow x=1+y\)
thay x=y+1 vào pt2 ta được
\(\left(y+1\right)^2+y^2-2y\left(y+1\right)-3\left(y+1\right)+2=0\)
\(\left(y^2+y^2-2y^2\right)+\left(2y-2y-3y\right)+\left(1-3+2\right)=0\)
\(-3y=0\Leftrightarrow y=0\)
thay \(y=0\)
Bài 2:a. 2x^2+2xy+y^2+96x-left|y+3right| Leftrightarrowleft|y+3right|6x-2x^2-2xy-y^2-9 Leftrightarrowleft|y+3right|-x^2-2xy-y^2-x^2+6x-9 Leftrightarrowleft|y+3right|-left(x+yright)^2-left(x-3right)^2 Leftrightarrowleft|y+3right|-left[left(x+yright)^2+left(x-3right)^2right] Có: left|y+3right|ge0 -left[left(x+yright)^2+left(x-3right)^2right]le0 Do đó: left|y+3right|-left[left(x+yright)^2+left(x-3right)^2right]0 Leftrightarrowhept{begin{cases}y+30x+y0x-30end{cases}}Leftrightarrowhept{begin{cases}x3...
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Bài 2:
a. \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow\left|y+3\right|=6x-2x^2-2xy-y^2-9\)
\(\Leftrightarrow\left|y+3\right|=-x^2-2xy-y^2-x^2+6x-9\)
\(\Leftrightarrow\left|y+3\right|=-\left(x+y\right)^2-\left(x-3\right)^2\)
\(\Leftrightarrow\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\)
Có: \(\left|y+3\right|\ge0\)
\(-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\le0\)
Do đó: \(\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}y+3=0\\x+y=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
b. \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+\left[2\left(x^2-5x-2012\right)\right]^2=0\)
\(\Leftrightarrow\left(2x^2+x-2013-2x^2+10x+4024\right)^2=0\)
\(\Leftrightarrow\left(11x+2011\right)^2=0\)
\(\Leftrightarrow11x+2011=0\)
\(\Leftrightarrow x=-\frac{2011}{11}\)
Số nghiệm của hệ phương trình \(\left\{{}\begin{matrix}\left(2x-\left|y\right|-1\right)\left(x+2y-1\right)=0\\\left(2x-\left|y\right|-2\right)\left(x+2y-3\right)=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}2x-\left|y\right|-1=0\\x+2y-3=0\end{matrix}\right.\)
- Với \(y\ge0\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x+2y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) (t/m)
- Với \(y< 0\Leftrightarrow\left\{{}\begin{matrix}2x+y=1\\x+2y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{1}{3}\\y=\frac{5}{3}>0\end{matrix}\right.\) (ko thoả mãn)
TH2: \(\left\{{}\begin{matrix}x+2y-1=0\\2x-\left|y\right|-2=0\end{matrix}\right.\)
- Với \(y\ge0\Rightarrow\left\{{}\begin{matrix}x+2y=1\\2x-y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\) (t/m)
- Với \(y< 0\Rightarrow\left\{{}\begin{matrix}x+2y=1\\2x+y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\) ko thoả
Vậy hệ có 2 cặp nghiệm
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Tìm cap số (x,y) thoa man
\(x^2+y^2=0\)
\(x^2+2y^2+2y\left(1-x\right)=-1\)
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(4x^2+y^2-2\left(2x+y-1\right)=0\)
\(2x^2\left(1-y\right)+y\left(y+xy-2x\right)=0\)
\(x^2+y^2=0\)
Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)
(Dấu "="\(\Leftrightarrow x=y=0\))
\(x^2+2y^2+2y\left(1-x\right)=-1\)
\(\Leftrightarrow x^2+2y^2+2y-2xy+1=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y+1\right)^2=0\)
Mà \(\left(x-y\right)^2+\left(y+1\right)^2\ge0\)
(Dấu "="\(\Leftrightarrow x=-1;y=-1\)
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Cho ba số x,y,z khác 0 thỏa mãn x+y+z=0.Tính \(\dfrac{2x\left(x+y\right)\left(z+x\right)+y\left(x+y\right)\left(y+z\right)}{z\left(x+z\right)\left(y+z\right)}\)
giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3-30=0\\x^2y+x\left(1+y+y^2\right)+y-11=0\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}3xy+2y=5\\2xy\left(x+y\right)+y^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)
TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)
Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)
TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)
2 câu dưới hình như em hỏi rồi?
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Giải hệa) left{{}begin{matrix}x^2left(y^2+1right)+2yleft(x^2+x+1right)3left(x^2+xright)left(y^2+yright)1end{matrix}right.b) left{{}begin{matrix}left(6x+5right)sqrt{2x+1}-2y-3y^30y+sqrt{x}sqrt{2x^2+4x-23}end{matrix}right.Giải bất ptdfrac{9}{left|x-5right|-3}geleft|x-2right|
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Giải hệ
a) \(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)
Giải bất pt
\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)