Giải pt :
\(\left\{{}\begin{matrix}2x^2y^2+x^2+2x=2\\2x^2y-x^2y^2+2xy=1\end{matrix}\right.\)
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 3 2021
a. ĐKXĐ: ..
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}-\sqrt{2\left(x+y\right)}=4\\x+2y+\dfrac{2\sqrt{\left(x+y\right)\left(2x+5y\right)}}{3}=24\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}=a\ge0\\\sqrt{2\left(x+y\right)}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=4\\\dfrac{a^2+b^2}{6}+\dfrac{ab}{3}=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left(a+b\right)^2=144\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left[{}\begin{matrix}a+b=12\\a+b=-12\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(8;4\right)\\\left(a;b\right)=\left(-4;-8\right)\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\left(2x+5y\right)=64\\2\left(x+y\right)=16\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 3 2021
b.
Thế pt trên xuống dưới:
\(x^4+6y^4=\left(x+2y\right)\left(x^3+3y^3-2xy^2\right)\)
\(\Leftrightarrow2x^3y-2x^2y^2-xy^3=0\)
\(\Leftrightarrow xy\left(2x^2-2xy-y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-\left(1+\sqrt{3}\right)x\\y=\left(-1+\sqrt{3}\right)x\end{matrix}\right.\)
Thế vào pt đầu ...
Đề cho hơi xấu, nếu pt đầu là \(x^3+3y^3-2x^2y=1\) thì đẹp hơn nhiều
29 tháng 7 2021
ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
b.
Với \(xy=0\) không là nghiệm
Với \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y^2+1\right)=y\left(5-x^2\right)\\y^2+1=y\left(5-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)
\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)
\(\Leftrightarrow5-x^2=5x-2x^2\)
\(\Leftrightarrow...\)
7 tháng 10 2021
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
7 tháng 10 2021
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
15 tháng 12 2021
Từ 2 PT ta được:
\(\Leftrightarrow x^2-x^2y+y^2-y^2x=x-2xy+y\\ \Leftrightarrow\left(x+y\right)^2-xy\left(x+y\right)-\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)\left(x+y-xy-1\right)=0\\ \Leftrightarrow\left(x+y\right)\left(1-y\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y=0\\y=1\\x=1\end{matrix}\right.\)
Với \(x+y=0\Leftrightarrow x=-y\Leftrightarrow-y+2y^2+y=3\Leftrightarrow y^2=\dfrac{3}{2}\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{\sqrt{6}}{2}\Leftrightarrow x=-\dfrac{\sqrt{6}}{2}\\y=-\dfrac{\sqrt{6}}{2}\Leftrightarrow x=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)
Với \(y=1\Leftrightarrow x-2x+1=3\Leftrightarrow x=-2\)
Với \(x=1\Leftrightarrow1-2y+y=3\Leftrightarrow y=-2\)
Vậy \(\left(x;y\right)\in\left\{\left(-2;1\right);\left(1;-2\right);\left(\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right);\left(-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\right)\right\}\)
H
25 tháng 6 2019
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
T
3 tháng 11 2019
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
13 tháng 4 2017
\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{2x+y+1}+2\sqrt[3]{7x+12y+8}=2xy+y+5\end{matrix}\right.\)
Xét \(pt\left(1\right)\) dễ dàng suy ra \(x+y\ge0\)
\(VT=\sqrt{\left(x-y\right)^2+\left(2x+y\right)^2}+\sqrt{\left(x-y\right)^2+\left(2y+x\right)^2}\)
\(\ge\left|2x+y\right|+\left|2y+x\right|\ge3\left(x+y\right)\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=y\\x,y\ge0\end{matrix}\right.\)
Thay vào \(pt\left(2\right)\) ta được:
\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)
\(\Leftrightarrow\left[\sqrt{3x+1}-\left(x+1\right)\right]+2\left[\sqrt[3]{19x+8}-\left(x+2\right)\right]=2x^2-2x\)
\(\Leftrightarrow\left(x-x^2\right)\left[\dfrac{1}{\sqrt{3x+1}+x+1}+2\cdot\dfrac{x+7}{\sqrt[3]{\left(19x+8\right)^2}+\left(x+2\right)\sqrt[3]{19x+8}+\left(x+2\right)^2}+2\right]=0\)
Do \(x;y\ge0\) nên pt trong ngoặc luôn dương
\(\Rightarrow x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Mà \(x=y\)\(\Rightarrow\left[{}\begin{matrix}x=y=0\\x=y=1\end{matrix}\right.\) là nghiệm của hpt
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
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?
Nhận thấy \(x=0\) không phải nghiệm, chia vế cho vế ta được:
\(\frac{2xy^2+x+2}{2xy-xy^2+2y}=2\Leftrightarrow2xy^2+x+2=4xy-2xy^2+4y\)
\(\Leftrightarrow4xy^2-2\left(2x+2\right)y+x+2=0\)
\(\Delta'=\left(2x+2\right)^2-4x\left(x+2\right)=4\)
\(\Rightarrow\left\{{}\begin{matrix}y=\frac{2x+2+2}{4x}=\frac{x+2}{2x}\\y=\frac{2x+2-2}{4x}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}xy=\frac{x+2}{2}\\y=\frac{1}{2}\end{matrix}\right.\)
Thay vào pt ban đầu
\(\Rightarrow\left[{}\begin{matrix}2\left(\frac{x+2}{2}\right)^2+x^2+2x=2\\2x^2.\frac{1}{4}+x^2+2x=2\end{matrix}\right.\) \(\Leftrightarrow...\)