Giải hệ phương trình:
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{x-y}=4\\x^2+y^2=128\end{matrix}\right.\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
10 tháng 1 2021
1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
29 tháng 4 2023
4. Đk: \(x,y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
2 tháng 2 2021
\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
2 tháng 2 2021
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
HP
18 tháng 5 2022
\(\left\{{}\begin{matrix}4\sqrt{x}-3\sqrt{y}=4\\2\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4\sqrt{x}-3\sqrt{y}=4\\4\sqrt{x}+2\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y}=0\\2\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=0\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
18 tháng 5 2022
\(\Leftrightarrow\left\{{}\begin{matrix}4\sqrt{x}-3\sqrt{y}=4\\4\sqrt{x}+2\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=1\end{matrix}\right.\)
24 tháng 1 2021
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
PA
16 tháng 4 2021
Đặt \(x+y=a\) ; \(\sqrt{x+1}=b\)
Ta được hpt sau:
\(\left\{{}\begin{matrix}2a+b=4\\a-3b=-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Leftrightarrow x=3\)
\(\Rightarrow y=-2\)
10 tháng 12 2023
a: \(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\sqrt{15}x-2\sqrt{3}\cdot y=2\sqrt{15}\left(\sqrt{3}-1\right)\\2\sqrt{15}x+15y=21\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2\sqrt{3}y-15y=2\sqrt{45}-2\sqrt{15}-21\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-2\sqrt{3}-15\right)=-15\sqrt{5}-2\sqrt{15}\\2\sqrt{3}\cdot x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{15\sqrt{5}+2\sqrt{15}}{2\sqrt{3}+15}=\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\2\sqrt{3}x=21-3\sqrt{5}\cdot\sqrt{5}=21-15=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\x=\dfrac{6}{2\sqrt{3}}=\sqrt{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y=19\\4,2x+10y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y+4,2x+10y=19,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12,7x=19,8\\2,1x+5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{198}{127}\\5y=0,4-2,1x=-\dfrac{365}{127}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{198}{127}\\y=-\dfrac{73}{127}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
30 tháng 5 2021
Lời giải:
Lấy PT (2) trừ đi 2* PT(1) ta có:
$-(2y-y^2)-2(y-1)^2=-6$
$\Leftrightarrow y^2-2y-4=0$
Vì $y^2-2y=4$ nên $2\sqrt{x-2}=2+(2y-y^2)=2-(y^2-2y)=2-4=-2<0$ (vô lý)
Do đó hpt vô nghiệm.
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{x-y}=4\left(1\right)\\x^2+y^2=128\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^2+y^2=128\)
\(\Leftrightarrow2x^2+2y^2=256\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy+y^2=256\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-y\right)^2=256\)
Hệ phương trình tương đương:
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{x-y}=4\\\left(x+y\right)^2+\left(x-y\right)^2=256\end{matrix}\right.\)(*)
Đặt \(\sqrt{x+y}=a;\sqrt{x-y}=b\), (*) trở thành:
\(\left\{{}\begin{matrix}a+b=4\\a^4+b^4=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left(a^2+b^2\right)^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left(16-2ab\right)^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\256-64ab+4\left(ab\right)^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\2\left(ab\right)^2-64ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\2ab\left(ab-32\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left[{}\begin{matrix}ab=0\\ab=32\end{matrix}\right.\end{matrix}\right.\)
Tới đây Vieta đảo làm tới thôi :)