Giải HPT: \(\left\{{}\begin{matrix}2\sqrt{x}\left(1+\dfrac{1}{x+y}\right)=3\\2\sqrt{y}\left(1-\dfrac{1}{x+y}+1\right)\end{matrix}\right.\)
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Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
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a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c: ĐKXĐ: y>2
\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)
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Đặt \(\left(x-2\right)^2=a;\dfrac{1}{\sqrt{y+5}}=b\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+b=3\\a-2b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1\\y+5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;1\right\}\\y=-4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
ĐK: \(x>\dfrac{1}{4};y>-\dfrac{1}{4}\), đặt \(a=\dfrac{1}{\sqrt{12x-3}};b=\dfrac{1}{\sqrt{4y+1}}\)với a,b>0
khi đó, ta có hệ phương mới \(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}80a+40b=8\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45a=3\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35.\dfrac{1}{15}+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\b=\dfrac{1}{15}\end{matrix}\right.\)
thay \(\dfrac{1}{\sqrt{12x-3}}=a\) hay \(\dfrac{1}{\sqrt{12x-3}}=\dfrac{1}{15}\Rightarrow\sqrt{12x-3}=15\Leftrightarrow12x-3=225\Leftrightarrow12x=228\Leftrightarrow x=19\left(TMĐK\right)\) thay \(\dfrac{1}{\sqrt{4y+1}}=b\) hay
\(\dfrac{1}{\sqrt{4y+1}}=\dfrac{1}{15}\Rightarrow\sqrt{4y+1}=15\Leftrightarrow4y+1=225\Leftrightarrow4y=224\Leftrightarrow y=56\left(TMĐK\right)\)
Vậy (x;y)=(9;56) là nghiệm duy nhất của hệ phương trình đã cho.
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
ĐK: x,y#0, khi đó \(\dfrac{1}{x}+\dfrac{1}{y}=4\Rightarrow x+y=4xy\)
Do đó \(x\left(1+4y\right)+y=2\Leftrightarrow x+4xy+y=2\Leftrightarrow x+x+y+y=2\Leftrightarrow2\left(x+y\right)=2\Leftrightarrow x+y=1\)
Mà \(4xy=x+y\Leftrightarrow4xy=1\Leftrightarrow xy=\dfrac{1}{4}\)
Vậy \(x+y=1;xy=\dfrac{1}{4}\)
Do đó x,y là nghiệm của phương trình:
\(t^2-t+\dfrac{1}{4}=0\)
\(\Delta=b^2-4ac=1-4.1.\dfrac{1}{4}=0\)
Phương trình có nghiêm kép \(x_1=x_2=-\dfrac{b}{2a}=-\dfrac{-1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=y=\dfrac{1}{2}\left(nhận\right)\)
Vậy (x;y)=\(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) là nghiệm duy nhất của hệ phương trình đã cho.
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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
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...
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\(1,HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)+\left(\dfrac{1}{y}-\dfrac{1}{x}\right)=0\\2y=x^3+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\dfrac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow2y=y^3+1\Leftrightarrow y^3-2y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;0\right);\left(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1+\sqrt{5}}{2}\right);\left(\dfrac{-1-\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right)\)
\(2,HPT\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x^2+y^2\right)}+2\sqrt{xy}=16\\x+y+2\sqrt{xy}=16\end{matrix}\right.\\ \Leftrightarrow\sqrt{2\left(x^2+y^2\right)}=x+y\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\\ \Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\)
Vậy \(\left(x;y\right)=\left(4;4\right)\)
\(3,\text{Sửa: }\left\{{}\begin{matrix}\sqrt{x^2+3}+\left|y\right|=\sqrt{3}\left(1\right)\\\sqrt{y^2+5}+\left|x\right|=\sqrt{x^2+5}\left(2\right)\end{matrix}\right.\)
Ta thấy \(\sqrt{x^2+3}\ge\sqrt{3};\left|y\right|\ge0\Leftrightarrow VT\left(1\right)\ge\sqrt{3}=VP\left(1\right)\)
Dấu \("="\Leftrightarrow x=y=0\)
Thay vào \(\left(2\right)\Leftrightarrow\sqrt{5}+0=\sqrt{5}\left(tm\right)\)
Vậy \(\left(x;y\right)=\left(0;0\right)\)
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1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
PT (2) là \(2\sqrt{y}\left(1-\dfrac{1}{x+y}\right)=1\) ạ !!!!!!!