This is my problem :)))
\(\left\{{}\begin{matrix}x^2-y^2+4=2\left(\sqrt{y}-\sqrt{x+2}-2x\right)\\4\sqrt{x+2}+\sqrt{28-3y}=y^2-4x+4\end{matrix}\right.\)
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AH
Akai Haruma
Giáo viên
6 tháng 1 2020
Lời giải:
PT $(1)$:
\(\Leftrightarrow (x^2+4x+4)-y^2=2(\sqrt{y}-\sqrt{x+2})\)
\(\Leftrightarrow (x+2)^2-y^2=2(\sqrt{y}-\sqrt{x+2})(*)\)
Nếu $\sqrt{y}+\sqrt{x+2}=0\Rightarrow y=x+2=0$
$\Rightarrow y=0; x=-2$. Thay vào PT $(2)$ thấy không thỏa mãn (loại)
Nếu $\sqrt{y}+\sqrt{x+2}>0$:
$(*)\Leftrightarrow (x+2-y)(x+2+y)=2.\frac{y-(x+2)}{\sqrt{y}+\sqrt{x+2}}$
$\Leftrightarrow (x+2-y)\left[x+2+y+\frac{2}{\sqrt{y}+\sqrt{x+2}}\right]=0$
Dễ thấy với mọi $\sqrt{y}+\sqrt{x+2}$ thì biểu thức trong ngoặc vuông luôn lớn hơn $0$
Do đó $x+2-y=0\Rightarrow x+2=y$
Thay vào PT $(2)$:
$4\sqrt{x+2}+\sqrt{22-3x}=x^2+8$
\(\Leftrightarrow 4\sqrt{x+2}+\sqrt{22-3x}=x^2+8\)
\(\Leftrightarrow 4(\sqrt{x+2}-2)+(\sqrt{22-3x}-4)=x^2-4\)
\(\Leftrightarrow 4.\frac{x-2}{\sqrt{x+2}+2}-\frac{3(x-2)}{\sqrt{22-3x}+4}=(x-2)(x+2)\)
\(\Leftrightarrow (x-2)\left[\frac{4}{\sqrt{x+2}+2}-\frac{3}{\sqrt{22-3x}+4}-(x+2)\right]=0\)
\(\Leftrightarrow (x-2)\left[\frac{4}{\sqrt{x+2}+2}-\frac{4}{3}-(\frac{3}{\sqrt{22-3x}+4}-\frac{1}{3})-(x+1)\right]=0\)
\(\Leftrightarrow (x-2)\left[\frac{-4(x+1)}{3\sqrt{x+2}+2)(\sqrt{x+2}+1)}-\frac{3(x+1)}{3(\sqrt{22-3x}+4)(5+\sqrt{22-3x})}-(x+1)\right]=0\)
\(\Leftrightarrow (x-2)(x+1)\left[\frac{-4}{.....}-\frac{3}{.....}-1\right]=0\)
Dễ thấy biểu thức trong ngoặc vuông luôn âm nên $(x-2)(x+1)=0\Rightarrow x=2$ hoặc $x=-1$
Với $x=2\rightarrow y=x+2=4$
Với $x=-1\rightarrow y=x+2=1$
25 tháng 11 2023
a:
ĐKXĐ: y+1>=0
=>y>=-1
\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}+7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-2x\right)+2\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x^2-2x\right)=-7\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x=-1\\3\cdot\left(-1\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1=0\\2\sqrt{y+1}=-3+7=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\sqrt{y+1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=0\\y+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\left(nhận\right)\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\cdot\sqrt{\left(2x-2\right)^2}+5\cdot\sqrt{\left(y+2\right)^2}=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}20\left|x-1\right|-12\left|y+2\right|=28\\20\left|x-1\right|+25\left|y+2\right|=65\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-37\left|y+2\right|=-37\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|y+2\right|=1\\4\left|x-1\right|=13-5=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|y+2\right|=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1\in\left\{2;-2\right\}\\y+2\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;-1\right\}\\y\in\left\{-1;-3\right\}\end{matrix}\right.\)
c: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-1\\y< >-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=3-4=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=2-9=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=-1+2\cdot\dfrac{19}{11}=\dfrac{27}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x=\dfrac{2}{9}\end{matrix}\right.\)(nhận)
d:
ĐKXĐ: x<>1 và y<>-2
\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)
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 12 2022
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
Lời giải:
ĐKXĐ:......
\(\left\{\begin{matrix} x^2-y^2+4=2(\sqrt{y}-\sqrt{x+2}-2x)(1)\\ 4\sqrt{x+2}+\sqrt{28-3y}=y^2-4x+4(2)\end{matrix}\right.\)
Từ \((1)\Leftrightarrow x^2+4x-y^2+4=2(\sqrt{y}-\sqrt{x+2})\)
\(\Leftrightarrow (x+2)^2-y^2=2(\sqrt{y}-\sqrt{x+2})\)
\(\Leftrightarrow (x+2-y)(x+2+y)=2(\sqrt{y}-\sqrt{x+2})\)
\(\Leftrightarrow (\sqrt{x+2}-\sqrt{y})(\sqrt{x+2}+\sqrt{y})(x+2+y)+2(\sqrt{x+2}-\sqrt{y})=0\)
\(\Leftrightarrow (\sqrt{x+2}-\sqrt{y})[(\sqrt{x+2}+\sqrt{y})(x+2+y)+2]=0\)
Với \(x+2,y\geq 0\), hiển nhiên biểu thức trong ngoặc vuông luôn lớn hơn $0$
Do đó \(\sqrt{x+2}-\sqrt{y}=0\)
\(\Rightarrow x+2=y\)
Thay \(x=y-2\) vào PT(2) ta có:
\(4\sqrt{y}+\sqrt{28-3y}=y^2-4(y-2)+4\)
\(\Leftrightarrow 4\sqrt{y}+\sqrt{28-3y}=y^2-4y+12\)
\(\Leftrightarrow 4(\sqrt{y}-2)+(\sqrt{28-3y}-4)=y^2-4y\)
\(\Leftrightarrow \frac{4(y-4)}{\sqrt{y}+2}-\frac{3(y-4)}{\sqrt{28-3y}+4}=y(y-4)\)
\(\Leftrightarrow (y-4)\left[y+\frac{3}{\sqrt{28-3y}+4}-\frac{4}{\sqrt{y}+2}\right]=0\)
\(\Leftrightarrow \left[\begin{matrix} y=4\\ y+\frac{3}{\sqrt{28-3y}+4}=\frac{4}{\sqrt{y}+2}(*)\end{matrix}\right.\)
Xét $(*)$
Nếu \(y>1\Rightarrow \text{VT}>\frac{4}{3}; \text{VP}< \frac{4}{3}\) (loại)
Nếu \(y<1 \Rightarrow \text{VT}< \frac{4}{3}: \text{VP}> \frac{4}{3}\) (loại)
Nếu $y=1$ thì thỏa mãn
Vậy \(\left[\begin{matrix} y=4\rightarrow x=2\\ y=1\rightarrow x=-1\end{matrix}\right.\)