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a. \(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=x-3\sqrt{x} +2\sqrt{x}-6=x-\sqrt{x}-6\)
b. \(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=x-y\)
c. \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right).\sqrt{3}\)
\(=\left(\dfrac{5}{\sqrt{3}}-\dfrac{7}{\sqrt{3}}+\sqrt{3}\right).\sqrt{3}=\dfrac{5}{3}-\dfrac{7}{3}+9=\dfrac{25}{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{5}\right)\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{3}\right)^2-5=1+2\sqrt{3}+3-5=2\sqrt{3}-1\)
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a, Với y >= 0
hpt có dạng \(\left\{{}\begin{matrix}2x+y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=9\\y=x-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)(ktmđk)
Với y < 0 hpt có dạng
\(\left\{{}\begin{matrix}2x-y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3-6=-9\end{matrix}\right.\)(tm)
b, bạn tự làm
c, đk : x>= 3
\(\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\\sqrt{x+3}-3\left|y-2\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\2\sqrt{x+3}-6\left|y-2\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\left|y-2\right|=1\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y-2=\dfrac{1}{7}\\y-2=-\dfrac{1}{7}\end{matrix}\right.\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\)
bạn tự giải nốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\sqrt{2x\left(x+y\right)^3}+y\sqrt{2\left(x^2+y^2\right)}\)
\(=\sqrt{\left(2x^2+2xy\right)\left(x^2+2xy+y^2\right)}+\sqrt{2}y.\sqrt{x^2+y^2}\)
\(\le\sqrt{\left(2x^2+2xy+2y^2\right)\left(x^2+2xy+y^2+x^2+y^2\right)}=2\left(x^2+xy+y^2\right)\)
\(\Rightarrow3\left(x^2+y^2\right)\le2\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left(x-y\right)^2\le0\)
\(\Rightarrow x=y\)
Thế vào pt đầu:
\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)
Đặt \(\sqrt{x^2+1}=t\Rightarrow t^2-\left(x+3\right)t+3x=0\)
\(\Delta=\left(x+3\right)^2-12x=\left(x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3-\left(x-3\right)}{2}=3\\t=\dfrac{x+3+x-3}{2}=x\end{matrix}\right.\)
\(\Rightarrow...\)
2. 4 biến xét dài quá, để người khác
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gợi ý nè
1) \(ab+c=ab+c\left(a+b+c\right)\)....
2) nhiều cách lắm nhưng tớ chỉ đưa ra 2 cách ...có vẻ hay
đặt \(\sqrt{x}=a,\sqrt{y}=b\)
=>a3+b3=a4+b4=a5+b5
c1: ta có: \(\left(a^3+b^3\right)\left(a^5+b^5\right)=\left(a^4+b^4\right)^2\)......
c2: a5+b5=(a+b)(a4+b4)-ab(a3+b3)
=> 1=(a+b)-ab .......
3) try use UCT
4) tính sau =))
x+y=0