Bạn tham khảo ở đây:
ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{1-x}=b\end{matrix}\right.\) ta được hpt:
\(\Rightarrow\left\{{}\begin{matrix}2a^3=\left(a+b\right)\left(2-ab\right)\\a^2+b^2=2\end{matrix}\right.\) thế pt dưới vào pt trên ta được:
\(2a^3=\left(a+b\right)\left(a^2+b^2-ab\right)\)
\(\Rightarrow2a^3=a^3+b^3\Rightarrow a^3=b^3\)
\(\Rightarrow a=b\Rightarrow\sqrt{x+1}=\sqrt{1-x}\)
\(\Rightarrow x+1=1-x\) \(\Rightarrow x=0\)Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
Rút gọn biểu thức :
\(B=\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\left(1+x\right)\sqrt{1+x}-\left(1-x\right)\sqrt{1-x}\right]}{x\left(2+\sqrt{1-x^2}\right)}\)
Giúp mình với các cao nhân
1. Rút gọn
P=\(2\sqrt{1+\frac{1}{4}\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}:\left[\sqrt{1+\frac{1}{4}\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}-\frac{1}{2}\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2\right]\)
rút gọn biểu thức P=\(\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\left(1+x\right)\sqrt{1+x}-\left(1-x\right)\sqrt{1-x}\right]}{x\left(2+\sqrt{1-x^2}\right)}\)
Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
Rút gọn các biểu thức sau:
a) \(\left(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)\)
b) \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\) với x>0
1) Rút gọn biểu thức
P=\(\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right).\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
\(\frac{\sqrt{1-\sqrt{1-x^2}}.\left(\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\right)}{2-\sqrt{1-x^2}}\)
Giải:
\(2\left(x+1\right)\sqrt{x+1}=\left(\sqrt{x+1}+\sqrt{1-x}\right)\left(2-\sqrt{1-x^2}\right)\)
