Violympic toán 9
Các câu hỏi tương tự
Giải hệ pt
1/left{{}begin{matrix}4xsqrt{y+1}+8xleft(4x^2-4x-3right)sqrt{x+1}dfrac{x}{x+1}+x^2left(y+2right)sqrt{left(x+1right)left(y+1right)}end{matrix}right.
2/left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.
3/left{{}begin{matrix}left(2x+y-1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end...
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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!!
25 tháng 4 2020 lúc 11:56
Gpt: a) sqrt[4]{3left(x+5right)}-sqrt[4]{11-x}sqrt[4]{13+x}-sqrt[4]{3left(3-xright)}
b) frac{1+2sqrt{x}-xsqrt{x}}{3-x-sqrt{2-x}}2left(frac{1+xsqrt{x}}{1+x}right) c) sqrt{x+1}+frac{4left(sqrt{x+1}+sqrt{x-2}right)}{3left(sqrt{x-2}+1right)^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+xsqrt{x^2+2}+left(x+1right)sqrt{x^2+2x+2}0
f) sqrt{2x+3}cdotsqrt[3]{x+5}x^2+x-6
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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) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
3 tháng 3 2022 lúc 21:02
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=\(\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\left(1+x\right)\sqrt{1+x}-\left(1-x\right)\sqrt{1-x}\right]}{x\left(2+\sqrt{1-x^2}\right)}\)
Cho bt: \(M=\left(\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x}+\sqrt{x+1}-\sqrt{x+2}\right)\left(\sqrt{x}-\sqrt{x+1}+\sqrt{x+2}\right)\left(-\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}\right)\) với x là số tự nhiên khác 0
Cm: M là số tự nhiên
Giải các phương trình sau:
a, sqrt{x^2-6x+9}+sqrt{2x^2+8x+8}sqrt{x^2-2x+1}
b, sqrt{x-3-2sqrt{x-4}}+sqrt{x-4sqrt{x-1}}1
c. sqrt{x+8-6sqrt{x-1}}4
d, sqrt{xleft(x-3right)}+sqrt{xleft(x-4right)}2sqrt{x^2}
e, sqrt{left(x+3right)left(x+2right)}+sqrt{left(x+3right)left(x-1right)}2sqrt{left(x+3right)^2}
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Giải các phương trình sau:
a, \(\sqrt{x^2-6x+9}+\sqrt{2x^2+8x+8}=\sqrt{x^2-2x+1}\)
b, \(\sqrt{x-3-2\sqrt{x-4}}+\sqrt{x-4\sqrt{x-1}}=1\)
c. \(\sqrt{x+8-6\sqrt{x-1}}=4\)
d, \(\sqrt{x\left(x-3\right)}+\sqrt{x\left(x-4\right)}=2\sqrt{x^2}\)
e, \(\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}=2\sqrt{\left(x+3\right)^2}\)
RG
A = \(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
B = \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(2\left(x+1\right)\sqrt{x+1}=\left(\sqrt{x+1}+\sqrt{1-x}\right)\left(2-\sqrt{1-x^2}\right)\)
Cho bt:
\(M=\left(\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x}+\sqrt{x+1}-\sqrt{x+2}\right)\left(\sqrt{x}-\sqrt{x-1}+\sqrt{x+2}\right)\left(-\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}\right)\) với x là số tự nhiên khác 0
Chm M là số tự nhiên