Violympic toán 9
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Rút gọn
\(P=\frac{1}{3}\left(\sqrt[3]{2}+1\right)\left(\sqrt{12\sqrt[3]{2}-15}+2\sqrt{3\sqrt[3]{4}-3}\right)\)
\(P=\frac{1}{3}\left(\sqrt[3]{2}+1\right)\left(\sqrt{12\sqrt[3]{2}-15}+2\sqrt{3\sqrt[3]{4}-3}\right)\)
\(P=\frac{1}{3}\left(\sqrt[3]{2}+1\right)\left(\sqrt{12\sqrt[3]{2}-15}+2\sqrt{3\sqrt[3]{4}-3}\right)\)
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!!
\(\sqrt{10}\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)}\)
Rút gọn biểu thức:
a)\(\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
b)\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
c)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
d)\(\left(1+tan^2a\right)\left(1-sin^2a\right)+\left(1+cotan^2a\right)\left(1-cos^2a\right)\)
29 tháng 12 2020 lúc 20:46
Cho
\(\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{3}\)
\(\sqrt{\left(a+2b\right)\left(a+2c\right)}+\sqrt{\left(b+2a\right)\left(b+2c\right)}+\sqrt{\left(c+2a\right)\left(c+2b\right)}=3\)
Hãy tính \(\left(2\sqrt{a}+3\sqrt{b}-4\sqrt{c}\right)^2\)
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
Đọc tiếp
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) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
\(\left(3\sqrt{3}+1\right)^2-2\sqrt{3}\left(3-\sqrt{3}\right)\)
\(\sqrt{2}\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
tính \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+2\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
1)tính
a)\(\left(\dfrac{1}{5}\sqrt{500}-3\sqrt{45}+5\sqrt{20}\right):\sqrt{5}\)
b)\(\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right).\sqrt{\dfrac{1}{48}}\)
c)\(\left(\dfrac{2\sqrt{3}+3}{\sqrt{3}+2}+\dfrac{2\sqrt{2}}{\sqrt{2}+1}\right):\left(\sqrt{12}+\sqrt{18}\right)\)
CMR:
a,\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}=8}\)
b,\(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)