Violympic toán 9
Các câu hỏi tương tự
\(\sqrt{10}\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)}\)
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...
Đọc tiếp
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 :
\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
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)\(\left(1+tan^2a\right)\left(1-sin^2a\right)+\left(1+cotan^2a\right)\left(1-cos^2a\right)\)
Tính:
\(a)A=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\\ b)\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
Thực hiện phép tính :
a, \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
b, \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
c, \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+3\right)\)
d,
1. Rút gọn Afrac{sqrt{14+6sqrt{5}}-sqrt{14-6sqrt{5}}}{sqrt{left(sqrt{5}+1right)cdotsqrt{6-2sqrt{5}}}}
2.Tính a) Bleft(sqrt[3]{2}+1right)^3cdotleft(sqrt[3]{2}-1right)^3
b)Tìm Ca^3b-ab^3 với afrac{6}{2sqrt[3]{2}-2+sqrt[3]{4}}; bfrac{2}{2sqrt[3]{2}+2+sqrt[3]{4}}
3. Giải left|x^2-x+1right|-left|x-2right|6
Đọc tiếp
1. Rút gọn \(A=\frac{\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}}{\sqrt{\left(\sqrt{5}+1\right)\cdot\sqrt{6-2\sqrt{5}}}}\)
2.Tính a) \(B=\left(\sqrt[3]{2}+1\right)^3\cdot\left(\sqrt[3]{2}-1\right)^3\)
b)Tìm C=\(a^3b-ab^3\) với \(a=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\); \(b=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
3. Giải \(\left|x^2-x+1\right|-\left|x-2\right|=6\)
\(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\left(-\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\)
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) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)