Các câu hỏi tương tự
Biết \(0< x\le y\)và \(\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+2\left(x+2y\right)}\right)+\left(\frac{y}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)=\frac{5}{3}\)
Tính \(\frac{x}{y}\)
Cho biết\(\left(\sqrt{x^2+5}+x\right)\left(\sqrt{y^2+5}+y\right)=5\). tính x+y?
Aleft(sqrt{5}-sqrt{2}right)^2-frac{9}{sqrt{10}-1}+sqrt{90}Bsqrt{2}left(3sqrt{2}+sqrt{3-sqrt{5}}right)-sqrt{5}Cleft(frac{5-sqrt{5}}{sqrt{5}-1}-frac{sqrt{5}+1}{5+sqrt{5}}right):frac{sqrt{5}+1}{sqrt{5}}Dfrac{xsqrt{y}-ysqrt{x}+sqrt{x}-sqrt{y}}{1+sqrt{xy}}:frac{x+2sqrt{xy}+y}{left(sqrt{x}+sqrt{y}right)^3left(x+yright)}vớix,y0TÍNH HOẶC RÚT GỌN
Đọc tiếp
\(A=\left(\sqrt{5}-\sqrt{2}\right)^2-\frac{9}{\sqrt{10}-1}+\sqrt{90}\)\(B=\sqrt{2}\left(3\sqrt{2}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\)\(C=\left(\frac{5-\sqrt{5}}{\sqrt{5}-1}-\frac{\sqrt{5}+1}{5+\sqrt{5}}\right):\frac{\sqrt{5}+1}{\sqrt{5}}\)\(D=\frac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}:\frac{x+2\sqrt{xy}+y}{\left(\sqrt{x}+\sqrt{y}\right)^3\left(x+y\right)}vớix,y>0\)
TÍNH HOẶC RÚT GỌN
H24
26 tháng 10 2021 lúc 17:16
Ghpt:
a) \(\left\{{}\begin{matrix}\left(4x^2+1\right).x+\left(y-3\right)\sqrt{5-2y}=0\\4x^2+y^2+2\sqrt{3-4x}=7\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+y^2=5\\\sqrt{y-1}\left(x+y-1\right)=\left(y-2\right)\sqrt{x+y}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{x+2}\left(x+3\right)=\sqrt{y}\left[\sqrt{y\left(x+2\right)}+1\right]\\x^2+\left(y+1\right)\left(2x-y+5\right)=x+16\end{matrix}\right.\)
Cho\(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\)Tính x + y
thực hiện phép tínha)dfrac{3}{5}-dfrac{1}{2}sqrt{1dfrac{11}{25}}b)(5+2sqrt{6})(5-2sqrt{6})c)sqrt{left(2-sqrt{3}right)^2}+sqrt{4-2sqrt{3}}d)dfrac{left(xsqrt{y}+ysqrt{x}right)left(sqrt{x}-sqrt{y}right)}{sqrt{xy}}(với x,y0)
Đọc tiếp
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
cho các số thực x,y thỏa mãn \(\left(x+2+\sqrt{x^2+4x+5}\right)\left(y-1+\sqrt{y^2-2y+2}\right)=1\).
Tính P=x+y
a)3sqrt{40sqrt{12}}+4sqrt{sqrt{75}}-5sqrt{5sqrt{48}}b)sqrt{8sqrt{3}}+3sqrt{20sqrt{3}}-2sqrt{45sqrt{3}}c)left(sqrt{x}-1right).left(x+sqrt{x}+1right)left(xge0;yge0right)d)left(sqrt{x}+1right)left(x+1-sqrt{x}right)left(xge0;yge0right)e)left(sqrt{x}+yright)left(x+y^2-ysqrt{2}right)left(xge0;yge0right)
Đọc tiếp
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right)\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)