ĐKXĐ: \(x>0\)
\(\Leftrightarrow2+\sqrt{x}\le6\sqrt{x}\)
\(\Leftrightarrow5\sqrt{x}\ge2\)
\(\Leftrightarrow\sqrt{x}\ge\frac{2}{5}\)
\(\Rightarrow x\ge\frac{4}{25}\)
ĐKXĐ: \(x>0\)
\(\Leftrightarrow2+\sqrt{x}\le6\sqrt{x}\)
\(\Leftrightarrow5\sqrt{x}\ge2\)
\(\Leftrightarrow\sqrt{x}\ge\frac{2}{5}\)
\(\Rightarrow x\ge\frac{4}{25}\)
Tìm x:
a) \(\frac{2+\sqrt{x}}{2\sqrt{x}+1}\le3\)
Tìm x
a)\(\sqrt{x-1}=2\left(x\ge1\right)\)
b)\(\sqrt{3-x}=4\left(x\le3\right)\)
c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)
d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)
e)\(\sqrt{x-1}-3=1\)
f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)
Cho x, y, z > 0 thỏa mãn \(x+y+z\le3\).Tìm GTLN :
\(A=\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
A=\(\frac{3\sqrt{x}}{-2.(\sqrt{x}+2)}\)
tìm x để A + 1<0
Rút gọn:
\(A=\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right):\left(\frac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) với \(x\ge0;x\ne1\)
\(B=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right):\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\) với \(x>0;x\ne1\)
tính giá trị biểu thức
1) A = \(\frac{15\sqrt{x}-11}{x-2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\) tại \(x=3-2\sqrt{2}\)
2) \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\) tại \(x=7-2\sqrt{6}\)
3) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) tại \(x=7-4\sqrt{3}\)
A = (\(\sqrt{x}-\frac{x+2}{x-1}\)) : ( \(\frac{\sqrt{x}}{\sqrt{x}+1}\)- \(\frac{\sqrt{x}-4}{1-x}\))
B= (\(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\)) :\(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\))
C=(\(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)) .(\(\frac{1-x}{\sqrt{2}}\))2
Mọi ng giúp em nhanh với ạ, em cảm ơn ạ.
Giải hộ mình với
1 chứng minh đẳng thức:
a) \(\frac{\sqrt{a^2+x^2}+\sqrt{a^2+x^2}}{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}}=\frac{a^2}{x^2}\)với \(\left|a\right|\)>\(\left|x\right|\)
b) \(\left(\frac{5+2\sqrt{6}}{\sqrt{x}+\sqrt{2}}\right)^2-\left(\frac{5-2\sqrt{6}}{\sqrt{3}-\sqrt{6}}\right)^2=4\sqrt{6}\)
2.
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
a) Rút gọn A nếu \(x\ge0\)và \(x\ne4\)
b) Tìm x để A-2