Ta có: \(\sqrt{x+1}=\sqrt{5}-3\)
\(\Leftrightarrow x+1=\left(\sqrt{5}-3\right)^2\)
\(\Leftrightarrow x+1=5+9-2\cdot3\cdot\sqrt{5}\)
\(\Leftrightarrow x+1=14-6\sqrt{5}\)
\(\Leftrightarrow x=14-6\sqrt{5}-1\)
hay \(x=13-6\sqrt{5}\)
Vậy: \(x=13-6\sqrt{5}\)
Rút gọn: \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x-2}}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\). Tìm các giá trị nguyên của x để Q âm
Tính
3) \(\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{2x-\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{3x\sqrt{x}-2x+\sqrt{x}-3}{x\sqrt{x}+1}\)
4) \(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
5)\(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-5\sqrt{x}+6}\)
Help !!! Mk đang cần gấp ,thank các ben
Giải phương trình \(\dfrac{3\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\dfrac{4\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{5}\right)}+\dfrac{5\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)
P=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
1. Tính P khi x=\(7+2\sqrt{3}\)
2. Tìm x để P<1
1/ Tính:
a) \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)
b) \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
2/ Rút Gọn: với a ≥ 0, a ≠ 1
B=\(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)
3/ Cho biểu thức: A = \(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)
a) Tìm điều kiện xác định của A
b) Rút gọn A
c) Tìm x để A < -1
Giải phương trình:
\(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\)
\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\)
Giải phương trình:
\(a)\sqrt{x^2+2x+4}\ge x-2\\ b)x=\sqrt{x-\frac{1}{x}}+\sqrt{x+\frac{1}{x}}\\ c)\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2\sqrt{2x-5}}\\ d)x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ e)\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
Cho A = \(\frac{2x+15\sqrt{x}+18}{x+3\sqrt{x}-18}+\frac{3x+4\sqrt{x}+1}{2x-3\sqrt{x}-5}-\frac{8x-15\sqrt{x}}{2x\sqrt{x}-11x+5\sqrt{x}}\)
Tính A tại \(x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(1,x=\sqrt{3-x}.\sqrt{4-x}+\sqrt{4-x}.\sqrt{5-x}+\sqrt{5-x}.\sqrt{3-x}\)
2) \(\dfrac{x^2-4}{x}+\dfrac{y^2-4}{y}+8=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)
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) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
