Tính và rút gọn:
C=\(\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
Giai các pt sau:
\(\sqrt{3x+1}-\sqrt{x-1}\)=2
b, \(\sqrt{x-5}\)=x-7
rút gọn biểu thức sau:
\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\left(x\ge0,x\ne1\right)\)
\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{3x+8\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{20\sqrt{x}+6x-10}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
Đề sai không vậy?
Ta có: \(N=\dfrac{3x+5\sqrt{x}}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{7}{\sqrt{x}+3}\)
\(=\dfrac{3x+5\sqrt{x}+3x+9\sqrt{x}-\sqrt{x}-3+7\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6x+20\sqrt{x}-10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
giải pt :
a, (x+5)(2-x)=3\(\sqrt{x^2+3x}\)
b, \(\sqrt[3]{\dfrac{2x}{x+1}}+\sqrt[3]{\dfrac{1}{2}+\dfrac{1}{2x}}=2\)
c,\(\sqrt[5]{\dfrac{16x}{x-1}}+\sqrt[5]{\dfrac{x-1}{16x}}=\dfrac{5}{2}\)
d, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
e, \(\sqrt{2x^2+4x+1}=1-2x-x^2\)
Rút gọn các biểu thức sau:
a) \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\), \(x\ge0\)
b) \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\), \(x>0\)
c) \(C=\dfrac{y}{2}+\dfrac{3}{4}\sqrt{1+9y^2-6y}-\dfrac{3}{2}\), \(y\le\dfrac{1}{3}\)
a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)
\(=14\sqrt{2x}+30\)
b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)
\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)
\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)
Câu 1 :Tính : B = ( 3 - \(\sqrt{5}\)) ( \(\sqrt{5}\) + 3 )
Câu 2 : Rút gọn : \(\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\)
Câu 3: \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a, Rút gọn biểu thức a
b, Tính giá trị của A khi x + \(\dfrac{2}{2+\sqrt{3}}\)
\(1,B=9-5=4\\ 2,\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(\sqrt{5}+1\right)\left(3+2\sqrt{2}\right)-\sqrt{10}\left(\sqrt{5}+2\right)+3\sqrt{2}-3\sqrt{5}\\ =3\sqrt{5}+2\sqrt{10}+3+2\sqrt{2}-5\sqrt{2}-2\sqrt{10}+3\sqrt{2}-3\sqrt{5}=3\)
\(3,\\ a,\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\left(x,y\ge0;xy\ne1\right)\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\\ =\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}+\sqrt{y}-y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{1+x+y+xy}\\ =\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+x\right)+y\left(1+x\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+y\right)\left(1+x\right)}\)
\(b,x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{1}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
Thay vào BT
\(=\dfrac{2\left(\sqrt{3}-1+\sqrt{y}\right)}{\left(1+y\right)\left(1+4-2\sqrt{3}\right)}=\dfrac{2\sqrt{3}-2+2\sqrt{y}}{\left(1+y\right)\left(3-2\sqrt{3}\right)}\\ =\dfrac{2\sqrt{3}-2+2\sqrt{y}}{3-2\sqrt{3}+3y-2y\sqrt{3}}\)
Rút gọn các biểu thức sau:
A=\(\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)
B=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
C=\(\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\)
\(A=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\left(x>0,x\ne1\right)\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
\(=\dfrac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\)
\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4;9\right)\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(C=\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)
\(=\dfrac{x+\sqrt{x}-1-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\dfrac{3\sqrt{x}-2}{x+\sqrt{x}+1}\)
7) cho biểu thức: P=\(\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{5}{x+\sqrt{x}-2}\right)\div\left(1+\dfrac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
a) rút gọn P
b) tính P khi \(x=6-2\sqrt{5}\)
c) tính giá trị của x để P= \(\dfrac{1}{\sqrt{x}}\)
giúp mk vs ah mk cần gấp lắm
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
tính
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
rút gọn biểu thức
A=\(\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
a, \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\left|2-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot3+3^2}\)
\(=\sqrt{5}-2+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(=\sqrt{5}-2+\left|\sqrt{5}-3\right|\)
\(=\sqrt{5}-2+3-\sqrt{5}\)
\(=1\)
b, (ĐKXĐ: x ≥ 0; x ≠ 1)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{x-\sqrt{x}+3\sqrt{x}-3}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-5}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
#\(Toru\)
a: \(=\sqrt{5}-2+3-\sqrt{5}=3-2=1\)
b:
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{x-5+\sqrt{x}-1+2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+\sqrt{x}-6+2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
Rút gọn :
a) \(\dfrac{3\sqrt{2-6}}{\sqrt{2-1}}\)
b) \(\dfrac{3\sqrt{5}+5\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)
c) \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\)
Câu a bạn xem lại đề nhé vì \(\sqrt{2-6}=\sqrt{-4}\left(VLý\right)\)
b) \(\dfrac{3\sqrt{5}+5\sqrt{3}}{\sqrt{3}+\sqrt{5}}=\dfrac{\sqrt{3.5}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}=\sqrt{15}\)
c) \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
Rút gọn các biểu thức sau:
\(a.A=2\sqrt{3}-\sqrt{75}+2\sqrt{12}\)
\(b.B=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(c.C=\left(\dfrac{x+2\sqrt{x}}{x-2\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}+1}\) (x > 0;x ≠ 4)