\(K=\dfrac{\sqrt x-11}{x+2\sqrt x-3}+\dfrac{3\sqrt x}{1-\sqrt x}-\dfrac {2\sqrt x+3}{\sqrt x+3}\)
Rút gọn K
Tìm x khi K=1/2
Tìm x để K có nghĩa
Cho biểu thức:
\(K=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a,Tìm x để K có nghĩa
b,Rút gọn K
c,Tìm x khi \(K=\dfrac{1}{2}\)
a ,rút gọn P (dkxd x\(\ge0,x\ne0\)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x-1}\right)}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
..............=\(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
a: ĐKXĐ: x>=0; x<>1
b: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
c: Để K=1/2 thì \(\dfrac{-5x+5\sqrt{x}-8}{x+2\sqrt{x}-3}=\dfrac{1}{2}\)
=>\(-10x+10\sqrt{x}-16-x-2\sqrt{x}+3=0\)
=>\(-11x+8\sqrt{x}-13=0\)
hay \(x\in\varnothing\)
\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)
a,Rút gọn K
b,Tính K khi x=\(24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
c,Tìm x để \(\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}\)≥1
\(ĐKXĐ:x\ge0,x\ne1\)
\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)
\(K=\left[\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x+2\sqrt{x}-\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\right]\)
\(K=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\left[\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\dfrac{x+2\sqrt{x}+1-x-\sqrt{x}}{x-1}.\dfrac{x-1}{2\sqrt{x}}\)
\(K=\dfrac{\sqrt{x}+1}{x-1}.\dfrac{x-1}{2\sqrt{x}}\)
\(K=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
b.
Ta có: \(24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}\)
\(=24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=24+\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=24+\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=24+\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=24+\sqrt{\sqrt{5}-\sqrt{5}+1}=24+1=25\)
Thay \(x=25\) vào \(K\) ta được:
\(K=\dfrac{\sqrt{x}+1}{2\sqrt{x}}=\dfrac{\sqrt{25}+1}{2.\sqrt{25}}=\dfrac{6}{10}=\dfrac{3}{5}\)
c.
Ta có: \(\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}\ge1\)
\(\Rightarrow\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)
\(\Rightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)
\(\Rightarrow\dfrac{16\sqrt{x}}{8\sqrt{x}+8}-\dfrac{x+2\sqrt{x}+1}{8\sqrt{x}+8}-\dfrac{8\sqrt{x}+8}{8\sqrt{x}+8}\ge0\)
\(\Rightarrow\dfrac{16\sqrt{x}-x-2\sqrt{x}-1-8\sqrt{x}-8}{8\sqrt{x}+8}\ge0\)
\(\Rightarrow\dfrac{6\sqrt{x}-x-9}{8\sqrt{x}+8}\ge0\)
\(\Rightarrow\dfrac{-\left(\sqrt{x}-3\right)^2}{8\sqrt{x}+8}\ge0\)
Ta có: \(\left\{{}\begin{matrix}-\left(\sqrt{x}-3\right)^2\le0\\8\sqrt{x}+8\ge0\end{matrix}\right.\)
⇒ Không có \(x\) thỏa mãn
Cho biểu thức \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{8\sqrt{x}}{1-x}\), \(N=\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\)
a, Rút gọn M và N
b, Xét biểu thức K = M:N. Tính giá trị của K khi \(x=14-6\sqrt{5}\)
Cho biểu thức: K = \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a.Rút gọn K
b. Tìm giá trị lớn nhất của K
Cho biểu thức : M= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)
a) Rút gọn M
b) Tính M khi x= 11+\(6\sqrt{2}\)
c) tìm các giá trị x để M<1
a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)
c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>căn x-3<0
=>0<x<9
`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)
\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)
\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)
\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)
\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
`b,`Ta có :
\(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)
\(M=\dfrac{3\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(3+\sqrt{2}\right)^2}-3}\)
\(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}\)
\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)
\(M=\dfrac{6+9\sqrt{2}}{2}\)
`c,` Để `M<1` Ta có :
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )
\(\sqrt{x}< 3\)
\(x< 9\)
Đối chiếu ĐKXĐ ta có : `0<x<9`
Rút gọn K
\(K=\dfrac{15\sqrt[]{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x+3}}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x-5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
cho biểu thức P=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\) với x\(\ge\)0; x\(\ne\)9
1.tìm ĐKXĐ và rút gọn P
2.tính P khi x=7+2\(\sqrt{3}\)
3.tìm x để P<1
tìm điều kiện để biểu thức có nghĩa và rút gọn biểu thức
\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(x\ge0;x\ne3;x\ne-3;x\ne9;x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\\ =\dfrac{3}{\sqrt{x}-2}\)
Tick hộ nha 😘
P=\(\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a)tìm điều kiện để P có nghĩa
b)rút gọn P
c)tính giá trị của P với x=\(3+2\sqrt{2}\)
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1