\(1+\left(\dfrac{2x+\sqrt{x}-1}{1-x}-\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1-x\sqrt{x}}\right):\dfrac{x-\sqrt{x}}{2\sqrt{x}-1}\)
1 ) Rút gọn
2) Tìm x để P=3
3) Tính P tại x 13-\(4\sqrt{10}\)
\(\left(\dfrac{\sqrt{x}}{2\sqrt{x}-2}+\dfrac{3-\sqrt{x}}{2x-2}\right):\left(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{x\sqrt{x}-1}\right)\)
1) Rút gọn
2)Tìm x để P=3
3)Tính P tại x = 15+\(6\sqrt{6}\)
1) Ta có: \(P=\left(\dfrac{\sqrt{x}}{2\sqrt{x}-2}+\dfrac{3-\sqrt{x}}{2x-2}\right):\left(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{x\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{2\left(\sqrt{x}-1\right)}+\dfrac{3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x+\sqrt{x}+3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+3}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-\sqrt{x}+3}\)
\(=\dfrac{\left(x+3\right)\left(x+\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+3\right)}\)
P=\(\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\):\(\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
Rút gọnP
Tính giá trị của P với \(x=7-4\sqrt{3}\)
P =\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
Rút gọn P
Tìm x để P=3
Tính P tại x=7+\(2\sqrt{3}\)
tìm x để P >3
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2x-\sqrt{x}+2\sqrt{x}+2\)
\(=2-x\)
b) Để P=3 thì 2-x=3
hay x=-1(Không thỏa mãn ĐKXĐ)
Vậy: Không có giá trị nào của x để P=3
c) Thay \(x=7+2\sqrt{3}\) vào P, ta được:
\(P=2-7-2\sqrt{3}=-5-2\sqrt{3}\)
Vậy: Khi \(x=7+2\sqrt{3}\) thì \(P=-5-2\sqrt{3}\)
Rút gọn bt A=\(\left(\dfrac{1+\sqrt{x}}{x+1}-\dfrac{4-3\sqrt{x}}{x-4\sqrt{x}+4}\right):\left(\dfrac{x-\sqrt{x}}{x\sqrt{x}-2x+\sqrt{x}-2}\right)\)
Sau đó tìm x để A>1
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{x+1}-\dfrac{4-3\sqrt{x}}{x-4\sqrt{x}+4}\right):\left(\dfrac{x-\sqrt{x}}{x\sqrt{x}-2x+\sqrt{x}-2}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-4\sqrt{x}+4\right)+\left(3\sqrt{x}-4\right)\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)^2}:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(x+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x+4\sqrt{x}+x-4\sqrt{x}+4+3x\sqrt{x}+3\sqrt{x}-4x-4}{\left(x+1\right)\left(\sqrt{x}-2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(x+1\right)}{x-\sqrt{x}}\)
\(=\dfrac{4x\sqrt{x}-7x+3\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\cdot\left(4\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}-3}{\sqrt{x}-2}\)
Để A>1 thì A-1>0
\(\Leftrightarrow\dfrac{4\sqrt{x}-3-\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-1}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1\le0\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x\le\dfrac{1}{9}\\x>4\end{matrix}\right.\)
\(D=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
1, Rút gọn D
2, Tìm x để \(D=\left(4-\dfrac{x-\sqrt{x}+13}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}\)
1: \(D=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
2: \(\Leftrightarrow D=\dfrac{4\sqrt{x}+12-x+\sqrt{x}-13}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}=1\)
\(\Leftrightarrow\left(-x+5\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)
\(\Leftrightarrow-2x\sqrt{x}-x+10x+5\sqrt{x}-2\sqrt{x}-1=x\sqrt{x}+3x+x+3\sqrt{x}+\sqrt{x}+3\)
\(\Leftrightarrow-2x\sqrt{x}+9x-3\sqrt{x}-1=x\sqrt{x}+4x+4\sqrt{x}+3\)
\(\Leftrightarrow-3x\sqrt{x}+5x-7\sqrt{x}-4=0\)
Bạn xem lại đề nhé, nghiệm rất xấu
1.P= \(\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}-1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right)\):\(\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
a) Rút gọn P
b) Tính giá trị của P khi x=\(\dfrac{1}{2}\)\(\left(3+2\sqrt{2}\right)\)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}-1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}+1}-1\right):\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}+1\right)+\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}-1\right)-2x+1}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}:\left(\dfrac{2x-1+\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)-\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}\right)\)
\(=\dfrac{x\sqrt{2}+\sqrt{x}+\sqrt{2x}+1+2x-\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{2x-1}:\dfrac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{x}+2}{-2-2\sqrt{x}}\)
cho B=\(\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\div\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
a. rút gọn B
b. tính \(\sqrt{B}\) khi \(x=5+2\sqrt{3}\)
c. tìm x để B= \(\dfrac{1}{2x^3-x-1}\)
d. tìm giá trị của x để giá trị của B không lớn hơn giá trị biểu thức \(\dfrac{1}{x+2}\)
Lm nhanh giúp mk nhé mk đang cần gấp
a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{1}{x-1}\)
a) Ta có: \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
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
Cho biểu thức B =\(\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
a) Tìm điều kiện để B có nghĩa
b) Rút gọn B
c) Tính B với x =\(\dfrac{2-\sqrt{3}}{2}\)
a) ĐKXĐ : \(x\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)
b) \(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right).\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\left(x-2\sqrt{x}+1\right)\)
\(=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
c) Có : \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)
Khi đó B = \(\dfrac{\sqrt{3}-1}{2}-1=\dfrac{\sqrt{3}-3}{2}\)
\(a,\) B có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(b,B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}-x}{1+\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(x-1\right)-\left(x-1\right)}{1+\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}-1\)
\(c,x=\dfrac{2-\sqrt{3}}{2}\Rightarrow B=\sqrt{\dfrac{2-\sqrt{3}}{2}}-1\)
\(=\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}.\sqrt{2}}-\sqrt{2}\) (Nhân \(\sqrt{2}\) để khử căn dưới mẫu)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-2\sqrt{2}}{2}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-2\sqrt{2}}{2}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-2\sqrt{2}}{2}\)
\(=\dfrac{\sqrt{3}-1-2\sqrt{2}}{2}\)