Rút gọn biểu thức:
P=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)
Câu 1:
1.Rút gọn biểu thức:
P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\),với x>0;x\(\ne\)1
\(=>P=\left[\dfrac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\sqrt{x}-1+2}{x-1}\right]\)
\(P=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{x-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}}\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
Cho biểu thức:
P=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
a) Rút gọn P
b) Tìm x để P>0
Rút gọn biểu thức:P=\(\left(\dfrac{3\sqrt{x}}{\sqrt{x}-3}+\dfrac{4\sqrt{x}}{\sqrt{x}+3}+\dfrac{7x-3}{9-x}\right):\left(\dfrac{2\sqrt{x}-4}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{3x+9\sqrt{x}+4x-12\sqrt{x}-7x+3}{x-9}:\dfrac{2\sqrt{x}-4-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{3\sqrt{x}+3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
* Cho biểu thức:
P=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a. Nêu ĐKXĐ
b. Rút gọn P
c. Tìm x để P<\(\dfrac{-1}{2}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
b) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
c) Để \(P< -\dfrac{1}{2}\) thì \(P+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)
Cho biểu thức:
P = \(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne4,x\ne9\right)\)
a) Rút gọn P
b) Với \(x>9\), tìm GTNN của P
a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
1) Cho biểu thức:
P=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2.\left(x-2\sqrt{x}+1\right)}{x-1}\)
a) Rút gọn P
b) Tìm x nguyên để P có giá trị nguyên
Lời giải:ĐK: $x>0; x\neq 1$;
\(P=\left[\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}\right]:\frac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right):\frac{2(\sqrt{x}-1)}{\sqrt{x}+1}\)
\(=2:\frac{2(\sqrt{x}-1)}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\). Để $P$ nguyên thì $\sqrt{x}-1$ là ước nguyên của $2$
$\Rightarrow \sqrt{x}-1\in\left\{\pm 1;\pm 2\right\}$
$\Rightarrow x\in\left\{0; 2; 9\right\}$
Kết hợp với ĐKXĐ suy ra $x\in\left\{2;9\right\}$
cho biểu thức:
P=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\right)\)\(:\left(\dfrac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\dfrac{1}{x+1}\right)\)
với x\(\ge\)0;x\(\ne\)1
1)Rút gọn P
2)Tìm x để P<\(\dfrac{1}{2}\)
3) tìm m để phương trình (\(\sqrt{x}+1\))P= m-x có nghiệm x
1: \(P=\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{x+1}\cdot\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}=\dfrac{\sqrt{x}-1}{x+1}\)
2: P<1/2
=>P-1/2<0
=>\(2\sqrt{x}-2-x-1< 0\)
=>-x+2căn x-1<0
=>(căn x-1)^2>0(luôn đúng)
rút gọn biểu thức:p=\(\dfrac{8-x}{2+\sqrt[3]{x}}:\left(2+\dfrac{\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\left(\sqrt[3]{x}+\dfrac{2\sqrt[3]{x}}{\sqrt[3]{x}-2}\right).\left(\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\right)\)
\(\dfrac{8-x}{2+\sqrt[3]{x}}:\left(2+\dfrac{\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\left(\sqrt[3]{x}+\dfrac{2\sqrt[3]{x}}{\sqrt[3]{x}-2}\right).\left(\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\right)\)
\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}:\left(\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\left(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\right)\)
\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\dfrac{\sqrt[3]{x}-2}{\sqrt[3]{x}}\)
\(=\sqrt[3]{x}-2+\sqrt[3]{x}=2\sqrt[3]{x}-2\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)