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B = \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}+...+\dfrac{1}{\sqrt{x+2015}+\sqrt{x+2016}}\)
B = \(\dfrac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+...+\dfrac{\sqrt{x+2015}-\sqrt{x+2016}}{x+2015-x-2016}\)
B = \(\dfrac{\sqrt{x}-\sqrt{x+1}}{-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}+...+\dfrac{\sqrt{x+2015}-\sqrt{x+2016}}{-1}\)
B = \(-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-...-\sqrt{2015}+\sqrt{2016}\)
B = \(-\sqrt{x}+\sqrt{2016}\)
Khi x = 2017
B = \(-\sqrt{2017}+\sqrt{2016}=\sqrt{2016}-\sqrt{2017}\)
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)
= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{x+2}{\sqrt{x}-1}\)
A = (15/√x) - (11x + 2√x - 3) - (3√x - 2√x - 1) - (2√x + 3√x - 3)
Tiếp theo, kết hợp các thành phần tương tự:
A = 15/√x - 11x - 2√x + 3 + 3√x - 2√x + 1 - 2√x - 3√x + 3
Đơn giản hóa biểu thức:
A = -11x + 15/√x + 4
Để tìm giá trị lớn nhất của A, ta có thể tìm điểm đạt cực đại của hàm số A(x). Tuy nhiên, để làm điều này, cần biết thêm về giá trị của x.
Sửa đề: (3căn x-2)/căn x-1-(2căn x+3)/(căn x+3)\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}==-5+\dfrac{17}{\sqrt{x}+3}< =\dfrac{17}{3}-5=\dfrac{2}{3}\)
Dấu = xảy ra khi x=0
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
a) \(\sqrt{\left|x-1\right|-3}\)
Với \(x\ge1\) thì
\(\sqrt{x-1-3}=\sqrt{x-4}\) được xác định khi:
\(x\ge4\)
Với \(x< 1\) thì
\(\sqrt{-\left(x-1\right)-3}=\sqrt{-x+1-3}=\sqrt{-x-2}\) được xác đinh khi:
\(x\le-2\)
\(a,\sqrt{\left|x-1\right|-3}\) xác định \(\Leftrightarrow\left|x-1\right|-3\ge0\Leftrightarrow\left|x-1\right|\ge3\)
\(TH_1:x\ge1\\ x-1\ge3\Leftrightarrow x\ge4\left(tm\right)\\ TH_2:x< 1\\ x-1\ge-3\\ \Leftrightarrow x\ge-2\left(tm\right)\)
Vậy căn thức trên xác định \(\Leftrightarrow x\ge4\)
\(b,\sqrt{x-2\sqrt{x-1}}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}\le\dfrac{x}{2}\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-1\le\dfrac{x^2}{4}\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x-4-x^2\le0\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-\left(x^2-4x+4\right)\le0\\x\ge1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2\ge0\left(LD\right)\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)
Vậy căn thức trên xác định \(\Leftrightarrow x\ge1\)
\(c,\dfrac{1}{\sqrt{9-12x+4x^2}}=\dfrac{1}{\sqrt{\left(3-2x\right)^2}}=\dfrac{1}{3-2x}\) xác định \(\Leftrightarrow3-2x\ne0\Leftrightarrow x\ne\dfrac{3}{2}\)
Vậy căn thức trên xác định \(\Leftrightarrow x\ne\dfrac{3}{2}\)
\(B=B_1+B_2+...+B_{2016}\)
\(B_1=\dfrac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}\)
\(B_1=\sqrt{x+1}-\sqrt{x}\)
\(B_2=\sqrt{x+2}-\sqrt{x+1}\)
\(B_3=\sqrt{x+3}-\sqrt{x+2}\)
...
\(B_{2015}=\sqrt{x+2015}-\sqrt{x+2014}\)
\(B_{2016}=\sqrt{x+2016}-\sqrt{x+2015}\)
\(B=\sqrt{x+2016}-\sqrt{x}\)
\(B\left(2017\right)=\sqrt{2017+2016}-\sqrt{2017}\)