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Bài 2 :
b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)
ĐKXĐ : \(x\ge1\)
Pt(1) tương đương :
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)
Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)
\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\) ( Thỏa mãn )
Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)
Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\Leftrightarrow2=2\) ( Luôn đúng )
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)
Bài 1 :
a) ĐKXĐ : \(-1\le a\le1\)
Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)
\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)
\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)
\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)
Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)
b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :
\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)
Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+\left(\sqrt{x}-10\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+x-2\sqrt{x}-\sqrt{x}+2+\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{2x-8}{x-4}\)
\(=\frac{2\left(x-4\right)}{x-4}\)
\(=2\)
1,
\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\frac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\frac{1}{\sqrt{h-1}+1}+\frac{1}{\sqrt{h-1}-1}\)
\(=\frac{\sqrt{h-1}-1+\sqrt{h-1}+1}{h-1-1}\)
\(=\frac{2\sqrt{h-1}}{h-2}\)
Thay \(h=3\)vào D ta có:
\(D=\frac{2\sqrt{3-1}}{3-2}=2\sqrt{2}\)
Vậy với \(h=3\)thì \(D=2\sqrt{2}\)
2,
a, \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)
Vậy PT có nghiệm là \(x=2\)
b, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)(ĐK: \(-\sqrt{2}\le x\le\sqrt{2}\))
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0=-3\)(vô lí)
Vậy PT đã cho vô nghiệm.
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
Bài 1:
a) Ta có: \(5\sqrt{12}-\sqrt{45}-3\sqrt{48}+\sqrt{75}\)
\(=5\cdot2\cdot\sqrt{3}-\sqrt{3}\cdot\sqrt{15}-3\cdot\sqrt{3}\cdot4+5\sqrt{3}\)
\(=10\sqrt{3}-3\sqrt{5}-12\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}-3\sqrt{5}\)
b) Ta có: \(\left(1+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\frac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(=\left(\frac{1-\sqrt{5}+5-\sqrt{5}}{1-\sqrt{5}}\right)\cdot\left(\frac{5+\sqrt{5}+1+\sqrt{5}}{1+\sqrt{5}}\right)\)
\(=\frac{6-2\sqrt{5}}{1-\sqrt{5}}\cdot\frac{6+2\sqrt{5}}{1+\sqrt{5}}\)
\(=\frac{6^2-\left(2\sqrt{5}\right)^2}{1^2-\left(\sqrt{5}\right)^2}=\frac{36-20}{1-5}=\frac{16}{-4}=-4\)
2)
a) Ta có: \(P=x-\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\frac{x-4}{\sqrt{4x}}\)
\(=x-\left(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=x-\frac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=x-\frac{2x}{2\sqrt{x}}\)
\(=x-\sqrt{x}\)
b) Ta có: \(x=7-\sqrt{48}\)
\(=\frac{14-2\sqrt{48}}{2}=\frac{8-2\cdot2\sqrt{2}\cdot\sqrt{6}+6}{2}\)
\(=\frac{\left(2\sqrt{2}-\sqrt{6}\right)^2}{2}=\frac{\left[\sqrt{2}\cdot\left(2-\sqrt{3}\right)\right]^2}{2}\)
\(=\frac{2\cdot\left(2-\sqrt{3}\right)^2}{2}=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(P=x-\sqrt{x}\), ta được:
\(P=\left(2-\sqrt{3}\right)^2-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=7-4\sqrt{3}-\left|2-\sqrt{3}\right|\)
\(=7-4\sqrt{3}-\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=7-4\sqrt{3}-2+\sqrt{3}\)
\(=5-3\sqrt{3}\)
c) Ta có: \(P=x-\sqrt{x}\)
\(=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)
hay \(x=\frac{1}{4}\)(nhận)
Vậy: Giá trị nhỏ nhất của biểu thức \(P=x-\sqrt{x}\) là \(-\frac{1}{4}\) khi \(x=\frac{1}{4}\)