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Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
Bài 1: Bạn đã post 1 lần
Bài 2:
\(C=\sqrt{(x-3)-2\sqrt{x-3}+1}-\sqrt{(x-3)-4\sqrt{x-3}+4}\)
\(=\sqrt{(\sqrt{x-3}-1)^2}-\sqrt{(\sqrt{x-3}-2)^2}\)
\(=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|\)
Áp dụng BĐT dạng $|a|-|b|\leq |a-b|(*)$ thì:
$C\leq |\sqrt{x-3}-1-(\sqrt{x-3}-2)|$ hay $C\leq 1$
Vậy $C_{\max}=1$
Mặt khác, vẫn áp dụng BĐT $(*)$:
\(|\sqrt{x-3}-1|=|(\sqrt{x-3}-2-(-1)|\geq |\sqrt{x-3}-2|-|-1|\)
\(=|\sqrt{x-3}-2|-1\Rightarrow C\geq -1\)
Vậy $C_{\min}=-1$
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
\(a,\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
\(b,A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\)
\(\Rightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\left(\sqrt{a}-5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{a+5\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\sqrt{a}-25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{a-10\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{a}-5\right)^2}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)
a: \(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
b: \(A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}=\dfrac{\left(\sqrt{a}-5\right)^2}{a-25}=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)
(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)
= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)
= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\sqrt{3}+2+\sqrt{2}-\dfrac{1}{2-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}+2+\sqrt{2}\right)\left(2-\sqrt{3}\right)-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)
= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)
= \(\dfrac{-2\left(-2+\sqrt{3}\right)}{-2}\) = \(\sqrt{3}-2\)
bài 2)
a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
a)\(\sqrt{\dfrac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\dfrac{\sqrt{b}-1}{\sqrt{a}+1}}\)
=\(\dfrac{\sqrt{\sqrt{a}-1}}{\sqrt{\sqrt{b}+1}}.\dfrac{\sqrt{\sqrt{a}+1}}{\sqrt{\sqrt{b}-1}}\)
=\(\sqrt{\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)
=\(\sqrt{\dfrac{a-1}{b-1}}\)(1)
Thay a=7,25, b=3,25 vào(1)
=>=\(\sqrt{\dfrac{7,25-1}{3,25-1}}=\sqrt{\dfrac{6,25}{2,25}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b) =\(\sqrt{\left(\sqrt{15}a+4\right)^2}=\sqrt{15}a+4\)
Thay a
=>\(\sqrt{15}.\left(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\right)+4=3+5+4=12\)
c)
=\(\sqrt{\left(a^2-1\right)+2\sqrt{a^2-1}+1}-\sqrt{\left(a^2-1\right)-2\sqrt{a^2-1}+1}\)
=\(\sqrt{\left(a^2-1+1\right)^2}-\sqrt{\left(a^2-1-1\right)^2}\)
=\(a^2-\left(a^2-2\right)\)
=\(2\)