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Lời giải:
a. ĐKXĐ: $a\geq 0; a\neq 1$
b.
\(P=\left[\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}+1\right].\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-1\right].\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}\)
\(=(\sqrt{a}+1)(\sqrt{a}-1).\sqrt{2}=\sqrt{2}(a-1)\)
c.
\(P=\sqrt{2}(\sqrt{2+\sqrt{2}}-1)=\sqrt{4+2\sqrt{2}}-\sqrt{2}\)
a. ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne1\\\sqrt{a}\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
b. \(P=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right).\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right].\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right].\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right).\sqrt{2}=2\left(a-1\right)=2a-2\)
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)
\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)
b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)
\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)
\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)
c:
\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)
d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
e:
\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)
f:
\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)
\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)
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 $C_{\min}=-1$
Ta có: \(b=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
Ta có: \(a=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
=2
Thay a=2 và \(b=\dfrac{2}{3}\) vào M, ta được:
\(M=\dfrac{1+2\cdot\dfrac{2}{3}}{2+\dfrac{2}{3}}-\dfrac{1-2\cdot\dfrac{2}{3}}{2-\dfrac{2}{3}}\)
\(=\dfrac{7}{8}+\dfrac{1}{4}\)
\(=\dfrac{7}{8}+\dfrac{2}{8}=\dfrac{9}{8}\)
a: \(=\dfrac{2\left(\sqrt{2}+1\right)}{2-1}-\sqrt{\dfrac{3}{4}:\dfrac{3}{2}}+2\sqrt{2}\)
\(=2\sqrt{2}+2+2\sqrt{2}-\sqrt{\dfrac{1}{2}}\)
\(=4\sqrt{2}+2-\dfrac{\sqrt{2}}{2}=\dfrac{7}{2}\sqrt{2}+2\)
b: \(B=\left|3x\right|+x+\sqrt{x}\)
\(=3x+x+\sqrt{x}=4x+\sqrt{x}\)
\(a,A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dk:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Thay \(x=4\) vào A :
\(A=\dfrac{\sqrt{4}-1}{\sqrt{4}+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)
(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) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
Với `x >= 0,x ne 4` có:
`M=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[3x-6\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]=[3\sqrt{x}]/[\sqrt{x}+2]`
____________
`N=(1/[\sqrt{a}-1]-1/\sqrt{a}):([\sqrt{a}+1]/[\sqrt{a}-2]-[\sqrt{a}+2]/[\sqrt{a}-1])`
- Biểu thức `N` là như vầy?
Với `a > 0,a ne 1,a ne 4` có:
`N=[\sqrt{a}-\sqrt{a}+1]/[\sqrt{a}(\sqrt{a}-1)]:[(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}+2)(\sqrt{a}-2)]/[(\sqrt{a}-2)(\sqrt{a}-1)]`
`N=1/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-2)(\sqrt{a}-1)]/[a-1-a+4]`
`N=[\sqrt{a}-2]/[3\sqrt{a}]`
Với \(x\ge0;x\ne4\)
Khi đó:
\(M=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x-4}+\dfrac{2x-4\sqrt{x}}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\\ =\dfrac{3x-6\sqrt{x}}{x-4}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Với \(a>0;a\ne1;a\ne4\)
Khi đó:
\(N=(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\dfrac{\sqrt{a}}{a-\sqrt{a}}-\dfrac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\dfrac{1}{a-\sqrt{a}}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right).3}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Lời giải:
$a+b=\frac{\sqrt{6}+\sqrt{2}+\sqrt{6}-\sqrt{2}}{2}=\sqrt{6}$
$ab=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2.2}=\frac{6-2}{4}=1$
Khi đó:
$S=\frac{1}{a^7}+\frac{1}{b^7}=\frac{a^7+b^7}{a^7b^7}$
$=\frac{a^7+b^7}{(ab)^7}=\frac{a^7+b^7}{1}=a^7+b^7$
$=(a^3+b^3)(a^4+b^4)-a^3b^3(a+b)$
$=(a^3+b^3)(a^4+b^4)-(a+b)$
Ta có:
$a^3+b^3=(a+b)^3-3ab(a+b)=(\sqrt{6})^3-3\sqrt{6}=6\sqrt{6}-3\sqrt{6}=3\sqrt{6}$
$a^4+b^4=(a^2+b^2)^2-2a^2b^2=(a^2+b^2)^2-2$
$=[(a+b)^2-2ab]^2-2=(6-2)^2-2=14$
$S=3\sqrt{6}.14-\sqrt{6}=41\sqrt{6}$
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{-\left(\sqrt{a}-1\right)}+\sqrt{2}\right)\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\sqrt{a}\right)\)
\(=\left(-\sqrt{a}+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{a}\right)\)
\(=2-a\)
lấy\(\sqrt{ }\) a và \(\sqrt{ }\)2 ra ngoài , trong ngoặc sẽ là :
(\(\sqrt{ }\)a +\(\sqrt{ }\)2)(\(\sqrt{ }\)2 +\(\sqrt{ }\)a) = (\(\sqrt{ }\)a+\(\sqrt{ }\)2)2