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1/ Ta có √(14 - 6√5) = √(9 - 6√5 +5) = 3 - √5
Từ đó a + b = 2
2/ Đề sai sửa lại là
√(15 - 6√6) = √(9 - 6√6 + 6) = (3 - √6)
Vậy a = 3; b = -1
=> a + b = 2
1.Nếu $\sqrt{55-6\sqrt{6}}=a+b\sqrt{6}$√55−6√6=a+b√6 với $a,b\in Z$a,b∈Z thì a-b=?
2. Nếu $\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=a+b\sqrt{6}$√15−6√6+√33−12√6=a+b√6 với $a,b\in Z$a,b∈Z thì a+b=?
Ta có :
a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
a) Ta có: \(\sqrt{96}\cdot\sqrt{125}\)
\(=\sqrt{16}\cdot\sqrt{6}\cdot\sqrt{25}\cdot\sqrt{5}\)
\(=20\cdot\sqrt{30}\)
b) Ta có: \(\sqrt{a^4\cdot6^5}\)
\(=a^2\cdot36\cdot\sqrt{6}\)
c) Ta có: \(\sqrt{a^6\cdot b^{11}}\)
\(=\sqrt{a^6}\cdot\sqrt{b^{11}}\)
\(=\left|a^3\right|\cdot\left|b^5\right|\cdot\sqrt{b}\)
\(=a^3b^5\cdot\sqrt{b}\)
d) Ta có: \(\sqrt{a^3\left(1-a\right)^4}\)
\(=\sqrt{a^3}\cdot\sqrt{\left(1-a\right)^4}\)
\(=a\sqrt{a}\cdot\left(1-a\right)^2\)
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
Lời giải:
a)
\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)
\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)
\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)
b)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
c)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(\sqrt{3}+1)^2+2}\)
\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)
\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)
1/ đkxđ: a > 0; a khác 1
a/ A= (\(\dfrac{\sqrt{a}}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\))\(\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{-4a}{a-1}=-\dfrac{2\sqrt{a}}{a-1}=\dfrac{2\sqrt{a}}{a+1}\)
b/+) A = 4
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}=4\)\(\Leftrightarrow2\sqrt{a}=4a+4\)
=> Không có gt a nào t/m
+) \(A>-6\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}>-6\)
\(\Leftrightarrow2\sqrt{a}>-6a-6\)
\(\Leftrightarrow6a+2\sqrt{a}+6>0\) (luôn đúng vì a > 0)
=> bpt có nghiệm với mọi a > 0
vậy........
c/ \(a^2-3=0\Leftrightarrow\left[{}\begin{matrix}a=\sqrt{3}\left(tm\right)\\a=-\sqrt{3}\left(ktmđkxđ\right)\end{matrix}\right.\)
Với a = \(\sqrt{3}\) ta có:
\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2}=\sqrt{3}\left(\sqrt{3}-1\right)=3-\sqrt{3}\)
C1: Bình phương 2 vế ta có: \(55-6\sqrt{6}=\left(a+b\sqrt{6}\right)^2\)
<=> \(55-6\sqrt{6}=a^2 +6b^2+2ab\sqrt{6}\)
=> a2 + 6b2 = 55 và 2ab = - 6
=> a2 + 6b2 = 55 (1) và ab = -3 => a = -3/b (2)
thế (2) vào (1) ta được : \(\left(-\frac{3}{b}\right)^2+6b^2=55\) => \(9+6b^4=55b^2\)
=> 6b4 - 55b2 + 9 = 0 => 6b4 - 54b2 - b2 + 9 =0 <=> 6b2.(b2 - 9) - (b2 - 9) = 0 <=> (6b2 - 1).(b2 - 9 ) = 0
<=> b2 = 1/6 (Loại; vì b nguyên ) hoặc b2 = 9
+) b2 = 9 => a2 = 1 => a = 1 hoặc - 1 ; b = 3 hoặc - 3
Do \(a+b\sqrt{6}\) > 0 và a; b trái dấu nên a = -1; b = 3 => a+ b = 2
Vậy a + b = 2
C2: \(\sqrt{55-6\sqrt{6}}=\sqrt{\left(3\sqrt{6}\right)^2-2.3\sqrt{6}.1+1}=\sqrt{\left(3\sqrt{6}-1\right)^2}\)
= \(\left|3\sqrt{6}-1\right|=3\sqrt{6}-1\)
=> a = -1; b = 3 => a + b = 2