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c) \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-144.10}\)
\(=\sqrt{144.91-144.10}=\sqrt{144.\left(91-10\right)}=12\sqrt{81}=12.9=108\)
\(\sqrt{117.5^2-26.5^2-1440}\)
\(=\sqrt{\left(117.5-26.5\right)\left(117.5+26.5\right)-1440}\)
\(=\sqrt{91\cdot144-1440}\)
\(=12\cdot9=108\)
\(=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\\ =\sqrt{91\cdot144-1440}=\sqrt{144\left(91-10\right)}\\ =\sqrt{144\cdot81}=\sqrt{144}\cdot\sqrt{81}=12\cdot9=108\)
a) \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)
\(=\sqrt{91.144-1440}=\sqrt{144\left(91-10\right)}=\sqrt{12^2.9^2}=12.9=108\)
b) \(\sqrt{146,5^2-109,5^2+27.256}=\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)
\(=\sqrt{37.256+27.256}=\sqrt{256\left(37+27\right)}=\sqrt{256.64}=\sqrt{16^2.8^2}=16.8=128\)
\(\sqrt{117,5^2-26,5^2}-1440=-202475\)
\(\sqrt{146,5^2-109,5^2+27,256=}-11816494\)
\(\sqrt{117,5^2-26,5^2-1440}\)
\(=\sqrt{\left(117,5+26,5\right)\left(117,5-26,5\right)-144.10}\)
\(=\sqrt{144.91-144.10}\)
\(=\sqrt{144\left(91-10\right)}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=12.9=108\)
\(\sqrt{117,5^2-26,5^2-1440}\)\(=\sqrt{\left(117,5-26.5\right)\left(117.5+26,5\right)-144\cdot10}\)\(=\sqrt{91\cdot144-144\cdot10}\)
\(=\sqrt{144\cdot\left(91-10\right)}\)
\(=\sqrt{144\cdot81}\)
\(=\sqrt{144}\cdot\sqrt{81}\)
\(=12\cdot9=108\)
\(a=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\left(10\right)}=\sqrt{36}=6\)
a) \(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}\)
=\(\sqrt{3,6.10}=\sqrt{36}=6\)
b)\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\)
=\(\sqrt{3,6.40}=\sqrt{144}=12\)
c)\(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)
=\(\sqrt{91.144-1440}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=108\)
d)\(\sqrt{146,5^2-109,5^2+27.256}\)=\(\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)
=\(\sqrt{37.256+\sqrt{27.256}}=\sqrt{64.256}=\sqrt{64}.\sqrt{256}=128\)
Bài 5:
a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:
\(E=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1-3}=\dfrac{\sqrt{3}}{\sqrt{3}-2}=-3-2\sqrt{3}\)
b: Để E<1 thì E-1<0
\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
c: Để E nguyên thì \(4⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)
hay \(x\in\left\{16;25;49\right\}\)
Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
Thay \(x=\sqrt{3}-1\) vào \(B\), ta được
\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)
b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)
Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)
Vậy \(B_{min}=-2\) khi \(x=0\)
1) Cho biểu thức A =