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Bài 2 :
a) Sửa đề :
\(A=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(A=\sqrt{3}-1-\sqrt{3}\)
\(A=-1\)
b) \(B=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(B=\sqrt{2}+1-\sqrt{2}+1\)
\(B=2\)
c) \(C=\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+2+\sqrt{3}\)
\(C=4\)
d) \(D=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(D=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(D=4+\sqrt{7}-\sqrt{7}\)
\(D=4\)
Bài 1 :
a) Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) có nghĩa
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)
TH1 :\(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow x\ge3}\)
TH2 : \(\hept{\begin{cases}x-1\le0\\x-3\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\le3\end{cases}\Leftrightarrow}x\le1}\)
Vậy để biểu thức có nghĩa thì \(\orbr{\begin{cases}x\ge3\\x\le1\end{cases}}\)
b) Để \(\sqrt{\frac{1-x}{x+2}}\)có nghĩa
\(\Leftrightarrow\frac{1-x}{x+2}\ge0\)
TH1 : \(\hept{\begin{cases}1-x\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-2\end{cases}\Leftrightarrow}-2\le x\le1}\)
TH2 : \(\hept{\begin{cases}1-x\le0\\x+2\le0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\le-2\end{cases}\Leftrightarrow x\in\varnothing}\)
Vậy để biểu thức có nghĩa thì \(-2\le x\le1\)
Bài 1 :
a)\(\sqrt{-2\text{x}+3}\) <=> -2x+3 \(\ge\)0 <=> -2x \(\ge\) -3 <=> x\(\le\) \(\frac{3}{2}\)
b)\(\sqrt{\frac{4}{x+3}}< =>x+3>0< =>x>-3\)
Bài 2 :
a)\(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)
b)\(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}=2\sqrt{3}+\left|2-\sqrt{3}\right|=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)
c) \(\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
Bài 3 :
a) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
VT = \(\sqrt{5-2.2.\sqrt{5}+2^2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{5}\right)^2-4\sqrt{5}+2^2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
=|\(\sqrt{5-2}\)| -\(\sqrt{5}\)
= \(\sqrt{5}-2-\sqrt{5}\)
= -2 = VP
b)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)
VT = \(\sqrt{7+2.4.\sqrt{7}+4^2}-\sqrt{7}\)
= \(\sqrt{\left(\sqrt{7}+4\right)^2}-\sqrt{7}\)
= |\(\sqrt{7}+4\)| -\(\sqrt{7}\)
=\(\sqrt{7}+4-\sqrt{7}\)
= 4 =VP
c) \(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
VT = \(16-8\sqrt{7}+7\)
= 23 - \(8\sqrt{7}\) = VP
Bài 4:
a)\(\frac{x^2-5}{x+\sqrt{5}}=\frac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\frac{\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
Tương tự
Bài 5 :
a) \(\sqrt{x^2+6\text{x}+9}=3\text{x}-1\)
=> \(\sqrt{\left(x+3^2\right)}\) = 3x-1
=> x+3 = 3x-1
+) x+3 =3x-1 => x= 2
+)x+3=-3x-1 => x= \(\frac{-1}{2}\) ( không tmđk)
b)+c) Tương tự
mình giúp bài 3 cho
\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\left(ĐKXĐ:x\ge5\right)\)
\(< =>\sqrt{25\left(x-5\right)}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=6\)
\(< =>\sqrt{25}.\sqrt{x-5}-3\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-3.\frac{\sqrt{x-5}}{3}-\frac{1}{3}.3.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\)
\(< =>3\sqrt{x-5}=6< =>\sqrt{x-5}=2\)
\(< =>x-5=4< =>x=4+5=9\left(tmđk\right)\)