Sửa đề:
a, \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}=\dfrac{3+\sqrt{2}+3-\sqrt{2}}{9-2}=\dfrac{6}{7}\)
b, \(\sqrt{3}\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)\)
\(=\sqrt{3}.4\sqrt{3}=12\)
Câu 2
\(\sqrt{x-1}+\sqrt{4x-1}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)
Vậy...
Câu 1:
a) \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}=\dfrac{3-\sqrt{2}}{9-4}+\dfrac{3+\sqrt{2}}{9-4}=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{5}=\dfrac{6}{5}\)
Bạn nên xem lại câu 1) b) thử có đúng đề không
câu 1:
a,Đặt \(A=\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
suy ra \(A=\dfrac{3-\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+\dfrac{3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{3^2-\sqrt{2}^2}\)
\(\Leftrightarrow A=\dfrac{6}{7}\)
b, Đặt B=\(\sqrt{3}\left(\sqrt{12}+\sqrt{27-\sqrt{3}}\right)\)
\(\Leftrightarrow B=\sqrt{3.12}+\sqrt{3.27-3\sqrt{3}}\)
\(\Leftrightarrow B=6+\sqrt{81-3\sqrt{3}}\)
câu 2:
ta có \(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
ĐK:\(\sqrt{x-1}\ge0\Leftrightarrow x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\)
Vậy nghiệm của pt S=\(\left\{2\right\}\)