\(\sqrt{28-10\sqrt{3}}=\sqrt{\left(5-\sqrt{3}\right)^2}=\left|5-\sqrt{3}\right|=5-\sqrt{3}\)
\(\sqrt{5+10\sqrt{3}}\sqrt{10\sqrt{3}-5}=\sqrt{\left(10\sqrt{3}\right)^2-5^2}=5\sqrt{11}\)
\(x^2+2x+6=\left(x+1\right)^2+5>0;\forall x\in R\) nên \(\sqrt{x^2+2x+6}\) xác định với mọi x thuộc R
\(\sqrt{23+\sqrt{11+\sqrt{6+\sqrt{9}}}}=\sqrt{23+\sqrt{11+\sqrt{6+3}}}=\sqrt{23+\sqrt{11+\sqrt{9}}}=\sqrt{23+\sqrt{11+3}}\)
\(=\sqrt{23+\sqrt{14}}\)
\(-x^2\le0;\forall x\Rightarrow-x^2+25\le25\Rightarrow\sqrt{-x^2+25}\le\sqrt{25}=5\) (câu này cả 4 đáp án đều sai)
Bài 1:
\(P=\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}=\dfrac{1}{\sqrt{x}+1}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-x-1}{x-1}=\dfrac{x+1}{1-x}\)
b.
\(x=\dfrac{1}{\sqrt{2}}\Rightarrow P=\dfrac{\dfrac{1}{\sqrt{2}}+1}{1-\dfrac{1}{\sqrt{2}}}=\dfrac{\sqrt{2}+1}{1-\sqrt{2}}=\dfrac{\left(\sqrt{2}+1\right)^2}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\dfrac{3+2\sqrt{2}}{-1}=-3-2\sqrt{2}\)
Bài 2:
ĐKXĐ: \(x\ne\left\{-1;0;1\right\}\)
\(A=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{x+2003}{x}\)
\(=\left(\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{x+2003}{x}\)
\(=\left(\dfrac{x^2-1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{x+2003}{x}=\left(\dfrac{x^2-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
\(=\dfrac{x+2003}{x}\)
b.
\(A=\dfrac{x+2003}{x}=1+\dfrac{2003}{x}\)
\(A\in Z\Leftrightarrow\dfrac{2003}{x}\in Z\)
\(\Rightarrow x=Ư\left(2003\right)\)
\(\Rightarrow x=\left\{-2003;-1;1;2003\right\}\)
Bài 3.
\(\sqrt{3x^2-2x+10}=\sqrt{x^2+14x-9}\)
\(\Rightarrow3x^2-2x+10=x^2+14x-9\)
\(\Leftrightarrow2x^2-16x+19=0\)
\(\Delta'=64-2.19=26>0\Rightarrow\) pt có 2 nghiệm phân biệt:
\(\left[{}\begin{matrix}x=\dfrac{8-\sqrt{26}}{2}\\x=\dfrac{8+\sqrt{26}}{2}\end{matrix}\right.\)
Thay vào pt ban đầu thấy thỏa mãn.
Vậy \(S=\left\{\dfrac{8-\sqrt{26}}{2};\dfrac{8+\sqrt{26}}{2}\right\}\)
giải giúp em ý nào cũng được ạ, giải chi tiết ạ
