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b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)
\(=\sqrt{1+\sqrt{2}+\sqrt{3}-\left(\sqrt{3}+\sqrt{2}\right)}=1\)
b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)
\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)
Th1: x-3 < 0
\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)
Th2: x-3 > 0
\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)
c)
Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\dfrac{\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
Th1: \(x-2\ge0\Leftrightarrow x\ge2\)
\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)
Th2: \(x-2\le0\Leftrightarrow x\le2\)
kết hợp với đk, ta được: 1 \< x \< 2
\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)
d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
chẳng biết có sai sót gì 0 nữa, xin lỗi tớ 0 xem lại đâu vì chán quá!
\(A=\dfrac{\sqrt{x-9+6\sqrt{x-9}+9}+\sqrt{x-9-6\sqrt{x-9}+9}}{\sqrt{\left(1-\dfrac{9}{x}\right)^2}}\)
\(A=\dfrac{\left(\sqrt{x-9}+3\right)+|3-\sqrt{x-9}|}{|1-\dfrac{9}{x}|}\) cũng khas dễ hiểu nếu a nguyên thì 3<=\(\sqrt{x-9}\)
\(A=\dfrac{2x\sqrt{x-9}}{x-9}=\dfrac{2x}{\sqrt{x-9}}\) đến đây khá dễ rồi bạn nhỉ
Bài 1 :
Câu a : \(B=5x+\sqrt{x^2+6x+9}=5x+\sqrt{\left(x+3\right)^2}=5x+x+3=6x+3\)
Câu b : \(B=-9\Leftrightarrow6x+3=-9\Leftrightarrow6x=-12\Leftrightarrow x=-2\)
Bài 1:
Ta có: \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\)
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=3^2\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của PT \(S=\left\{11\right\}\)
a)\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+x-3\)
\(=2x\)
b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\sqrt{\left(x+2\right)^2}-x\)
\(=x+2-x\)
=2
c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)
\(=\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(=\sqrt{x-1}\)