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#)Giải :
\(A=\frac{\sqrt{x+1}}{\sqrt{x-2}}+\frac{2\sqrt{x}}{\sqrt{x+2}}+\frac{2+5\sqrt{x}}{4-x}\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2+5\sqrt{x}}{x-4}\)
\(A=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Vậy \(A=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
\(A-1=\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{(\sqrt{x}-1)^2}{\sqrt{x}}\ge0\)\(\Rightarrow A\ge1\)
\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{6+x-x^2}.\)
\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{-x^2+x+6}\)
\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{-x^2-3x+2x+6}\)\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{-\left(x^2+3x-2x-6\right)}\)
\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{-\left[x\left(x+3\right)-2\left(x+3\right)\right]}\)
\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{-\left(x+3\right)\left(x-2\right)}\)
\(6x+\sqrt{x+2}+2\sqrt{3-x}=8\sqrt{\left(3-x\right)\left(x-2\right)}\)
Từ đây giải tiếp ạ.
\(\sqrt{5-3x}=\sqrt{2x+8}\)
\(\Leftrightarrow5-3x=2x+8\)
\(\Leftrightarrow-3x-2x=8-5\)
\(\Leftrightarrow-5x=3\)
\(\Leftrightarrow x=\frac{-3}{5}\)
P/S" ko chắc
Mk sửa đề lại 1 chút ( chả bt mk nhìn thế nào mak vt lộn hết cả đề )......
BÀI 1: Rút gọn
\(C=a\sqrt{\frac{4a^2-4ab+b^2}{a^2}}-2a-b\)
\(ĐKXĐ:x^2-12\ge0\Rightarrow x^2\ge12\Rightarrow x\ge-2\sqrt{3}\)
\(\sqrt{x^2-12}=2\)
\(\Leftrightarrow x^2-12=4\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
\(\frac{\sqrt{2}+\sqrt{3}}{2+\sqrt{6}}=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}\)
\(=\frac{2\sqrt{2}-2\sqrt{3}+2\sqrt{3}-3\sqrt{2}}{4-6}\)
\(=\frac{-\sqrt{2}}{-2}\)
\(=\frac{\sqrt{2}}{2}\)
a)\(\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{3-2\sqrt{6}+2}\)
\(=\sqrt{3-2\sqrt{2}\sqrt{3}+2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(\left|\sqrt{3}-\sqrt{2}\right|\)
\(a,\sqrt{5-2\sqrt{6}}=\left(\sqrt{2}-\sqrt{3}\right)^2=|\sqrt{2}-\sqrt{3}|=\sqrt{3}-\sqrt{2}\)
\(b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-\left(20-10\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
\(c,\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-7\right)^2}-\sqrt{\left(3\sqrt{5}+7\right)^2}\)
\(=|3\sqrt{5}-7|-|3\sqrt{5}+7|\)
\(=7-3\sqrt{5}-3\sqrt{5}-7\)
\(=-6\sqrt{5}\)