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\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
Mk làm như này, k biết có sai chỗ nào k. Nếu sai thì bạn sửa nhé.
A=\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{3+\sqrt{x}}\)
A=\(\frac{15\sqrt{x}-11-\left(3x-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
A=\(\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Lời giải:
ĐK: $x\geq 0; x\neq 1$
Đặt biểu thức là $P$
\(P=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{15\sqrt{x}-11-(3x+7\sqrt{x}-6)-(2x+\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{-5x+7\sqrt{x}-2}{(\sqrt{x}+3)(\sqrt{x}-1)}=\frac{(\sqrt{x}-1)(-5\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
\(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)ĐK : \(x\ge0;x\ne1\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}-3}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}-3}{x+2\sqrt{x}-3}\)
\(=\frac{7\sqrt{x}-8-5x}{x+2\sqrt{x}-3}=\frac{-5x+7\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)