Tháng Tư 4, 2026

Thẻ: Toán lớp 9

Hỏi Đáp

Phân tích các đa thức sau thành nhân tử: \( \eqalign{& a)\,\,2x – 7\sqrt x – 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d)\,\,3x + 2\sqrt x – 5 \cr & b)\,\,x – 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,e)\,\,\,4\sqrt x – x – 4 \cr & c)\,\,x\sqrt x – 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f)\,\,x + \sqrt x – 6 \cr} \)

adminby admin

Phân tích các đa thức sau thành nhân tử: \( \eqalign{& a)\,\,2x – 7\sqrt x – 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d)\,\,3x + 2\sqrt x – 5 \cr & b)\,\,x …

Hỏi Đáp

Tính: a) \(\sqrt{18}-\frac{1}{2}\sqrt{48}-\sqrt{8}+\frac{4-5\sqrt{2}}{5-2\sqrt{2}}\) b) \(\sqrt{{{(2-\sqrt{7})}^{2}}}-\sqrt{\frac{2}{8-3\sqrt{7}}}\) c) \(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}\)

adminby admin

Tính: a) \(\sqrt{18}-\frac{1}{2}\sqrt{48}-\sqrt{8}+\frac{4-5\sqrt{2}}{5-2\sqrt{2}}\) b) \(\sqrt{{{(2-\sqrt{7})}^{2}}}-\sqrt{\frac{2}{8-3\sqrt{7}}}\) c) \(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}\) Phương pháp giải: Phương pháp: +) Câu a: Khai căn thức bậc hai dựa vào công thức: \(\sqrt{{{A}^{2}}.B}=\left| …

Hỏi Đáp

Thực hiện phép tính: 1)\(A = \sqrt {12} – 2\sqrt {48} + \frac{7}{5}\sqrt {75} \) 2)\(B = \sqrt {14 – 6\sqrt 5 } + \sqrt {{{\left( {2 – \sqrt 5 } \right)}^2}} \) A \(\begin{array}{l}1)\,\,\sqrt 3 \\2)\,\,2\sqrt 5 + 1\end{array}\) B \(\begin{array}{l}1)\,\,\sqrt 2 \\2)\,\,2\sqrt 5 \end{array}\) C \(\begin{array}{l}1)\,\,\sqrt 3 \\2)\,\,2\sqrt 5 – 1\end{array}\) D \(\begin{array}{l}1)\,\,\sqrt 2 \\2)\,\,2\sqrt 5 – 1\end{array}\)

adminby admin

Thực hiện phép tính: 1)\(A = \sqrt {12} – 2\sqrt {48} + \frac{7}{5}\sqrt {75} \) 2)\(B = \sqrt {14 – 6\sqrt 5 } + \sqrt …

Hỏi Đáp

Tính giá trị của các biểu thức: \(a)\;A = 5\sqrt {27} – 5\sqrt 3 – 2\sqrt {12} \) \(b)\;B = \frac{{\sqrt {15} – \sqrt 3 }}{{\sqrt 5 – 1}} – \frac{{\sqrt {15} + \sqrt 3 }}{{\sqrt 5 + 1}}\) A \(\begin{array}{l}a)\,\,A = 6\sqrt 3 \\b)\,\,B = 0\end{array}\) B \(\begin{array}{l}a)\,\,A = 6\sqrt 3 \\b)\,\,B = \sqrt 3 \end{array}\) C \(\begin{array}{l}a)\,\,A = 3\sqrt 3 \\b)\,\,B = 0\end{array}\) D \(\begin{array}{l}a)\,\,A = 3\sqrt 3 \\b)\,\,B = \sqrt 3 \end{array}\)

adminby admin

Tính giá trị của các biểu thức: \(a)\;A = 5\sqrt {27} – 5\sqrt 3 – 2\sqrt {12} \) \(b)\;B = \frac{{\sqrt {15} – \sqrt 3 …

Hỏi Đáp

Đưa thừa số vào trong dấu căn \(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} & & & b)\,\,a\sqrt 2 & & & c)\,\, – \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\left( {a > 0,\,\,\,b > 0} \right)\\d)\,\,a\sqrt {\frac{3}{a}} & & & e)\,\,\frac{1}{{2x – 1}}\sqrt {5\left( {1 – 4x + 4{x^2}} \right)} .\end{array}\) A \(\begin{array}{l} a)\,\,\left\{ \begin{array}{l} \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ – \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0 \end{array} \right.\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\left\{ \begin{array}{l} \sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\ – \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2} \end{array} \right. \end{array}\) B \(\begin{array}{l} a)\,\,\left\{ \begin{array}{l} \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ – \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0 \end{array} \right.\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\, – \sqrt {\frac{a}{b}} \\ d)\,\,\left\{ \begin{array}{l} \sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\ – \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2} \end{array} \right. \end{array}\) C \(\begin{array}{l} a)\,\,\frac{{\sqrt 2 }}{2}\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\left\{ \begin{array}{l} \sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\ – \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2} \end{array} \right. \end{array}\) D \(\begin{array}{l} a)\,\,\frac{{\sqrt 2 }}{2}\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\sqrt 5 \end{array}\)

adminby admin

Đưa thừa số vào trong dấu căn \(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} & & & b)\,\,a\sqrt 2 & & & c)\,\, – \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\left( {a > 0,\,\,\,b …