Tháng Hai 4, 2026

Tìm điểm sai trong câu: “Dưới ánh nắng sương long lanh triệu cành hồng khoe sắc thắm”

by adminTháng Chín 29, 2021

Tìm điểm sai trong câu: “Dưới ánh nắng sương long lanh triệu cành hồng khoe sắc thắm”

Câu trả lời là Câu trả lời đúng là Khoe sắc khoé!

Câu trả lời là Câu trả lời đúng là Dùng ống hút.

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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 \(\begin{array}{l} a)\,\,\frac{{\sqrt 2 }}{2}\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\sqrt 5 \end{array}\)