Question

[pham lan phuong]

Bài 2: Rút gọn biểu thức

1) 2\(\sqrt{a^{2^{ }}}\) với a \(\ge\) 0

2) 3\(\sqrt{\left(a-2\right)^{2_{ }}}\) với a<2

3) \(\sqrt{81a^{4^{ }}}\) + 3a²

4) \(\sqrt{64a^{2^{ }}}+2a\) (a\(\ge\) 0)

5) 3\(\sqrt{9a^{6^{ }}}-6a^3\) ( a bất kỳ)

6) \(\sqrt{a^{2^{ }}+6a+9}+\sqrt{a^{2^{ }}-6a+9}\) ( a bất kì)

7) \(\dfrac{\sqrt{1-2x+x^2}}{x-1}\)

8) A= \(\dfrac{\sqrt{9x^{2^{ }}-6x+1}}{9x^{2^{ }}-1}\)

9) B= 4-x- \(\sqrt{4-4x+x^2}\)

10) C= \(\sqrt{4x^{2^{ }}-4x+1}-\sqrt{4x^{2^{ }}+4x+1}\)

Answer 1

[Phùng Khánh Linh]

Làm nốt ::v

\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)

\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)

\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)

\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)

\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)

\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)

\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)

Answer 2

[Hắc Hường]

Mình làm ba câu mẫu, bạn theo đó mà làm các câu còn lại.

Giải:

1) \(2\sqrt{a^2}\)

\(=2\left|a\right|\)

\(=2a\left(a\ge0\right)\)

Vậy ...

5) \(3\sqrt{9a^6}-6a^3\)

\(=3\sqrt{\left(3a^3\right)^2}-6a^3\)

\(=3.3a^3-6a^3\)

\(=9a^3-6a^3\)

\(=3a^3\)

Vậy ...

10) \(C=\sqrt{4x^2-4x+1}-\sqrt{4x^2+4x+1}\)

\(\Leftrightarrow C=\sqrt{\left(2x-1\right)^2}-\sqrt{\left(2x+1\right)^2}\)

\(\Leftrightarrow C=2x-1^2-\left(2x+1^2\right)\)

\(\Leftrightarrow C=2x-1-2x-1\)

\(\Leftrightarrow C=-2\)

Vậy ...