Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Làm nốt ::v
\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)
\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)
\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)
\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)
\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)
\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)
\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)
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 ...
\(A=\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}\\ =\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}\\ \\ =a+3+3-a\\ =6\)
\(B=\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\\ =\sqrt{\left(a-1\right)+2\sqrt{a-1}+1}+\sqrt{\left(a-1\right)-2\sqrt{a-1}+1}\\ =\sqrt{\left(\sqrt{a-1}+1\right)^2}+\sqrt{\left(\sqrt{a-1}-1\right)^2}\\ =\sqrt{a-1}+1+1-\sqrt{a-1}\\ =2\)
\(a,\frac{a-4\sqrt{a}+4-1}{\sqrt{a}-3}=\frac{\left(\sqrt{a}-2\right)^2-1}{\sqrt{a}-3}.\)
\(=\frac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)
\(=\sqrt{a}-1\)
\(b,\frac{a+\sqrt{a^2-6a+9}}{2a-3}=\frac{a+\sqrt{\left(a-3\right)^2}}{2a-3}\)
\(=\frac{a+a-3}{2a-3}=\frac{2a-3}{2a-3}\)
\(=1\)
\(\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}\)
\(=\left|a+3\right|+\left|a-3\right|\)
Vì \(-3\le a\le3\)\(\Rightarrow\left|a+3\right|=a+3\)và \(\left|a-3\right|=-\left(a-3\right)=-a+3\)
\(\Rightarrow\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\left(a+3\right)+\left(-a+3\right)=6\)
a/ \(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b/ \(P< 1\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}< 1\Rightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
Xét 2 trường hợp:
- \(\hept{\begin{cases}2\sqrt{x}+3>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\hept{\begin{cases}2\sqrt{x}>-3\\\sqrt{x}< 3\end{cases}\Rightarrow}\hept{\begin{cases}\sqrt{x}>-\frac{3}{2}\\\sqrt{x}< 3\end{cases}}\Rightarrow-\frac{3}{2}< \sqrt{x}< 3}\)
\(\Rightarrow-\frac{9}{4}< x< 9\)
- \(\hept{\begin{cases}2\sqrt{x}+3< 0\\\sqrt{x}>3\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{x}< -\frac{3}{2}\\\sqrt{x}>3\end{cases}}}\) (vô lí)
Vậy -9/4 < x < 9
\(A=\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}\)
\(A=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}\)
\(A=\left|a+3\right|+\left|a-3\right|\)
có \(\hept{\begin{cases}a\ge-3\Rightarrow a+3\ge0\\a\le3\Rightarrow a-3̸\le0\end{cases}}\)
nên \(A=a+3+3-a=6\)