\(4x^2-3x+15\) có \(\Delta< 0;a=4>0\Rightarrow4x^2-3x+15>0\forall x\)

\(PT\Leftrightarrow\sqrt{4x^2-3x+15}=3x-1\Rightarrow x>\dfrac{1}{3}\)

BP 2 vế

\(\Leftrightarrow4x^2-3x+15=9x^2-6x+1\)

\(\Leftrightarrow5x^2-3x-14=0\)

\(\Rightarrow x_1=-\dfrac{7}{5}< \dfrac{1}{3}\) (loại); \(x_2=2>\dfrac{1}{3}\) (chọn)

\(\Leftrightarrow\sqrt{4x^2-3x+15}=3x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\4x^2-3x+15=\left(3x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\4x^2-3x+15=9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\5x^2-3x-14=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\\left[{}\begin{matrix}x=2\\x=-\dfrac{7}{5}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=2\)

Ta có: \(\sqrt{a^2-ab+b^2}=\sqrt{\left(a+b\right)^2-3ab}\ge\sqrt{\left(a+b\right)^2-\dfrac{3.\left(a+b\right)^2}{4}}=\sqrt{\dfrac{\left(a+b\right)^2}{4}}\)

Tương tự: \(\sqrt{b^2-bc+c^2}\ge\sqrt{\dfrac{\left(b+c\right)^2}{4}};\sqrt{c^2-ca+a^2}\ge\sqrt{\dfrac{\left(a+c\right)^2}{4}}\)

\(\Rightarrow M\ge\sqrt{\dfrac{\left(a+b\right)^2}{4}}+\sqrt{\dfrac{\left(b+c\right)^2}{4}}+\sqrt{\dfrac{\left(c+a\right)^2}{4}}=\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}\)

\(=\dfrac{2\left(a+b+c\right)}{2}=\dfrac{2.1}{2}=1\)

Dấu ''='' xảy ra khi \(a=b=c=\dfrac{1}{3}\)

\(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\sqrt{\left(1-6x\right)^2}=5\)

\(\Leftrightarrow\left|1-6x\right|=5\)

Với: \(\left[{}\begin{matrix}x\le\dfrac{1}{6}\Leftrightarrow1-6x=5\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\\x>\dfrac{1}{6}\Leftrightarrow6x-1=5\Leftrightarrow x=1\left(tm\right)\end{matrix}\right.\)

biểu thức trong căn được viết lại có dạng của hằng đẳng thức:

 1 - 2.1.6x + (6x)² = (1 - 6x)²

\(\sqrt{1-12x+36x^{2^{ }}}\) = 5  <=> \(\sqrt{\left(1-6x\right)^2}\) = 5

                                    <=>  | 1 - 6x | = 5       

                                    <=>   1 - 6x = 5  hoặc 1 - 6x = -5

                                     <=>         x = - 4/6 = - 2/3  hoặc x = 1

Vậy phương trình đã cho có 2 nghiệm là x = -2/3 và x = 1

Ta có: \(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\)

\(=1:\left(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)\)

\(=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=1:\left(\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left(\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)=1:\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Vậy \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Bạn viết lại đề bằng latex nhé, nhìn thế này mình không giúp được

a) \(x^2-11=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)

b) \(x-3\sqrt{x}+4=x-4\sqrt{x}+\sqrt{x}-4=\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)\)

c) \(x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)

d) \(x+5\sqrt{x}+6=\left(x+3\sqrt{x}\right)+\left(2\sqrt{x}+6\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)\)

a) \(^{x^2}\)- 11 = ( x - \(\sqrt{11}\) )(x + \(\sqrt{11}\) )

Ta có \(\dfrac{1}{\sqrt{n}+\sqrt{n+2}}=\dfrac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}+\sqrt{n}\right)\left(\sqrt{n+2}-\sqrt{n}\right)}\) \(=\dfrac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}\right)^2-\left(\sqrt{n}\right)^2}\) \(=\dfrac{\sqrt{n+2}-\sqrt{n}}{2}\)

Như vậy, ta có \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{23}+\sqrt{25}}\) 

\(=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+...+\dfrac{\sqrt{25}-\sqrt{23}}{2}\)

\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{25}-\sqrt{23}}{2}\) 

\(=\dfrac{\sqrt{25}-1}{2}=\dfrac{5-1}{2}=2\)

\(A=\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{23}+\sqrt{25}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}}{\left(\sqrt{3}-\sqrt{1}\right)\left(\sqrt{3}+1\right)}+\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+...+\dfrac{\sqrt{25}-\sqrt{23}}{\left(\sqrt{25}-\sqrt{23}\right)\left(\sqrt{25}+\sqrt{23}\right)}\)

\(=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+...+\dfrac{\sqrt{25}-\sqrt{23}}{2}\)

\(=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{5}}{2}-...-\dfrac{\sqrt{23}}{2}+\dfrac{\sqrt{25}}{2}\\ =\dfrac{1}{2}+\dfrac{\sqrt{25}}{2}=\dfrac{1}{2}+\dfrac{5}{2}=3\)