a: Sửa đề: \(ME\cdot MN=MF\cdot MP\)

Xét ΔMKN vuông tại K có KE là đường cao

nên \(ME\cdot MN=MK^2\left(1\right)\)

Xét ΔMKP vuông tại K có KF là đường cao

nên \(MF\cdot MP=MK^2\left(2\right)\)

Từ (1),(2) suy ra \(ME\cdot MN=MF\cdot MP\)

b: Xét tứ giác MEKF có \(\hat{MEK}=\hat{MFK}=\hat{FME}=90^0\)

nên MEKF là hình chữ nhật

=>MK=EF

Xét ΔMNP vuông tại M có MK là đường cao

nên \(MK^2=KN\cdot KP\)

=>\(EF^2=KN\cdot KP\)

d: Xét ΔMKP vuông tại K có KF là đường cao

nên \(FM\cdot FP=KF^2\)

=>\(FM\cdot FP=ME^2\)

\(4x^2-3x+15\) 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

closing

closing 

Công thức: Would you mind + V-ing

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

Bài 7:

a: \(x^3-4x^2-x+4=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>(x-4)(x-1)(x+1)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=1\\ x=-1\end{array}\right.\)

b: \(x^3+4x-5=0\)

=>\(x^3-x^2+x^2-x+5x-5=0\)

=>\(\left(x-1\right)\left(x^2+x+5\right)=0\)

=>x-1=0

=>x=1

c: \(x^4+2x^3+5x^2+4x-12=0\)

=>\(x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

=>\(\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

=>\(\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)

=>\(\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)

=>(x-1)(x+2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-2\end{array}\right.\)

d: \(x^3-7x^2+15x-25=0\)

=>\(x^3-5x^2-2x^2+10x+5x-25=0\)

=>\(\left(x-5\right)\left(x^2-2x+5\right)=0\)

=>x-5=0

=>x=5

Bài 5:

a: \(\left(x-5\right)\left(4x-7\right)+x^2-25=0\)

=>(x-5)(4x-7)+(x-5)(x+5)=0

=>(x-5)(4x-7+x+5)=0

=>(x-5)(5x-2)=0

=>\(\left[\begin{array}{l}x-5=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=\frac25\end{array}\right.\)

b: \(\left(x+7\right)\left(3x-5\right)-x^2+49=0\)

=>(x+7)(3x-5)-(x-7)(x+7)=0

=>(x+7)(3x-5-x+7)=0

=>(x+7)(2x+2)=0

=>2(x+1)(x+7)=0

=>(x+1)(x+7)=0

=>\(\left[\begin{array}{l}x+1=0\\ x+7=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-7\end{array}\right.\)

c: \(\left(3x+1\right)^2=\left(4x-5\right)^2\)

=>\(\left(4x-5\right)^2-\left(3x+1\right)^2=0\)

=>(4x-5-3x-1)(4x-5+3x+1)=0

=>(x-6)(7x-4)=0

=>\(\left[\begin{array}{l}x-6=0\\ 7x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=\frac47\end{array}\right.\)

b:

a: \(x^2-11x+30=0\)

=>\(x^2-5x-6x+30=0\)

=>x(x-5)-6(x-5)=0

=>(x-5)(x-6)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\end{array}\right.\)

b: \(10x^2-13x+3=0\)

=>\(10x^2-10x-3x+3=0\)

=>10x(x-1)-3(x-1)=0

=>(x-1)(10x-3)=0

=>\(\left[\begin{array}{l}x-1=0\\ 10x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=\frac{3}{10}\end{array}\right.\)

c: \(3x^2+10x-1=0\)

=>\(x^2+\frac{10}{3}x-\frac13=0\)

=>\(x^2+2\cdot x\cdot\frac53+\frac{25}{9}-\frac{28}{9}=0\)

=>\(\left(x+\frac53\right)^2=\frac{28}{9}\)

=>\(x+\frac53=\pm\frac{2\sqrt7}{3}\)

=>\(x=\pm\frac{2\sqrt7}{3}-\frac53\)

d: \(2x^2-3x+17=0\)

\(\Delta=\left(-3\right)^2-4\cdot2\cdot17=9-8\cdot17=9-136=-127<0\)

=>Phương trình vô nghiệm

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 + \(\sqrt{11}\) )

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

c: \(x-5=\left(\sqrt{x}\right)^2-\left(\sqrt5\right)^2=\left(\sqrt{x}-\sqrt5\right)\left(\sqrt{x}+\sqrt5\right)\)