thảo có ghi lộn đề k

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

(7x - 15) :3 = 2

7x - 15 = 2 x 3

7x - 15 = 6

7x = 6 + 15

7x = 21

x = 21 :7

x = 3

( 7x - 15 ) : 3 = 2

  7x - 15        =  2 . 3

  7x -15         =   6

          7x       =  6 + 15

            7x     =  21

              x     = 3

  vậy x = 3

1: Ta có: \(x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

2: Ta có: \(x^2+7x+12=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

3: Ta có: \(x^2+8x+15=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

4: Ta có: \(x^2+5x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)

\(\left(x+3\right)\left(x^2-4x+5\right)\)

x^3-x^2-7x+15=0

<=> x^3+3x^2-4x^2-12x+5x+15=0

<=> x^2(x+3)-4x(x+3)+5(x+3)=0

<=> (x+3)(x^2-4x+5)=0

<=> x+3=0 vì x^2-4x+5 khác 0

<=> x=-3

(x+3)(x²-4x+5)

Ta có $\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=2$

$=>2M=(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15})(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15})$

$=>2M=\sqrt{x^2-7x+19}^2-\sqrt{x^2-7x+15}^2$

$=>2M=(x^2-7x+19)-(x^2-7x+15)=4$

$=>M=2$

\(2.M=\left(x^2-7x+19\right)-\left(x^2-7x+15\right)=4\Rightarrow M=2\)

Đặt \(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=B\)  = B

   Xét tích \(AB=\left(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}\right)\left(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15}\right)\)

                   \(=x^2-7x+19-\left(x^2-7x+15\right)=x^2-7x+19-x^2+7x-15\)

                     \(=4\)

     Mà \(B=2\Leftrightarrow A=2\)