b: \(\Leftrightarrow\left(x-5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)

bạn làm đúng rồi nhé

chúc bạn học tốt@

CÂU 10:

a, -x - 84 + 214 = -16                                                b, 2x -15 = 40 - ( 3x +10 )

       x                = - ( -16 -214 + 84 )                            2x + 3x = 40 -10 +15

       x                = 16 + 214 - 84                                        5x    = 45

      x                 = 146                                                        x     = 9

c, \(|-x-2|-5=3\)                                                             d, ( x - 2)(2x + 1) = 0

    \(|-x-2|=8\)                                                                     =>  x - 2 = 0 hoặc 2x + 1 = 0

    => - x - 2 = 8 hoặc x + 2 = 8                                                                         \(\orbr{\begin{cases}x-2=0\\2x+1=0\end{cases}=>}\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)

\(\orbr{\begin{cases}-x-2=8\\x+2=8\end{cases}=>\orbr{\begin{cases}x=-10\\x=6\end{cases}}}\)

CÂU 11:

Ta có : VT = - ( - a + b + c ) + ( b + c -1 ) = a - b - c + b + c - 1 = a - 1

           VP = ( b - c + 6 ) - ( 7 - a + b ) + c = b - c + 6 - 7 + a - b + c = a - 1

=> VT = VP hay - ( -a + b +c ) + ( b + c -1 ) = ( b - c + 6 ) - ( 7 - a + b ) + c 

Câu 17:

Xét ΔADC có OE//DC

nên \(\dfrac{OE}{DC}=\dfrac{AO}{AC}\left(1\right)\)

Xét ΔBDC có OH//DC

nên \(\dfrac{OH}{DC}=\dfrac{BO}{BD}\left(2\right)\)

Xét ΔOAB và ΔOCD có

\(\widehat{OAB}=\widehat{OCD}\)(hai góc so le trong, AB//CD)

\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)

Do đó: ΔOAB đồng dạng với ΔOCD
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)

=>\(\dfrac{OC}{OA}=\dfrac{OD}{OB}\)

=>\(\dfrac{OC}{OA}+1=\dfrac{OD}{OB}+1\)

=>\(\dfrac{OC+OA}{OA}=\dfrac{OD+OB}{OB}\)

=>\(\dfrac{AC}{OA}=\dfrac{BD}{OB}\)

=>\(\dfrac{OA}{AC}=\dfrac{OB}{BD}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\dfrac{OE}{DC}=\dfrac{OH}{DC}\)

=>OE=OH

Câu 15:

a: \(3x\left(x-1\right)+x-1=0\)

=>\(3x\left(x-1\right)+\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(3x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b: \(x^2-6x=0\)

=>\(x\cdot x-x\cdot6=0\)

=>x(x-6)=0

=>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Câu 14:

a. $6.2^2-36:3^2=6.4-36:9=24-4=20$

b. $19.48+52.19-400=19(48+52)-400=19.100-400=1900-400=1500$

a, 3\(x\).(\(x\) - 1) + \(x\) - 1 = 0

         (\(x\) - 1).(3\(x\) + 1) = 0

          \(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b, \(x^2\) - 6\(x\) = 0

   \(x\).(\(x\) - 6) = 0

   \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Câu 2:

\(a,ĐK:x\ge-3\\ PT\Leftrightarrow6\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+3}=2\\ \Leftrightarrow2\sqrt{x+2}=2\\ \Leftrightarrow\sqrt{x+2}=1\\ \Leftrightarrow x+2=1\\ \Leftrightarrow x=-1\left(tm\right)\\ b,\Leftrightarrow\sqrt{\left(2x-3\right)^2}=2017\Leftrightarrow\left|2x-3\right|=2017\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=2017\\3-2x=2017\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1010\\x=-1007\end{matrix}\right.\)

Câu 3:

\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}< 0,\forall x\left(-3< 0;\sqrt{x}+3>0\right)\\ \Leftrightarrow x\in\varnothing\)

a. (x² - 4).(x+3/5) = 0

 TH1: x² - 4 = 0

          x²      = 4

          x²      = 2²

                       -2²

    =>  x        = 2

                       -2

Vậy x \(\in\){-2;2}