x.( 7x - 42 ) + 10. ( 7x - 42 ) = 0

=> ( 7x - 42 ). ( x + 10 ) = 0

=> \(\orbr{\begin{cases}x+10=0\\7x-42=0\end{cases}}\)

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

a,  7\(x\).(2\(x\) + 10) = 0

        \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\){-5; 0}

b, - 9\(x\) : (2\(x\) - 10) = 0

      - 9\(x\) = 0

           \(x\) = 0

c, (4 - \(x\)).(\(x\) + 3) = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

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

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

    \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-2023; 2024}

a, 7\(x\).(2\(x\) + 10) =0

    \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\) {-5; 0}

b, -9\(x\) : (2\(x\) - 10) = 0

    9\(x\)                   = 0 

     \(x\)                    = 0 

c, (4 - \(x\)).(\(x\) + 3)  = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

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

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

     \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\in\) {-2023; 2024}

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

\(\Leftrightarrow x^2+2x+5x+10=0\\ \Leftrightarrow\left(x+2\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-5\end{matrix}\right.\)

x²+2x+5x+10=0

x(x+2)+5x+5.2=0

x(x+2)+5(x+2)=0

(X+2)(x+5)=0

Th1: x+2=0

x=-2

Th2 : x+5=0

x=-5

a) \(\left(x^2-3x\right)\left(x^2+7x+10\right)=216\Rightarrow x\left(x-3\right)\left(x+2\right)\left(x+5\right)=216\)

\(\Rightarrow x\left(x+2\right)\left(x-3\right)\left(x+5\right)=216\Rightarrow\left(x^2+2x\right)\left(x^2+2x-15\right)=216\)

Đặt \(t=x^2+2x\Rightarrow\) pt trở thành \(t\left(t-15\right)=216\Rightarrow t^2-15t-216=0\)

\(\Rightarrow\left(t+9\right)\left(t-24\right)=0\Rightarrow\left[{}\begin{matrix}t=-9\\t=24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+2x=-9\\x^2+2x=24\end{matrix}\right.\)

\(TH_1:x^2+2x=-9\Rightarrow x^2+2x+9=0\Rightarrow\left(x+1\right)^2+8=0\) (vô lý)

\(TH_2:x^2+2x=24\Rightarrow x^2+2x-24=0\Rightarrow\left(x-4\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)

b) \(\left(2x^2-7x+3\right)\left(2x^2+x-3\right)+9=0\)

\(\Rightarrow\left(x-3\right)\left(2x-1\right)\left(x-1\right)\left(2x+3\right)+9=0\)

\(\Rightarrow\left(x-3\right)\left(2x+3\right)\left(x-1\right)\left(2x-1\right)+9=0\)

\(\Rightarrow\left(2x^2-3x-9\right)\left(2x^2-3x+1\right)+9=0\)

Đặt \(t=2x^2-3x-9\Rightarrow\) pt trở thành \(t\left(t+10\right)+9=0\)

\(\Rightarrow t^2+10t+9=0\Rightarrow\left(t+1\right)\left(t+9\right)=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-9\end{matrix}\right.\)

\(TH_1:t=-1\Rightarrow2x^2-3x-9=-1\Rightarrow2x^2-3x-8=0\)

\(\Delta=\left(-3\right)^2-4\left(-8\right).2=73\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{73}}{4}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{73}}{4}\end{matrix}\right.\)

\(TH_2:t=-9\Rightarrow2x^2-3x-9=-9\Rightarrow2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

a) x² - 7x + 10 = 0

x² - 5x - 2x + 10 = 0

(x² - 5x) - (2x - 10) = 0

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

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

\(\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}}\)

\(\orbr{\begin{cases}x=0+5\\x=0+2\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=2\end{cases}}\)

Vậy x = 5 hoặc x = 2

b) tương tự

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

=> x=2 , x=5

b) ( x+2)(7x-1) =0

=> x= -2 , x= 1/7

a) x.x - 7 x = - 10

x( x - 7 )    = - 10 

=> x ; x -7 thuộc ước của 10

Ta có:

x² + 7x + 10 = 0

<=> x^2 + 5x + 2x + 10 = 0

<=> x(x + 5) + 2(x + 5) = 0

<=> (x+2)(x+5) = 0

<=> x+2=0 hoặc x+5=0

<=> x= -2 hoặc x= -5

Vậy x = -2; -5.

\(x^2+7x+10=0\)

\(\Leftrightarrow\left(x^2+5x\right)+\left(2x+10\right)=0\)

\(\Leftrightarrow x\left(x+5\right)+2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+5=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=-2\end{cases}}}\)

a, x/4 - 3x + 11 = 5/6 - x +7x

\(\frac{44-11x}{4}=\frac{36x+5}{6}\Rightarrow\left(44-11x\right)6=4\left(36x+5\right)\)

\(\Rightarrow264-66x=144x+20\)

\(\Rightarrow-210x=-244\)

\(\Rightarrow x=\frac{122}{105}\)

b,x^2 - 2x = 0

=>x(x-2)=0

=>x=0 hoặc x-2=0

=>x=0 hoặc x=2

c, x^2 - 7x - 10 =0

đề có khi sai