\(d,x\left(x-3\right)-7x+21=0\)

\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)

\(a,2x\left(x-7\right)+5x-35=0\)

 \(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)

 \(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)

\(c,4x^2+12x+9=0\)

\(\Leftrightarrow4x^2+6x+6x+9=0\)

\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow x=-\frac{3}{2}\)

a) 2x(x-7)+5x-35=0
<=> 2x(x-7)+5(x-7)=0
<=>(2x+5)(x-7)=0
<=> (2x+5)=0 <=> x=-5/2
hoặc <=> x-7=0 <=> x=7

a) Ta có: \(7x\left(x-20\right)-x+20=0\)

\(\Leftrightarrow\left(x-20\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=\dfrac{1}{7}\end{matrix}\right.\)

b) Ta có: \(x^3-15x=0\)

\(\Leftrightarrow x\left(x-\sqrt{15}\right)\left(x+\sqrt{15}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² = 4

<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy S = { 5 ; 1 }

b) x² - 9 = 0

<=> x² = 9

<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy S = { 3 ; -3 }

c) x( x - 2x ) - x² - 8 = 0

<=> x² - 2x² - x² - 8 = 0

<=> -2x² - 8 = 0

<=> -2x² = 8

<=> x² = -4 ( vô lí )

<=> x = \(\varnothing\)

Vậy S = { \(\varnothing\)}

d) 2x( x - 1 ) - 2x² + x - 5 = 0

<=> 2x² - 2x - 2x² + x - 5 = 0

<=> -x - 5 = 0

<=> -x = 5

<=> x = -5

Vậy S = { -5 }

e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0 

<=> x² - 3x - ( x² - x - 2 ) = 0

<=> x² - 3x - x² + x + 2 = 0

<=> - 2x + 2 = 0

<=> -2x = -2

<=> x = 1

Vậy S = { 1 }

f) x( 3x - 1 ) - 3x² - 7x = 0

<=> 3x² - x - 3x² - 7x = 0

<=> -8x = 0

<=> x = 0

Vậy S = { 0 } 

d) 2x(x - 1) - 2x² + x - 5 = 0

=> 2x² - 2x - 2x² + x - 5 = 0

=> -x = 5

=> x = -5

e) x(x - 3) - (x + 1)(x - 2) = 0

=> x² - 3x - (x² - x - 2) = 0 

=> x² - 3x - x² + x + 2 = 0

=> -2x = - 2

=> x = 1

f) x(3x - 1) - 3x² - 7x = 0

=> 3x² - x - 3x² - 7x = 0

=> -8x = 0

=> x = 0

Trả lời

Bài 1:

a)(45-25).(-11)+29.(-3-17)

=20.(-11)+29.(-20)

=20.(-11)+(-29).20

=20.[(-11)+(-29)]

=20.(-30)

=-600

b)(36-6).(-5)+21.(-17-3)

=30.(-5)+21.(-20)

=(-150)+(-420)

=-570

trả lời:

a,(45-25).(-11)+29.(-3-17)

  =20.(-11)+29.(-20)

  =20.(-11)+(-29).20

  =20.[(-11)+(-29).20

  =20.(-30)

  =-600

học tốt

I

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a) -4.(2x+9)-(-8x+3)-(x+13)=0

     -8-36+8x-3-x-13=0

     -x-52=0

      x=-52

b) 7x.(2+x)-7x.(x+3)=14

     7x.(2+x-x-3)=14

     7x.(-1)=14

     7x=14:(-1)

     7x=-14

       x=(-14):7

       x=-2

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}

Trả lời:

a) \(x\left(x-3\right)+2x-6=0\Leftrightarrow x\left(x-3\right)+\left(2x-6\right)=0\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)b) \(x^2+7x-2\left(x+7\right)=0\Leftrightarrow\left(x^2+7x\right)-2\left(x+7\right)=0\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$