Bài 1:

a, Thay m=-1 vào (1) ta có:
\(x^2-2\left(-1+1\right)x+\left(-1\right)^2+7=0\\ \Leftrightarrow x^2+1+7=0\\ \Leftrightarrow x^2+8=0\left(vô.lí\right)\)

Thay m=3 vào (1) ta có:

\(x^2-2\left(3+1\right)x+3^2+7=0\\ \Leftrightarrow x^2-2.4x+9+7=0\\ \Leftrightarrow x^2-8x+16=0\\ \Leftrightarrow\left(x-4\right)^2=0\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\)

b, Thay x=4 vào (1) ta có:

\(4^2-2\left(m+1\right).4+m^2+7=0\\ \Leftrightarrow16-8\left(m+1\right)+m^2+7=0\\ \Leftrightarrow m^2+23-8m-8=0\\ \Leftrightarrow m^2-8m+15=0\\ \Leftrightarrow\left(m^2-3m\right)-\left(5m-15\right)=0\\ \Leftrightarrow m\left(m-3\right)-5\left(m-3\right)=0\\ \Leftrightarrow\left(m-3\right)\left(m-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=3\\m=5\end{matrix}\right.\)

c, \(\Delta'=\left[-\left(m+1\right)\right]^2-\left(m^2+7\right)=m^2+2m+1-m^2-7=2m-6\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2m-6\ge0\Leftrightarrow m\ge3\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2+7\end{matrix}\right.\)

\(x_1^2+x_2^2=0\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=0\\ \Leftrightarrow\left(2m+2\right)^2-2\left(m^2+7\right)=0\\ \Leftrightarrow4m^2+8m+4-2m^2-14=0\\ \Leftrightarrow2m^2+8m-10=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-5\left(ktm\right)\end{matrix}\right.\)

\(x_1-x_2=0\\ \Leftrightarrow\left(x_1-x_2\right)^2=0\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\\ \Leftrightarrow\left(2m+2\right)^2-4\left(m^2+7\right)=0\\ \Leftrightarrow4m^2+8m+4-4m^2-28=0\\ \Leftrightarrow8m=28=0\\ \Leftrightarrow m=\dfrac{7}{2}\left(tm\right)\)

Bài 2:

a,Thay m=-2 vào (1) ta có:

\(x^2-2x-\left(-2\right)^2-4=0\\ \Leftrightarrow x^2-2x-4-4=0\\ \Leftrightarrow x^2-2x-8=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

b, \(\Delta'=\left(-m\right)^2-\left(-m^2-4\right)\ge0=m^2+m^2+4=2m^2+4>0\)

Suy ra pt luôn có 2 nghiệm phân biệt

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=-m^2-4\end{matrix}\right.\)

\(x_1^2+x_2^2=20\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=20\\ \Leftrightarrow2^2-2\left(-m^2-4\right)=20\\ \Leftrightarrow4+2m^2+8-20=0\\ \Leftrightarrow2m^2-8=0\\ \Leftrightarrow m=\pm2\)

\(x_1^3+x_2^3=56\\ \Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=56\\ \Leftrightarrow2^3-3\left(-m^2-4\right).2=56\\ \Leftrightarrow8-6\left(-m^2-4\right)-56\\ =0\\ \Leftrightarrow8+6m^2+24-56=0\\ \Leftrightarrow6m^2-24=0\\ \Leftrightarrow m=\pm2\)

\(x_1-x_2=10\\ \Leftrightarrow\left(x_1-x_2\right)^2=100\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-100=0\\ \Leftrightarrow2^2-4\left(-m^2-4\right)-100=0\\ \Leftrightarrow4+4m^2+16-100=0\\ \Leftrightarrow4m^2-80=0\\ \Leftrightarrow m=\pm2\sqrt{5}\)

\(x^2=-1\)(vôlí)

Vì \(x^2\ge0\forall x;-1< 0\)

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) 2x² - 3x - 2 = 0.

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

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

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

<=> x = -1 hoặc x = 10/3

c) 2x² - 5x + 3 = 0.

<=> (x - 1)(2x - 3) = 0

<=> x = 1 hoặc x = 3/2

giúp mk vs , mk đang cần gấp 

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

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

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

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

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

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

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

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

\(x.\left(x-2\right)-5\left(x-2\right)=0\)

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

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

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

Tham khảo nhé~

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

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

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

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

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

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

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

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

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

\(\Rightarrow\hept{\begin{cases}x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=2\end{cases}}}\)\((\)tmđk\()\)

Chúc bạn học tốt :>

a, |1 - 9x| - 10 = 0

=> |1 - 9x| = 10

=> 1 - 9x = 10 hoặc 1 - 9x = -10

=> 9x = -9 hoặc 9x = 11

=>x = -1 hoặc x = 11/9

vậy_

b, |3 - 14x| - 13 = 0

=> |3 - 14x| = 13

=> 3 - 14x = 13 hoặc 3 - 14x = -13

=> 14x = -10 hoặc 14x = 16

=> x = -10/14 hoặc x = 16/14

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