\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)

\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)

Vậy \(x=\dfrac{-255}{2}\)

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

a) \(x\left(x-6\right)=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

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

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

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

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

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

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

\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)

Từ đó bạn tìm những giá trị của x nha!

a) \(\Rightarrow\left(2x-3\right)^2=49\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

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

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

a, ⇒ (2x - 3)² = 49

    ⇒  (2x - 3)² = \(\left(\pm7\right)^2\)

    ⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0

    ⇒ (x - 5).(2x + 7)  = 0

    ⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c, ⇒ x² - 5x + 2x - 10 = 0

    ⇒ (x² - 5x) + (2x - 10) = 0

    ⇒ x.(x - 5) +2.(x - 5)    = 0

    ⇒ (x - 5).(x + 2)=0

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

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+1\right)=4\)

\(\Leftrightarrow x^3-1-x^3-x=4\)

\(\Leftrightarrow-x=5\)

hay x=-5

c: Ta có: \(\left(2x-1\right)^3+\left(x+2\right)^3-9x\left(x+1\right)\left(x-1\right)=7\)

\(\Leftrightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8-9x^3+9x=7\)

\(\Leftrightarrow-6x^2+27x=0\)

\(\Leftrightarrow-3x\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{9}{2}\end{matrix}\right.\)

a: \(\left(2x-3\right)^2-49=0\)

\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)

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

a. (2x - 3)² - 49 = 0

<=> (2x - 3)² - 7² = 0

<=> (2x - 3 + 7)(2x - 3 - 7) = 0

<=> (2x + 4)(2x - 10) = 0

<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

b. 2x(x - 5) - 7(5 - x) = 0

<=> 2x(x - 5) + 7(x - 5) = 0

<=> (2x + 7)(x - 5) = 0

<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

c. x² - 3x - 10 = 0

<=> x² - 5x + 2x - 10 = 0

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

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

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

<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

a, (2x - 3)² - 49 = 0

(2x - 3)² - 7² = 0

(2x - 3 + 7)( 2x - 3 - 7) = 0

(2x + 4)( 2x - 10) = 0

=> 2x + 4 = 0                => 2x - 10 = 0

     2x       = - 4                   2x         = 10

       x       = - 2                     x         = 5

\(a,\Leftrightarrow3\left(x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=2\\6x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-2013\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=\dfrac{1}{4}\end{matrix}\right.\\ d,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)

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

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

- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .

a) (x-3)²-4=0

⇒ (x-3)²=4

⇒ hoặc x-3=2⇒x=5

hoặc x-3=-2⇒x=1

c) (x+4)²-(x+1)(x-1)=16

⇒ x²+8x+16-x²+1=16

⇒ 8x+17=16

⇒ 8x=-1

⇒ x=-1/8

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)( vô lý)

Vậy \(S=\varnothing\)

b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)