thif 1 trong 2 so = 0

Giúp tớ với tớ cần gấp

a: =>(x-5)(x+5)=0

=>x=5 hoặc x=-5

b: \(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow x\in\left\{1;-3\right\}\)

c: =>x-3=4

hay x=7

d: =>x+1=7

hay x=6

e: =>x=0 và y-2=0

hay x=0 và y=2

a) \(4x.\left(2-x\right)+\left(2x+1\right)^2=2\)

    \(8x-4x^2+4x^2+4x+1=2\)

                                         \(12x+1=2\)

                                                  \(12x=2-1\)

                                                  \(12x=1\)

                                                        \(x=\frac{1}{12}\)

b) \(\left(x+3\right)^2-5.\left(x+3\right)=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

(x+1)+(x+3)+(x+5)+....+(x+99)=0

=> (x+x+x...+x) + (1+3+5+...+99) = 0

=>  50x             + 2500                 = 0

=>   50x            = -2500

=> x                 = -50

Vậy x = -50

x²-x=0

x(x-1)=0

=>x=0 hoặc x-1=0

                   x=0+1=1

vậy x\(\in\){0;1}

\(x^2-x=0\)

\(x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy \(x\in\left\{0;1\right\}\)

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

<=> \(\hept{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)

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

học tốt

a. (x - 3) . (2x + 6) = 0

<=> \(\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\2x=-6\end{cases}}\)

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

Vậy x = + 3

(x-3)(2x+6)=0

x(2+6-3)=0

x.5=0

x=0:5=0

Chử đề sai câu 1:

1, x(x-2)+x-2=0

Bài 1

1.(x-3)(x+2)-x(x-7)=15

\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)

\(\Leftrightarrow-6+6x=15\)

\(\Leftrightarrow6x=15+6\) =21

\(\Rightarrow x=\dfrac{21}{6}=3,5\)

2.(x-5)(x+5)+x(3-x)=20

\(\Leftrightarrow x^2-25+3x-x^2=20\)

\(\Leftrightarrow-25+3x=20\)

\(\Leftrightarrow3x=20+25=45\)

\(\Rightarrow x=\dfrac{45}{3}=15\)

3.(x-7)²-x(2+x)=-7

\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)

\(\Leftrightarrow-16x+49=-7\)

\(\Leftrightarrow-16x=-7-49=-56\)

\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)

Tiếp bài 1

4.(x-4)²-(x+4)(x-4)=-16

\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)

\(\Leftrightarrow-8x=-16\)

\(\Rightarrow x=\dfrac{-16}{-8}=2\)

5.(x-5)(x+5)-x(2-3x)=4x²-7

\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)

\(\Leftrightarrow-2x=18\)

\(\Rightarrow x=\dfrac{18}{-2}=-9\)

B2

1. 2x(x-2)+3x-6=0

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

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

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

2.3x(x-5)-5x+25=0

=>3x(x-5)-5(x-5)=0

=>(x-5)(3x-5)=0

\(\Rightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\3x-5=0\Leftrightarrow x=\dfrac{5}{3}\end{matrix}\right.\)

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

=>x(x+7)-(x+7)=0

=>(x+7)(x-1)=0

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