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

a, x^2 - 25 = 0

=> x^2 = 25

Mà : 5^2=25

=> x^2=25

=> x^2=5^2

=> x=5

b, x^2 - x - 6 = 0

=> x^2-x=6

Ta có : 3^2=6

=> x = 3 

=> 3^2-3=6

cái này mk ko chắc lắm :v

c, x^2 - 8x + 15 lộn đề òi bạn

a) \(x^2-25=0\)

\(x^2=0+25=25\)

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

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

b) \(x^2-x-6=0\)

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

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

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

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)

Câu c bằng bao nhiêu bn ơi!!!!!!

a)x²-25=0

x²=25

x=5

8x²+30x+7=0

 8x²+16x+14x+7=0

8x(x+2) +7(x+2)=0

(8x+7)(x+2)=0

=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)

a)

4x²-8x+4=2(1-x)(x+1)

4x²-8x+4-2+2x²=0

6x²-8x+2=0

2(3x²-4x+1)=0

3x²-3x-x+1=0

3x(x-1) -(x-1)=0

(3x-1)(x-1)=0

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

d,

x²+3x-18=0

=> x²-3x+6x-18=8

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

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

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

`a, x^2-10x+25=0`

`<=>x^2 -2.x.5+5^2=0`

`<=>(x-5)^2=0`

`<=>x-5=0`

`<=>x=5`

__

`x^2 -8x+16=0`

`<=> x^2 - 2.x.4+4^2=0`

`<=>(x-4)^2=0`

`<=>x-4=0`

`<=>x=4`

__

`x^2-49=0`

`<=>x^2 - 7^2=0`

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

`<=>x-7=0` hoặc `x+7=0`

`<=> x=7` hoặc `x=-7`

__

`4x^2-25=0`

`<=> (2x)^2 -5^2=0`

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

`<=>2x-5=0` hoặc `2x+5=0`

`<=> 2x=5` hoặc `2x=-5`

`<=>x=5/2` hoặc `x=-5/2`

a: =>(x-5)^2=0

=>x-5=0

=>x=5

b: =>(x-4)^2=0

=>x-4=0

=>x=4

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

=>x-7=0 hoặc x+7=0

=>x=7 hoặc x=-7

d: =>(2x-5)(2x+5)=0

=>2x-5=0 hoặc 2x+5=0

=>x=5/2 hoặc x=-5/2

a: =>2x(x+4)=0

=>x=0 hoặc x=-4

b: =>(x-5)(2x+10)=0

=>x=5 hoặc x=-5

a: =>2x(x+4)=0

=>x=0 hoặc x=-4

b: =>(x-5)(2x+10)=0

=>x=5 hoặc x=-5

a, \(x\) + 99: 3 = 55

    \(x\) + 33 = 55

     \(x\)        = 55 - 33

      \(x\)       = 22

b, (\(x\) - 25):15 = 20

       \(x\) - 25  = 20 x 15

       \(x\) - 25  = 300

        \(x\)         = 300 + 25

          \(x\)        = 325

c, (3\(x\) - 15).7 = 42

      3\(x\) - 15 = 42:7

       3\(x\) - 15 = 6

        3\(x\)         = 6 + 15

         3\(x\)         = 21

            \(x\)         = 21: 3

             \(x\)         = 7

d, (8\(x\) - 16).(\(x\) -5) = 0

       \(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=16:8\\x=5\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) {2; 5}

\(a,8x-75=5x+21\)

    \(8x-5x=21+75\)

               \(3x=96\)

                  \(x=32\)

\(b,9x+25=-\left(2x-58\right)\)

    \(9x+25=-2x+58\)

    \(9x+2x=58-25\)

            \(11x=33\)

                 \(x=3\)

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

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

a) \(8x-75=5x+21\)

\(8x-5x=21+75\)

\(3x=96\)

\(x=32\)

vậy \(x=32\)

b) \(9x+25=-\left(2x-58\right)\)

\(9x+25=-2x+58\)

\(9x+2x=58-25\)

\(11x=33\)

\(x=3\)

vậy \(x=3\)

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

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

vậy....

d) \(3.\left|x+3\right|-4=-7-\left(-18\right)\)

\(3.\left|x+3\right|-4=-7+18\)

\(3.\left|x+3\right|=11+4\)

\(3.\left|x+3\right|=15\)

\(\left|x+3\right|=5\)

\(\Rightarrow\orbr{\begin{cases}x+3=5\\x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-8\end{cases}}\)

vậy..

<=> \(\left(x^2+8x+16\right)+9=0\)

<=> (x + 4)² = -9 (VN)

Vậy ko có gtri x thỏa mãn đề bài