a, 4x² - 49 = 0

⇔⇔ (2x)² - 7² = 0

⇔⇔ (2x - 7)(2x + 7) = 0

⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72

b, x² + 36 = 12x

⇔⇔ x² + 36 - 12x = 0

⇔⇔ x² - 2.x.6 + 6² = 0

⇔⇔ (x - 6)² = 0

⇔⇔ x = 6

e, (x - 2)² - 16 = 0

⇔⇔ (x - 2)² - 4² = 0

⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0

⇔⇔ (x - 6)(x + 2) = 0

⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2

f, x² - 5x -14 = 0

⇔⇔ x² + 2x - 7x -14 = 0

⇔⇔ x(x + 2) - 7(x + 2) = 0

⇔⇔ (x + 2)(x - 7) = 0

⇔{x+2=0x−7=0⇔{x=−2x=7

a,\(4x^2-49=0\)

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

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

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

b.\(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

c.\(\frac{1}{16x^2}-x+4=0\)

\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)

\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)

........

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

a,2x+5 = 0 hoặc 5-x=0 ( còn lại tự tính)

b,,x²-4=0 hoặc x²-36=0 ( còn lại tự tính)

tương tự như vậy làm câu c

d, bài này dài ( không làm )

e, ......( dài)

f, x={4;5;6} 

a) \(4x^2-49=0\)

<=> \(\left(2x-7\right)\left(2x+7\right)=0\)

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

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

b) x² + 36 = 12x

<=>x² + 36 - 12x=0

<=> (x-6)²=0

<=> x-6 =0

<=> x=6

d) x³ -3\(\sqrt{3}\) x²+9x - 3\(\sqrt{3}\) =0

<=> \(\left(x-\sqrt{3}\right)^3=0\)

<=> \(x-\sqrt{3}=0\)

<=> \(x=\sqrt{3}\)

1, \(x^2\) - 9 = 0

 (\(x\) - 3)(\(x\) + 3) = 0

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

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

 vậy \(x\) \(\in\) {-3; 3}

7, (3\(x\) + 1)² - 16 = 0

    (3\(x\) + 1 - 4)(3\(x\) + 1 + 4) = 0

    (3\(x\) - 3).(3\(x\) + 5) = 0

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

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

       \(\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{3}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {1; - \(\dfrac{5}{3}\)}

10, (\(x\) + 3)² - \(x^2\) = 45

      [(\(x\) + 3) - \(x\)].[(\(x\) + 3) + \(x\)] = 45

                 3.(2\(x\) + 3)  = 45

                     2\(x\) + 3   = 15

                     2\(x\)          = 12

                       \(x\)          = 6

1. \(4x^2-49=0\)

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

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

Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)

===========

2. \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x=6\)

Vậy: \(x=6\)

===========

3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)

\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)

1: Ta có: \(4x^2-49=0\)

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

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

2: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\)

hay x=6

a. 16a² - 49.( b - c )²

= ( 4a )² - 7².( b - c )²

= ( 4a )² - [ 7.( b - c ) ]²

= ( 4a )² - ( 7b - 7c )²

= ( 4a - 7b + 7c ).( 4a + 7b - 7c )

b. ( ax + by )² - ( ax - by )²

=( ax + by + ax - by ).( ax + by - ax + by )

= 2ax . 2by

= 2.( ax + by )

c.a⁶ - 1

= ( a³ )² - 1

= ( a³ - 1 ).( a³ + 1 )

= ( a - 1 ).( a² + a + 1 ).( a + 1 ).( a² - a + 1 )

d. a⁸ - b⁸

= ( a⁴ )² - ( b⁴ )²

= ( a⁴ - b⁴ ).( a⁴ + b⁴ )

= [ ( a² )² - ( b² )² ].( a⁴ + b⁴ )

= ( a² - b² ).( a² + b² ).( a⁴ + b⁴ )

= ( a - b ).( a + b ).( a² + b² ).( a⁴ + b⁴ )

B₂

( x - 4 )² - 36 = 0

\(\Leftrightarrow\) ( x - 4 )² = 36

\(\Leftrightarrow\) ( x - 4 )² = 6²

\(\Leftrightarrow\) x + 4 = \(\pm\) 6

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

Vậy x = 10 , x = -2

b. ( x - 8 )² = 121

\(\Leftrightarrow\) ( x - 8 )² = 11²

\(\Leftrightarrow\) x - 8 = \(\pm\)11

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=11\\x-8=-11\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=19\\x=-3\end{cases}}\)

Vậy x = 19 , x = -3

c. x² + 8x + 16 = 0

\(\Leftrightarrow\)x² + 2.4x + 4² = 0

\(\Leftrightarrow\) ( x + 4 )² = 0

\(\Leftrightarrow\) x + 4 = 0

\(\Leftrightarrow\) x = -4

Vậy x = -4

d. 4x² - 12x = - 9

\(\Leftrightarrow\)( 2x )² - 2.2.x.3 + 3² = 0

\(\Leftrightarrow\) ( 2x - 3 )² = 0

\(\Leftrightarrow\) 2x - 3 = 0

\(\Leftrightarrow\) 2x = 3

\(\Leftrightarrow\) \(x=\frac{3}{2}\)

Vậy x = \(\frac{3}{2}\)

\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)

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

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

Vậy \(S=\left\{3;-3\right\}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)

\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)

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

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

Vậy \(S=\left\{-2;5\right\}\)