Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

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

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

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

........

a) 2x (x-5) -(x²-10x +25)=0

\(\Leftrightarrow\)2x(x-5)-(x-5)²=0

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

\(\Leftrightarrow\)(x-5)(x+5)=0

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

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

b) x² - 9 +3x(x+3) = 0

\(\Leftrightarrow\)(x² - 9) +3x(x+3) =0

\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0

\(\Leftrightarrow\)(x+3)(x-3+3x)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)

c) x³ - 16x = 0

\(\Leftrightarrow\)x(x²-16)=0

\(\Leftrightarrow\)x(x-4)(x+4)=0

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

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

d) (2x+3)(x-2) - (x² -4x+4) = 0

\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)²=0

\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0

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

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

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

e) 9x² -(x² -2x +1)=0

\(\Leftrightarrow\)(3x)²-(x-1)²=0

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

\(\Leftrightarrow\)(2x+1)(4x-1)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

f)x³-4x² -9x +36 = 0

\(\Leftrightarrow\)(x³-9x)-(4x²-36)=0

\(\Leftrightarrow\)x(x²-9)-4(x²-9)=0

\(\Leftrightarrow\)(x-4)(x²-9)=0

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)

g) 3x - 6 = (x-1).(x-2)

\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)

\(\Leftrightarrow\)x-1=3

\(\Leftrightarrow\)x=4

i) (x-2).(x+2) +(2x+1)² =-5x.(x-3) =5 (?? đề sao vậy ??)

k) x² -1 = (x-1).(2x+3)

\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)

\(\Leftrightarrow\)x+1=2x+3

\(\Leftrightarrow\)x-2x=3-1

\(\Leftrightarrow\)-x=2

\(\Leftrightarrow\)x=-2

l) (2x-1)² +(x+3).(x-3) -5x(x-2)=6

\(\Leftrightarrow\)4x²-4x+1+x²-9-5x²+10x=6

\(\Leftrightarrow\)6x-8=6

\(\Leftrightarrow\)6x=14

\(\Leftrightarrow\)x=\(\frac{7}{3}\)

a) \(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

b) \(3x\left(x-2\right)-5x+10=0\)

\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)

c) \(4x\left(x+3\right)-x^2+9=0\)

\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)

f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)

g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)

a) 4x³ - 9x = 0

<=> x( 4x² - 9 ) = 0

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

<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x + 3 = 0

<=> x = 0 hoặc x = ±3/2

b) 3x( x - 2 ) - 5x + 10 = 0

<=> 3x( x - 2 ) - 5( x - 2 ) = 0

<=> ( x - 2 )( 3x - 5 ) = 0

<=> x - 2 = 0 hoặc 3x - 5 = 0

<=> x = 2 hoặc x = 5/3

c) 4x( x + 3 ) - x² + 9 = 0

<=> 4x( x + 3 ) - ( x² - 9 ) = 0

<=> 4x( x + 3 ) - ( x - 3 )( x + 3 ) = 0

<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0

<=> ( x + 3 )( 4x - x + 3 ) = 0

<=> ( x + 3 )( 3x + 3 ) = 0

<=> x + 3 = 0 hoặc 3x + 3 = 0

<=> x = -3 hoặc x= -1

d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )

<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0

<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0

<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0

<=> ( x - 4 ).3x = 0

<=> x - 4 = 0 hoặc 3x = 0

<=> x = 4 hoặc x = 0

e) 16x² - 25 = ( 4x - 5 )( 2x + 1 )

<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0

<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0

<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0

<=> ( 4x - 5 )( 2x + 4 ) = 0

<=> 4x - 5 = 0 hoặc 2x + 4 = 0

<=> x = 5/4 hoặc x = -2

f) ( x + 1/5 )² = 64/9

<=> ( x + 1/5 )² = ( ±8/3 )²

<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3

<=> x = 37/15 hoặc x = -43/15

g) 9( x + 2 )² = ( x + 3 )²

<=> 3²( x + 2 )² - ( x + 3 )² = 0

<=> [ 3( x + 2 ) ]² - ( x + 3 )² = 0

<=> ( 3x + 6 )² - ( x + 3 )² = 0

<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0

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

<=> ( 2x + 3 )( 4x + 9 ) = 0

<=> 2x + 3 = 0 hoặc 4x + 9 = 0

<=> x = -3/2 hoặc x = -9/4

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)