a) Ta có : (x - 5)² - 16

= (x - 5)² - 4²

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

= (x - 1)(x - 9)

b) 25 - (3 - x)²

= 5² - (3 - x)²

= (5 - 3 + x)(5 + 3 - x)

= (x + 2)(8 - x)

c) (7x - 4)² - (2x + 1)²

= (7x - 4 - 2x - 1)(7x - 4 + 2x + 1)

= (5x - 5)(9x - 3)

= 5(x - 1)3(3x - 1)

= 15(x - 1)(3x - 1)

\(49.\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=\left[7.\left(y-4\right)\right]^2-\left[3.\left(y+2\right)\right]^2\)

\(=\left[7.\left(y-4\right)-3.\left(y+2\right)\right].\left[7.\left(y-4\right)+3.\left(y+2\right)\right]\)

\(=\left(7y-28-3y-6\right).\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right).\left(10y-22\right)\)

\(=4.\left(y-17\right).\left(5y-11\right)\)

\(8x^3+\frac{1}{27}=\left(2x\right)^3+\left(\frac{1}{3}\right)^3=\left(2x+\frac{1}{3}\right).\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)\)

\(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right).\left(25+5x^2+x^4\right)=\left(\sqrt{5}-x\right).\left(\sqrt{5}+x\right).\left(25+5x^2+x^4\right)\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=15-8=7\)

\(\Leftrightarrow x=\frac{-7}{2}\)

Vậy \(x=\frac{-7}{2}\)

\(a,\Delta=4\left(m-1\right)^2-4\left(-2m-3\right)=4m^2-8m+4+8m+12\\ \Delta=4m^2+16>0\left(đpcm\right)\\ b,\Delta=\left(2m-1\right)^2-4\left(2m-2\right)=4m^2-4m+1-8m+8\\ \Delta=4m^2-12m+9=\left(2m-3\right)^2\ge0\left(đpcm\right)\\ c,Sửa:x^2-2\left(m+1\right)x+2m-2=0\\ \Delta=4\left(m+1\right)^2-4\left(2m-2\right)=4m^2+8m+4-8m+8\\ \Delta=4m^2+12>0\left(đpcm\right)\\ d,\Delta=4\left(m+1\right)^2-4\cdot2m=4m^2+8m+4-8m\\ \Delta=4m^2+4>0\left(đpcm\right)\\ e,\Delta=4m^2-4\left(m+7\right)=4m^2-4m+7=\left(2m-1\right)^2+6>0\left(đpcm\right)\\ f,\Delta=4\left(m-1\right)^2-4\left(-3-m\right)=4m^2-8m+4+12+4m\\ \Delta=4m^2-4m+16=\left(2m-1\right)^2+15>0\left(đpcm\right)\)

a,ĐKXĐ:\(x\ge2\)

\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

b,ĐKXĐ:\(x\in R\)

\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(x\ge0\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)

a. (x - 2²) - 1 = 0

<=> x - 4 - 1 = 0

<=> x = 5

b. 4 - (x - 2)² = 0

<=> 2² - (x - 2)² = 0

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

<=> x(4 - x) = 0

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

d. (3x - 2)² - (2x + 3)² = 5(x + 4)(x - 4)

<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x² - 16)

<=> (x - 5)(5x + 1) = 5x² - 80

<=> 5x² + x - 25x - 5 = 5x² - 80

<=> 5x² - 5x² + x - 25x = -80 + 5

<=> -24x = -75

<=> x = \(\dfrac{25}{8}\)

a)\(\left(x-2^2\right)-1=0\Rightarrow x-4-1=0\Rightarrow x=5\)

Bài 1 :

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

\(\Rightarrow\left(x-10\right)\left(x-2\right)=0\)\(\Rightarrow x\in\left\{10;2\right\}\)

\(b,\)\(\left(x+8\right)^2=121\)\(\Rightarrow\left(x+8\right)^2-11^2=0\)

\(\Rightarrow\left(x+8+11\right)\left(x+8-11\right)=0\)\(\Rightarrow\left(x+19\right)\left(x-3\right)=0\)\(\Rightarrow x\in\left\{-19;3\right\}\)

\(c,x^2+8x+16=0\)\(\Rightarrow\left(x+4\right)^2=0\)

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

\(d,4x^2-12x=-9\)\(\Rightarrow4x^2-12x+9=0\)

\(\Rightarrow\left(2x-3\right)^2=0\)\(\Rightarrow2x-3=0\)\(\Rightarrow x=\frac{3}{2}\)

Bài 1 a) (x-4)^2 -36=0

=>  (x-4)^2 = 36

=> x-4 = 6

=> x= 10

b) (x+8)^2 = 121

=> x+8 = 11

=> x=3

c) x^2 + 8x +16=0

=> (x+4)^2 =0

=> x+ 4 =0 => x= -4

d) 4x^2 - 12x= -9

=> 4x^2 -12x+9=0

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

=> 2x-3 =0

=> x= 3/2

Bài 1)

a) (x+4)² - 36 =0

(x+4)^2 - 6^2 =0

=> (x+2)(x-10) =0 (sử dụng hằng đẳng thức) 

=> x =-2 ; x= 10(tự xét)

b) (x+8)^2 =121

(x+8)^2 - 121 =0

(x+8)^2 - 11^2 = 0

=> (x+19)(x-3) = 0

=> x= -19 ; x =3

c) x^2 + 8x + 16 =0

x^2 + 8x + 4^2 = 0

=> (x+4)^2 =0

=> x = -4

d) 4x^2 -12x =-9

(2x)^2 - 12x -9 = 0(chuyển vế)

=> (2x)^2 -12x -(3)^2=0

= (2x-3)^2 =0

=> x = 1,5

Bài 2:

a) (n+2)^2 - ( n-2)^2 = ( n +2+n-2)(n + 2 -n+2)

=> 2n * 4 = 8n : 8 với mọi n

b) (n+7)^2 - ( n-5)^2 = (n+7+n-5)(n+7-n+5)

= (2n +2)*12 = 24(n+1) : 24 với mọi x