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 = \(\frac{3}{2}\)

\(X=\)\(-2\)

\(X=3\)

\(X=-4\)

\(X=1,5\)

a/ \(\left(x-4\right)^2-36=0\)

<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)

<=> \(\left(x-10\right)\left(x+2\right)=0\)

<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

b/ \(\left(x+8\right)^2=121\)

<=> \(\left(x+8\right)^2-121=0\)

<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)

<=> \(\left(x-3\right)\left(x+19\right)=0\)

<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)

d/ \(4x^2-12x+9=0\)

<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)

<=> \(\left(2x-3\right)^2=0\)

<=> \(2x-3=0\)

<=> \(x=\frac{3}{2}\)

\(c,x^2+8x+16=0\)

\(\Rightarrow x^2+4x+4x+16=0\)

\(\Rightarrow x.\left(x+4\right)+4.\left(x+4\right)=0\)

\(\Rightarrow\left(x+4\right)^2=0\)

\(\Rightarrow x=-4\)

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

\(4x^2-6x-6x+9=0\)

\(\Rightarrow\left(2.2x^2-2.3x\right)-\left(3.2x-3^2\right)=0\)

\(\Rightarrow2.\left(2x^2-3x\right)-3.\left(2x-3\right)=0\)

\(\Rightarrow2x.\left(2x-3\right)-3.\left(2x-3\right)=0\)

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

\(\Rightarrow2x-3=0\Rightarrow x=1,5\)

Hai phần đầu dễ bãn tự làm

Bài1:

\(â,\left(x-4\right)^2-36=0\\ \Leftrightarrow\left(x-4\right)^2=36\\ \Leftrightarrow x-4\in\left\{-6;6\right\}\\ \Leftrightarrow x\in\left\{-2;10\right\}\)

Vậy...

b<Tương tự

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

Vậy...

d,Tương tự

Bài2:

\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =100.50=5000\)

\(b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\)

Bài 1:

a) (x-4)^2 - 36 = 0

=> (x-4)^2 = 36

=> (x-4)^2 = 6^2

=> x-4 = 6

=>x = 2

b) (x-8)^2 = 121

=> (x-8)^2 = 11^2

=> x-8 = 11

=> x = 19

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

=> x( x +8) = -16

=> x = -4

d) 4x^2 - 12x = -9( Mk chưa nghĩ ra !)

Bài 2 :

a) 75^2 - 25^2

= (75-25)(75+25)

=50.100

=5000

b) 53^2 - 47^2

= (53-47)(53+47)

=6.100

=600

a.x = -3

b.\(\frac{13}{2}\)

c. (x+4)²

d. không bik

Bài làm:

a) \(\left(x+4\right)^2-1=0\)

\(\Leftrightarrow\left(x+4\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

b) \(\left(2x-3\right)^2=100\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=10\\2x-3=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=13\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{2}\\x=-\frac{7}{2}\end{cases}}\)

c) \(x^2+8x+16=0\)

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

\(\Rightarrow x+4=0\)

\(\Rightarrow x=-4\)

d) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Rightarrow2x-3=0\)

\(\Rightarrow x=\frac{3}{2}\)

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

\(\Leftrightarrow\left(x+4\right)^2=1^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

b) \(\left(2x-3\right)^2=100\)

\(\Leftrightarrow\left(2x-3\right)^2=10^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=10\\2x-3=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=13\\2x=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{2}\\x=-\frac{7}{2}\end{cases}}\)

c) \(x^2+8x+16=0\)

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

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

d) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

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

a) Theo mình thì chỉ min thôi nhé!

\(A=\frac{8x^2-1}{4x^2+1}+1+11=\frac{12x^2}{4x^2+1}+11\ge11\)

b)Bạn rút gọn lại giùm mìn, lười quy đồng lắm:(

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

ĐKXĐ; ...

a/ \(P=\frac{x^2}{x+4}\left[\frac{\left(x+4\right)^2}{x}\right]+9=x\left(x+4\right)+9=\left(x+2\right)^2+5\ge5\)

\(P_{min}=5\) khi \(x=-2\)

b/ \(Q=\left(\frac{\left(x+2\right)\left(x^2-2x+4\right).4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right).\frac{x\left(x-2\right)^3}{-16}\)

\(=\left(\frac{4\left(x^2-2x+4\right)-4x\left(x-2\right)}{\left(x-2\right)^2}\right).\frac{-x\left(x-2\right)^3}{16}\)

\(=\frac{16}{\left(x-2\right)^2}.\frac{-x\left(x-2\right)^3}{16}=-x\left(x-2\right)=-x^2+2x\)

\(=1-\left(x-1\right)^2\le1\)

\(Q_{max}=1\) khi \(x=1\)