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\(-1< x< 1\Leftrightarrow-1< 0< x^2< 1\Leftrightarrow1-x>0\) (*)

Ta co \(\left(3x-5\right)^2\ge0\forall x\)

Dau '' = '' xay ra \(3x-5=0\Leftrightarrow3x=5\Leftrightarrow x=\frac{5}{3}\) ma de ra \(-1< x< 1\Leftrightarrow\left(3x-5\right)^2\ge0\) (**)

Tu (*) va (**) \(\Leftrightarrow\frac{\left(3x-5\right)^2}{1-x^2}>0\) (Khong tim duoc MinA)

Khong biet do de bai sai hay toi sai nua @@

a) \(\left(a^2-4\right)\left(a^2+4\right)\)

\(=a^4-8\)

c) \(\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\)

=\(\left(a^2-b^2\right)\left(a^2+b^2\right)=a^4-b^4\)

d) \(\left(a-b+c\right)\left(a+b+c\right)\)

=\(a^2-\left(b+c\right)^2\)

e) \(\left(x+2-y\right)\left(x-2-y\right)\)

=\(x-\left(2-y\right)\)

mik lm tắt có gì sai cho mik xin lỗi

( a² - 4 )( a² + 4 ) = a⁴ - 16

( x³ - 3y )( x³ + 3y ) = x⁶ - 9y²

( a - b )( a + b )( a² + b² )( a⁴ + b⁴ ) = ( a² - b² )( a² + b² )( a⁴ + b⁴ ) = ( a⁴ - b⁴ )( a⁴ + b⁴ ) = a⁸ - b⁸

( a - b + c )( a + b + c ) = ( a + c )² - b² = a² - b² + c² + 2ac

( x + 2 - y )( x - 2 - y ) = ( x - y )² - 2² = x² - 2xy + y² - 4

\(B=\left(\frac{x-2}{2x-2}-\frac{3}{2-2x}-\frac{x+3}{2x+2}\right):\left(1-\frac{x-3}{x+1}\right)\)    \(ĐKXĐ:x\ne\pm1\)

\(B=\left(\frac{\left(x-2\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right)\)\(:\left(\frac{x+1-x+3}{x+1}\right)\)

\(B=\left(\frac{x^2-x-2+3x+3-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{4}{x+1}\right)\)

\(B=\left(\frac{4}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{4}{x+1}\right)\)

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

\(\frac{3^2-1}{5^2-1}\times\frac{7^2-1}{9^2-1}\times\frac{11^2-1}{13^2-1}\times...\times\frac{43^2-1}{45^2-1}\)

\(=\frac{2\times4}{4\times6}\times\frac{6\times8}{8\times10}\times\frac{10\times12}{12\times14}\times...\times\frac{42\times44}{44\times46}\)

\(=\frac{2\times4\times6\times8\times...\times42\times44}{4\times6\times8\times10\times...\times44\times46}\)

\(=\frac{2}{46}=\frac{1}{23}\)

\(8x-x^2-22=-\left(x^2-8x+22\right)=-\left(x^2-2.x.4+4^2+6\right)=-\left(x-4\right)^2-6< 0\)

với mọi \(x\).

Trả lời:

E = 1² - 2² + 3² - 4² + 5² - 6² + ... + 2011² - 2012² 

= ( 1² - 2² ) + ( 3² - 4² ) + ( 5² - 6² ) + ... + ( 2011² - 2012² )

= ( 1 - 2 )( 1 + 2 ) + ( 3 - 4 )( 3 + 4 ) + ( 5 - 6 )( 5 + 6 ) + ... + ( 2011 - 2012 )( 2011 + 2012 )

= ( - 1 ).3 + ( - 1).7 + ( - 1 ).11 + ... + ( - 1 ).4023 

= - 3 - 7 - 11 - ... - 4023 

= - ( 3 + 7 + 11 + ... + 4023 )

= - [ ( 3 + 4023 ).1006 : 2 ]         (1006 là số số hạng)

= - 2025078

Tính nhanh:E=1^2-2^2+3^2-4^2+5^2-6^2+...+2011^2-2012^2

= ( 1^2 - 2^2 ) + ( 3^2 - 4^2 ) + ......... ( 2011^2 - 2012^2 

= -3 + -7 + ..... + 4023

= 2025078

A = (x² + x)² + 2(x² + x) + 1 

= (x² + x + 1)²

B = (x - a)⁴ - (x + a)⁴ 

= [(x - a)²]² - [(x + a)²]² 

= [(x - a)² - (x + a)²].[(x - a)² + (x + a)²]

= (x² - 2ax + a² - x² - 2ax - a²)(x² - 2ax + a² + x² + 2ax + a²) 

= -4ax(2x² + 2a²) 

= -8ax(x² + a²) 

C = (x² - 10x + 25) - 4y² 

= (x - 5)² - (2y)² 

 = (x - 5 + 2y)(x - 5 - 2y)

D = x⁶ + 27 = (x²)³ + 3³ 

= (x² + 3)[x⁴ - 3x² + 9) 

\(A=\left(x^2+x\right)^2+2\left(x^2+x\right)+1=\left(x^2+x+1\right)^2\)

\(B=\left(x-a\right)^4-\left(x+a\right)^4=\left[\left(x-a\right)^2\right]^2-\left[\left(x+a\right)^2\right]^2\)

\(=\left[\left(x-a\right)^2-\left(x+a\right)^2\right]\left[\left(x-a\right)^2+\left(x+a\right)^2\right]\)

\(=\left(x-a-x-a\right)\left(x-a+x+a\right)\left(x^2-2xa+a^2+x^2+2ax+a^2\right)\)

\(=-2a.2x\left(2x^2+2a^2\right)=-8ax\left(x^2+a^2\right)\)

3. A = x³ - 64 - ( x³ - x² + x - 1 ) = x³ - 64 - x³ + x² - x + 1 = x² - x - 63

B = x³ + 8 - ( x³ - 8 ) = x³ + 8 - x³ + 8 = 16

C = x³ - 3x² + 3x - 1 - ( 4x² - 1 ) = x³ - 3x² + 3x - 1 - 4x² + 1 = x³ - 7x² + 3x

D = x( x² - 25 ) - ( x³ + 1 ) = x³ - 25x - x³ - 1 = -25x - 1

4. a) x² - 4x + 1 = 0 <=> ( x² - 4x + 4 ) - 3 = 0 <=> ( x - 2 )² - (√3)² = 0

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

b) 9x² - 6x - 8 = 0 <=> ( 9x² - 6x + 1 ) - 9 = 0 <=> ( 3x - 1 )² - 3² = 0

<=> ( 3x - 4 )( 3x + 2 ) = 0 <=> x = 4/3 hoặc x = -2/3

c) x³ - 3x² + 3x + 7 = 0 <=> ( x³ - 3x² + 3x - 1 ) + 8 = 0

<=> ( x - 1 )³ + 2³ = 0 <=> ( x + 1 )( x² - 4x + 7 ) = 0

<=> x + 1 = 0 <=> x = -1 ( vì x² - 4x + 7 = ( x - 2 )² + 3 > 0 )