a)

 Ta có

 \(\left(x-1\right)^3-\left(x-1\right)^3-\left(6x-1\right)=-10\)

\(\Leftrightarrow-6x+1=-10\)

\(\Leftrightarrow-6x=-11\)

\(\Leftrightarrow x=\frac{11}{6}\)

  Vậy \(x=\frac{11}{6}\)

a) ( x - 1 )³ - ( x - 1 )³ - ( 6x - 1 ) = -10

<=> -( 6x - 1 ) = -10

<=> -6x + 1 = -10

<=> -6x = -11

<=> x = 11/6

b) ( 2x - 1 )² + ( 2x - 1 )( 2x - 3 ) - ( 2x + 3 )² + ( 2x + 3 )( -3x ) - 24 = 4

<=> 4x² - 4x + 1 + 4x² - 8x + 3 - ( 4x² + 12x + 9 ) - 6x² - 9x - 24 = 4

<=> 4x² - 4x + 1 + 4x² - 8x + 3 - 4x² - 12x - 9 - 6x² - 9x - 24 = 4

<=> -2x² - 33x - 29 - 4 = 0

<=> -2x² - 33x - 33 = 0 ( muốn kết quả thì ib còn mình để là vô nghiệm vì nó có nghiệm vô tỉ )

=> Vô nghiệm 

\(63x^2-16x+1=0\)

\(\Leftrightarrow63x^2-9x-7x+1=0\)

\(\Leftrightarrow9x\left(7x-1\right)-\left(7x-1\right)=0\)

\(\Leftrightarrow\left(9x-1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}9x-1=0\\7x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=\frac{1}{7}\end{cases}}}\)

Bài làm:

Ta có: \(63x^2-16x+1=0\)

\(\Leftrightarrow\left(63x^2-9x\right)-\left(7x-1\right)=0\)

\(\Leftrightarrow9x\left(7x-1\right)-\left(7x-1\right)=0\)

\(\Leftrightarrow\left(7x-1\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x-1=0\\9x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=\frac{1}{7}\end{cases}}\)

Vì \(\left(x+1\right)^4\ge0\forall x\); \(\left(x-3\right)^4\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^4+\left(x-3\right)^4\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}\left(ktm\right)}\)

=> Pt vô nghiệm

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

Vì \(\left(x+1\right)^4\ge0\forall x\inℤ\)

     \(\left(x-3\right)^4\ge0\forall x\inℤ\)

 Nên \(\left(x+1\right)^4+\left(x-3\right)^4=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=3\end{cases}}}\)

Vậy .....

a) Ta có: \(\left(x+1\right)^4+\left(x-3\right)^4=0\)

Nhận thấy: \(\hept{\begin{cases}\left(x+1\right)^4\ge0\left(\forall x\right)\\\left(x-3\right)^4\ge0\left(\forall x\right)\end{cases}\Rightarrow}\left(x+1\right)^4+\left(x-3\right)^4\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\) (mâu thuẫn)

=> pt vô nghiệm

b) \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^3-8x^2\right)+\left(4x^2-8x\right)+\left(3x-6\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)

=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a,\(\left(x+1\right)^4+\left(x-3\right)^4=0\)

\(x^4-1+x^4-81=0\)

\(2x^4-82=0\)

\(2x^4=82\)

\(x^4=41\)

\(x=\sqrt[4]{41}\)

\(\Rightarrow\)vô nghiệm

help me, please

1. a . 3x² - 6x = 0

\(\Leftrightarrow3x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b. x³ - 13x = 0

\(\Leftrightarrow x\left(x^2-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)

c. 5x ( x - 2001 ) - x + 2001 = 0

<=> 5x ( x - 2001 ) - ( x - 2001 ) = 0

\(\Leftrightarrow\left(5x-1\right)\left(x-2001\right)=0\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2001=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2001\end{cases}}\)

a) x² - 4y² 

= x² - ( 2y )²

= ( x - 2y )( x + 2y )

b) x² + x - 12

= x² - 3x + 4x - 12

= x( x - 3 ) + 4( x - 3 )

= ( x - 3 )( x + 4 )

c) x² + 2xy + y² - 11

= ( x² + 2xy + y² ) - 11

= ( x + y )² - ( √11 )²

= ( x + y - √11 )( x + y + √11 )

d) x⁴ + 1 

= ( x⁴ + 2x² + 1 ) - 2x²

= ( x² + 1 )² - ( √2x )²

= ( x² - √2x + 1 )( x² + √2x + 1 )

a) \(x^2-4y^2\)

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

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

b) \(x^2+x-12\)

\(=x^2+4x-3x-12\)

\(=\left(x^2+4x\right)-\left(3x+12\right)\)

\(=x.\left(x+4\right)-3.\left(x+4\right)\)

\(=\left(x+4\right).\left(x-3\right)\)

c) \(x^2+2xy+y^2-11\)

\(=\left(x^2+2xy+y^2\right)-11\)

\(=\left(x+y\right)^2-11\)

\(=\left(x+y\right)^2-\left(\sqrt{11}\right)^2\)

\(=\left(x+y-\sqrt{11}\right).\left(x+y+\sqrt{11}\right)\)

a³ + b³ + c³ = \(\orbr{\begin{cases}\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\\\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\end{cases}}\)

Chứng minh từng cái một nhé :)

1/ ( a + b + c )( a² + b² + c² - ab - bc - ca ) + 3abc

= a( a² + b² + c² - ab - bc - ca ) + b( a² + b² + c² - ab - bc - ca ) + c( a² + b² + c² - ab - bc - ca )  + 3abc

= a³ + ab² + ac² - a²b - abc - a²c + a²b + b³ + c²b - ab² - b²c - abc + a²c + b²c + c³ - abc - c²b - c²a + 3abc

= a³ + b³ + c³ - 3abc + 3abc 

= a³ + b³ + c³ ( đpcm )

2/ ( a + b + c )³ - 3( a + b )( b + c )( c + a )

= [ ( a + b ) + c ]³ - ( 3a + 3b )( bc + ab + c² + ac )

= [ ( a + b )³ + 3( a + b )²c + 3( a + b )c² + c³ ] - ( 3abc + 3a²b + 3ac² + 3a²c + 3b²c + 3ab² + 3bc² + 3abc )

= [ a³ + 3a²b + 3ab² + b³ + 3a²c + 6abc + 3b²c + 3ac² + 3bc² + c³ ] - ( 3abc + 3a²b + 3ac² + 3a²c + 3b²c + 3ab² + 3bc² + 3abc )

= a³ + 3a²b + 3ab² + b³ + 3a²c + 6abc + 3b²c + 3ac² + 3bc² + c³ -  3abc - 3a²b - 3ac² - 3a²c - 3b²c - 3ab² - 3bc² - 3abc 

= a³ + b³ + c³ ( đpcm )

Cái HĐT này nó khá là khó khi phân tích từ VT , nên mình chỉ có thể khai triển từ VP thôi. Thông cam nhé =)