Bài 1.

a) 2x² + 3( x - 1 )( x + 1 ) - 5x( x + 1 )

= 2x² + 3( x² - 1 ) - 5x² - 5x

= 2x² + 3x² - 3 - 5x² - 5x

= -5x - 3 

b) 4( x - 1 )( x + 5 ) - ( x - 2 )( x + 5 ) - 3( x - 1 )( x + 2 )

= 4( x² + 4x - 5 ) - ( x² + 3x - 10 ) - 3( x² + x - 2 )

= 4x² + 16x - 20 - x² - 3x + 10 - 3x² - 3x + 6

= 10x - 4

Bài 2.

a) ( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) = 0

<=> -5x² - 2x + 16 + 4( x² - x - 2 ) + 2( x² - 4 ) = 0

<=> -5x² - 2x + 16 + 4x² - 4x - 8 + 2x² - 8 = 0

<=> x² - 6x = 0

<=> x( x - 6 ) = 0

<=> x = 0 hoặc x = 6

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

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 0

<=> x² + 5x + 6 - x² - 3x + 10 = 0

<=> 2x + 16 = 0

<=> 2x = -16

<=> x = -8

Bài 3.

A = ( n² + 3n - 1 )( n + 2 ) - n³ + 2

= n³ + 2n² + 3n² + 6n - n - 2 - n³ + 2

= 5n² + 5n

= 5n( n + 1 ) chia hết cho 5 ( đpcm )

B = ( 6n + 1 )( n + 5 ) - ( 3n + 5 )( 2n - 1 )

= 6n² + 30n + n + 5 - ( 6n² - 3n + 10n - 5 )

= 6n² + 31n + 5 - 6n² - 7n + 5

= 24n + 10

= 2( 12n + 5 ) chia hết cho 2 ( đpcm )

bài 1:a,\(2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)

\(=2x^2+3x^2-3-5x^2-5x\)

\(=-3-5x\)

b.\(4\left(x-1\right)\left(x+5\right)-\left(x-2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)

\(=4\left(x^2+4x-5\right)-\left(x^2+3x-10\right)-3\left(x^2+x-2\right)\)

\(=4x^2+16x-20-x^2-3x+10-3x^2-3x+6\)

\(=10x-4\)

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2+2x-2x-4\right)=0\)

\(-2x+16-5x^2+4x^2-4x-8+2x^2-8=0\)

\(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

a) ( x - 3)⁴ + ( x - 5)⁴ = 82

Đặt : x - 4 = a , ta có :

( a + 1)⁴ + ( a - 1)⁴ = 82

⇔ a⁴ + 4a³ + 6a² + 4a + 1 + a⁴ - 4a³ + 6a² - 4a + 1 = 82

⇔ 2a⁴ + 12a² - 80 = 0

⇔ 2( a⁴ + 6a² - 40) = 0

⇔ a⁴ - 4a² + 10a² - 40 = 0

⇔ a²( a² - 4) + 10( a² - 4) = 0

⇔ ( a² - 4)( a² + 10) = 0

Do : a² + 10 > 0

⇒ a² - 4 = 0

⇔ a = + - 2

+) Với : a = 2 , ta có :

x - 4 = 2

⇔ x = 6

+) Với : a = -2 , ta có :

x - 4 = -2

⇔ x = 2

KL.....

b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8

⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680

⇔ ( n² - 9n + 18)( n² - 9n + 20) = 1680

Đặt : n² - 9n + 19 = t , ta có :

( t - 1)( t + 1) = 1680

⇔ t² - 1 = 1680

⇔ t² - 41² = 0

⇔ ( t - 41)( t + 41) = 0

⇔ t = 41 hoặc t = - 41

+) Với : t = 41 , ta có :

n² - 9n + 19 = 41

⇔ n² - 9n - 22 = 0

⇔ n² + 2n - 11n - 22 = 0

⇔ n( n + 2) - 11( n + 2) = 0

⇔ ( n + 2)( n - 11) = 0

⇔ n = - 2 hoặc n = 11

+) Với : t = -41 ( giải tương tự )

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