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\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)\left(x-3\right)\left(x+3\right)=\left(x-3\right)^2\left(x+3\right)\)
\(\left(3x-2\right)\left(3x+2\right)-\left(4-3x\right)^2=9x^2-4-16+24x-9x^2=24x-20=4\left(6x-5\right)\)
\(\left(2x+5\right)^2-\left(1+2x\right)\left(2x-1\right)=-3\)
\(4x^2+20x+25-4x^2+1=-3\)
\(20x=-3-25-1\)
\(20x=-29\)
\(x=-\frac{29}{20}\)
\(x^2\left(x-2x^3\right)=x^3-2x^5\)
\(\left(x^2+1\right)\left(5-x\right)=5x^2-x^3+5-x\)
\(\left(x-2\right)\left(x^2+3x-4\right)=x^3+3x^2-4x-2x^2-6x+8=x^3+x^2-10x+8\)
\(\left(x+3\right)\left(x^2+3x-5\right)=x^3+3x^2-5x+3x^2+9x-15=x^3+6x^2+4x-15\)
\(\left(x^2-1\right)\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
Bài 1.
[ 4( x - y )5 + 2( x - y )3 - 3( x - y )2 ] : ( y - x )2 < sửa một lũy thừa rồi nhé >
= [ 4( x - y )5 + 2( x - y )3 - 3( x - y )3 ] : ( x - y )2
Đặt t = x - y
bthuc ⇔ ( 4t5 + 2t3 - 3t2 ) : t2
= 4t5 : t2 + 2t3 : t2 - 3t2 : t2
= 4t3 + 2t - 3
= 4( x - y )3 + 2( x - y ) - 3
Bài 2.
5x( x - 2 ) + 3x - 6 = 0
⇔ 5x( x - 2 ) + 3( x - 2 ) = 0
⇔ ( x - 2 )( 5x + 3 ) = 0
⇔ x - 2 = 0 hoặc 5x + 3 = 0
⇔ x = 2 hoăc x = -3/5
Bài 3.
A = x2 - 6x + 2023
= ( x2 - 6x + 9 ) + 2014
= ( x - 3 )2 + 2014 ≥ 2014 ∀ x
Dấu "=" xảy ra khi x = 3
=> MinA = 2014 <=> x = 3
Bài 4.
B = ( 3x + 5 )2 + ( 3x - 5 )2 - 2( 3x + 5 )( 3x - 5 )
= [ ( 3x + 5 ) - ( 3x - 5 ) ]2
= ( 3x + 5 - 3x + 5 )2
= 102 = 100
Vậy B không phụ thuộc vào x ( đpcm )
Bài 6.
C = 12 - 22 + 32 - 42 + 52 - 62 + ... + 20132 - 20142 + 20152
= ( 20152 - 20142 ) + ... + ( 52 - 42 ) + ( 32 - 22 ) + 1
= ( 2015 - 2014 )( 2015 + 2014 ) + ... + ( 5 - 4 )( 5 + 4 ) + ( 3 - 2 )( 3 + 2 ) + 1
= 4029 + ... + 9 + 5 + 1
= \(\frac{\left(4029+1\right)\left[\left(4029-1\right)\div4+1\right]}{2}\)
= 2 031 120
2)a)3x(x-5)-(x-1)(2+3x)=30
<=>3x2-15x-3x2+x+2=30
<=>-14x+2=30
<=>-14x=30-2
<=>-14x=28
<=>x=-2
b)(x+2)(x+3)-(x-2)(x+5)=0
<=>x2+5x+6-x2-3x+10=0
<=>2x+16=0
<=>2x=-16
<=>x=-8
c)(3x+2)(2x+9)-(x+2)(6x+1)=9
<=>6x2+31x+18-6x2-13x-2=9
<=>18x+16=9
<=>18x=9-16
<=>18x=-7
<=>x=-7/18
1; \(x^2\) + 3\(x^2\) + 3\(x\) = 4\(x^2\) + 3\(x\) (1)
Thay \(x=99\) vào (1) ta có:
4.992 + 3.99 = 4.9801 + 297 = 39204 + 297 = 39501
\(\left(3x-2\right)^2-2\left(x-1\right)^2\)
\(=9x^2-18x+4-2\left(x^2-2x+1\right)\)
\(=9x^2-18x+4-2x^2+4x-2\)
\(=7x^2-14x+2\)
\(\left(3x-2\right)^2-2\left(x-1\right)^2=9x^2-12x+4-2x^2+4x-2=7x^2-8x+2\)