2x^4+x^3-5x^2-3x-3 2x^2+x+1 x^2-3 2x^4 -6x^2 - x^3+x^2-3x-3 x^3 -3x - x^2-3 x^2-3 x^2-3 - 0

x^3-3x^2+x-3 x-3 x^2+1 x^3-3x^2 - - x+3 x+3 - 0

1: \(\Leftrightarrow-4x^2+3x-4x^2+8x=10\)

=>-8x^2+11x-10=0

=>\(x\in\varnothing\)

2: \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)

=>-14x+5=x-2

=>-15x=-7

=>x=7/15

3: \(\Leftrightarrow12x^2-12x^2+20x=10x-17\)

=>10x=-17

=>x=-17/10

4: \(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)

=>18x+3=7x-3

=>11x=-6

=>x=-6/11

5: \(\Leftrightarrow-3x+15+5x-5+3x^2=4-x\)

\(\Leftrightarrow3x^2+2x+10-4+x=0\)

=>3x^2+3x+6=0

hay \(x\in\varnothing\)

\(\left(9x-1\right)^2-2\left(9x-1\right)\left(5x-1\right)+\left(5x-1\right)^2=\left(9x-1-5x+1\right)^2=\left(14x\right)^2=196x^2\)

\(5^4.3^4-\left(15^4-1\right)=15^4-15^4+1=1\)

mình chỉ viết đáp án thôi nhé! còn nếu ý nào bạn cần lời giải chi tiết mình sẽ giải cho!

a) S= { -2/3;-3/2}

b) S= {-5;1}

c) S= {-1/2;1}

d) S= {3/7;4}

e) S= {3;5}

NHỚ BẤM ĐÚNG CHO MÌNH NHÉ!

cho mk lời giải chi tiết đi

OIMEOI

may lam dai the thi cho no giai ho a 

a, \(x^6-x^4+2x^3+2x^2=x^2\left(x^4-x^2+2x+2\right)=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)

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

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

b, \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

\(=2y\left(2x^2+2xy+x^2-2xy+y^2\right)\)\(=2y\left(3x^2+y^2\right)\)

c, \(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)

\(=\left(x-1-y\right)\left[x^2-2x+1+xy-y+y^2\right]\)

d, \(x^5+x^4+1=x^5+x^4+x^3-x^3+1\)

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

e, \(4x^4+81=\left(2x^2\right)^2+36x^2+9^2-36x^2=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2+9+6x\right)\left(2x^2+9-6x\right)\)

d, \(64x^4+y^4=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)

tự làm đi dễ vầy

a du

M = ( x + 4 )( x - 4 ) - 2x( 3 + x ) + ( x + 3 )²

= x² - 16 - 6x - 2x² + x² + 6x + 9

= -7 ( đpcm )

N = ( x² + 4 )( x + 2 )( x - 2 ) - ( x² + 3 )( x² - 3 )

= ( x² + 4 )( x² - 4 ) - ( x⁴ - 9 )

= x⁴ - 16 - x⁴ + 9

= -7 ( đpcm )

P = ( 3x - 2 )( 9x² + 6x + 4 ) - 3( 9x³ - 2 )

= 27x³ - 8 - 27x³ + 6

= -2 ( đpcm )

Q = ( 3x + 2 )² + ( 6x + 10 )( 2 - 3x ) + ( 2 - 3x )²

= 9x² + 12x + 4 + 12x - 18x² + 20 - 30x + 4 - 12x + 9x²

= -18x + 28 ( có phụ thuộc vào biến )

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)

\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)

\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)

\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)

\(\Leftrightarrow-25x=-13\)

\(\Leftrightarrow x=\dfrac{13}{25}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

gắp cái gì