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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
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 \(x\in\varnothing\)
\(\left(9x-1\right)^2-2\left(9x-1\right)\left(5x-1\right)+\left(5x-1\right)^2=\left(9x-1-5x+1\right)^2=\left(14x\right)^2=196x^2\)
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!
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, \(64x^4+y^4=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)
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 \(S=\left\{\dfrac{13}{25}\right\}\)
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 )
\(\left(3x-4\right)\left(3x+4\right)+\left(x-5\right)\left(x-3\right)=10x^2+3\)
\(\Leftrightarrow9x^2-16+x^2-8x+15-10x^2-3=0\)
\(\Leftrightarrow-8x=4\)
hay \(x=-\dfrac{1}{2}\)