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\(a,\left(5x-2y\right)\left(x^2-xy+1\right)=5x^3-5x^2y+5x-2x^2y-2xy^2-2y=5x^3-7x^2y-2xy^2+5x-2y\)\(b\left(x-1\right)\left(x+1\right)\left(x-2\right)=\left(x^2-1\right)\left(x+2\right)=x^3+2x^2-x-2\)\(c,\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
\(\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{1}{2}x+3\)\(b,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(c,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)
(8 - 5x).(x + 2) + 4.(x - 2)(x - 1) + 2.(x - 2)(x + 2) + 10
= (8x + 16 - 5x2 - 10x) + 4.(x2 - 3x + 2) + 2.(x2 - 4) + 10
= 8x + 16 - 5x2 - 10x + 4x2 - 12x + 8 + 2x2 - 8 + 10
= (8x - 10x - 12x) + (-5x2 + 4x2 + 2x2) + (16 + 8 - 8 + 10)
= -14x + x2 + 26
câu b ko ghi lại đề bài
\(\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow1.\left[\left(x-1\right)+\left(x+2\right)\right]\)
\(=1.\left(2.x+1\right)\)
\(=2x+1\)
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
\(ĐKXĐ:x\ne3;x\ne-1\)
Nếu x=0 là nghiệm của phương trình
Nếu x khác 0 ta có:
\(\frac{1}{2\left(x-3\right)}+\frac{1}{2\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x-1+x-3}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{2x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow2x-4=4\)
\(\Leftrightarrow x=4\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne-1;x\ne3\right)\)
<=> \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x2-6x=0
<=> 2x(x-3)=0
<=> \(\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
ĐCĐK x khác -1 và x khác 3 => x=0
Vậy x=0 là nghiệm của phương trình
(x2-1)(x2+x+1)(x2-x+1)
=(x2-1)[(x2+1)2-x2]
=(x2-1)[x4+2x2+1-x2]
=(x2-1)(x4+x2+1)
=(x2)3-1
=x6-1
( x2 - 1 )( x2 + x + 1 )( x2 - x + 1 )
= ( x2 - 1 )[ ( x2 + 1 ) + x ][ ( x2 + 1 ) - x ]
= ( x2 - 1 )[ ( x2 + 1 )2 - x2 ]
= ( x2 - 1 )( x4 + 2x2 + 1 - x2 )
= ( x2 - 1 )( x4 + x2 + 1 )
= x6 + x4 + x2 - x4 - x2 - 1
= x6 - 1