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\(x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=x.5x-x.3-x^2.x+x^2.1+x.x^2-x.6x-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2+10+3x\)
\(=-10\)
Biểu thức trên kết quả là -10 => ĐPCM
\(x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
=\(5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)
=\(\left(x^3-x^3\right)+\left(5x^2+x^2-6x^2\right)+\left(-3x+3x\right)-10\)
=-10
=> ĐPCM
A/ x(5x-3)-x^2(x-1)+x(x^2-6x)-10+3x
=> A=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x
=> A=(x^3-x^3)+(5x^2+x^2-6x^2)+(3x-3x)-10
=> A= 0 + 0 + 0 -10
=> A=-10
Vậy giá trị ko phụ thuộc vào biến.
B/x(x^2+x+1)-x^2(x+1)-x+5
=> B=x^3+x^2+x-x^3-x^2-x+5
=> B= 0 +5
=> B= 5.
UNDERSTAND !!!
( x - 1 )3 - ( x - 1 )( x2 + x + 1 ) - 3( 1 - x )x < đã sửa đề >
= x3 - 3x2 + 3x - 1 - ( x3 - 1 ) + 3x2 - 3x
= x3 - 1 - x3 + 1
= 0 ( đpcm )
\(=\left[\left(x^2-3x+5\right)-\left(x^2-3x-1\right)\right]^2\)
\(=\left(x^2-3x+5-x^2+3x+1\right)^2\)
\(=6^2\)
\(=36\)
\(\left(x^2-3x+5\right)-2\left(x^2-3x+5\right)\left(x^2-3x-1\right)+\left(x^2-3x-1\right)^2\)
\(=\left[\left(x^2-3x+5\right)-\left(x^2-3x-1\right)\right]^2\)
\(=\left(x^2-3x+5-x^2+3x+1\right)^2\)
\(=6^2=36\)ko phụ thuộc vào biến (đpcm)
Giải:
\(\left(x-3\right)\left(x+2\right)+\left(x-1\right)\left(x+1\right)-\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)-x^2\)
\(=x^2-x-6+x^2-1^2-\left(x-\dfrac{1}{2}\right)^2-x^2\)
\(=x^2-x-6+x^2-1-\left(x^2-x+\dfrac{1}{4}\right)-x^2\)
\(=x^2-x-6+x^2-1-x^2+x-\dfrac{1}{4}-x^2\)
\(=-6-1-\dfrac{1}{4}\)
\(=-\dfrac{29}{4}\)
Vậy ...
\(\dfrac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}=\dfrac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)
\(=\dfrac{29x^2+29}{x^2+1}=\dfrac{29\left(x^2+1\right)}{x^2+1}=29\)
Vậy.....
Ta có: \(\dfrac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}\)
\(=\dfrac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)
\(=\dfrac{29x^2+29}{x^2+1}=29\)
\(\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)
\(=\left[x.\left(2x^2-3x+4\right)+2.\left(2x^2-3x+4\right)\right]-\left[x.\left(2x+1\right)-1.\left(2x+1\right)\right]\)
\(=\left(2x^3-3x^2+4x+4x^2-6x+8\right)-\left(2x^3+x-2x-1\right)\)
\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x+2x+1\)
\(=9\)
\(x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)
\(=x.x^2+x.x+x.1-x^2.x-x^2.1-x+5\)
\(=x^3+x^2+x-x^3-x^2-x+5\)
\(=5\)
Biểu thức kết quả là 5 => ĐPCM
\(x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5=x^3+x^2+x-x^3-x^2-x+5=5\)