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Làm này mới đúng chứ:
(3x-5)(x+1) - (x-1)(x+2)= (2x-3)(x+2) + 1
=> (3x-5)(x+1) = (2x-3)(x+2) + (x-1)(x+2) + 1
=> (3x - 4 - 1)(x + 1) = (x+2) [(2x-3) + (x-1)] + 1
=> (3x-4)(x+1) - x - 1 = (x+2).(3x-4) + 1
=> (3x-4)(x+1) - (x+2).(3x-4) = x + 1 + 1
=> (3x-4).[(x+1) - (x+2)] = x + 2
=> (3x-4).(-1) = x + 2
=> - 3x + 4 = x + 2
=> 3x + x = 4 + 2
=> 4x = 6
=> x = 6 : 4
=> x = 3/2
a) \(\left(x^2+2x+1\right)\left(x+1\right)\)
\(=x^3+x^2+2x^2+2x+x+1\)
\(=x^3+3x^2+3x+1\)
b) Ta có: \(\left(x^3-x^2+2x-1\right)\left(5-x\right)\)
\(=5x^3-x^4-5x^2+x^3+10x-2x^2-5+5x\)
\(=-x^4+6x^3-7x^2+15x-5\)
Ta có: \(\left(x-5\right)\left(x^3-x^2+2x-1\right)\)
\(=-\left(5-x\right)\left(x^3-x^2+2x-1\right)\)
\(=x^4-6x^3+7x^2-15x+5\)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
(x-1)(x\(^5\)+x\(^4\)+x\(^3\)+x\(^2\)+x+1)
=x(\(x^5+x^4+x^3+x^2+x+1\))-1(\(x^5+x^4+x^3+x^2+x+1\))
= x.\(x^5+x\cdot x^4+x\cdot x^3+x\cdot x^2+x\cdot x+x\cdot1\)-1.\(x^5-1\cdot x^4-1\cdot x^3-1\cdot x^2-1\cdot x-1\cdot1\)
=\(x^6\)+\(x^5\)\(+x^4\)+\(x^3\)+\(x^2\)+1x -1\(x^5\)-1\(x^4\)-1\(x^3\)-1\(x^2\)-1x -1
=\(x^6\)+(\(x^5\)-1\(x^5\))+(x\(^4\)-1\(x^4\))+(\(x^3\)-1\(x^3\))+(x\(^2\)-1\(x^2\))+(1x-1x)-1
=x\(^6\)-1
a) (x-1)*(x+2)-(x-3)*(-x+4)=19
\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)
\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)
\(\Leftrightarrow2x^2-3x+7-19=0\)
\(\Leftrightarrow2x^2-3x-12=0\)
Đề sai??
b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)
\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)
\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)
\(\Leftrightarrow-30x^2+7x-4+17=0\)
\(\Leftrightarrow-30x^2+7x+13=0\)
???
a. 2x^2 ( 5x^3 - 4x^2 - 7xy + 1 )
= 10x^5 - 8x^4 - 14x^3y + 1
b. ( x - 5 ) ( x + 3 )
= x^2 + 3x - 5x - 15
= x^2 - 2x - 15
c. ( x - 1 ) ( x + 2 )
= x^2 + 2x - x - 2
= x^2 - x - 2
d. ( 2x + y ) ( 2x - y )
= 4x^2 - 2xy + 2xy - y^2
= 4x^2 - y^2
a) \(2x^2\left(5x^3-4x^2.g-7xy+1\right)\)
\(=10x^5-8x^4.g-14x^3y+2x^2\)
b) \(\left(x-5\right)\left(x+3\right)\)
\(=x^2+3x-5x-15\)
\(=x^2-2x-15\)
c) \(\left(x-1\right)\left(x+2\right)\)
\(=x^2+2x-x-2\)
\(=x^2+x-2\)
d) \(\left(2x+y\right)\left(2x-y\right)\)
\(=4x^2-y^2\)
(x3 - 2x2 + x - 1)(5 - x) = (x3 - 2x2 + x - 1).5 - [ (x3 - 2x2 + x - 1).x ] = 5x3 - 10x2 + 5x - 5 - (x4 - 2x3 + x2 - x)
= 5x3 - 10x2 + 5x - 5 - x4 + 2x3 - x2 + x = x4 + (5x3 + 2x3) + (-10x2 - x2) + (5x + x) - 5 = x4 + 7x3 - 11x2+ 6x - 5