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a, \(x\left(x^2+x+1\right)-x^2\left(x+1\right)=2x+5\)
\(\Rightarrow x^3+x^2+x-x^3-x^2-2x=5\)
\(\Rightarrow-x=5\Rightarrow x=-5\)
b, \(\left(x-3\right)\left(x-2\right)-\left(x+1\right)\left(x-5\right)=0\)
\(\Rightarrow x^2-2x-3x+6-\left(x^2-5x+x-5\right)=0\)
\(\Rightarrow x^2-5x+6-x^2+4x+5=0\)
\(\Rightarrow-x=-5-6\Rightarrow x=11\)
c, \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Rightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)
\(\Rightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\)
\(\Rightarrow-6x=3,5+4,5\Rightarrow-6x=8\Rightarrow x=-\dfrac{4}{3}\)
Chúc bạn học tốt!!!
Bạn ơi ở câu a bạn làm sai rùi
\(\left(-x^2\right).\left(-1\right)=+x^2\)chứ sao lại \(-x^2\)
a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )
2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)
\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)
\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )
Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))
1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
=> \(-4x^2+28x+4x^3-20x=28x^2-13\)
=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)
=> \(-4x^2+4x^3+8x-28x^2+13=0\)
=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)
=> \(-32x^2+4x^3+8x+13=0\)
=> vô nghiệm
2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)
=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)
=> \(-14x^2-56x+12=0\)
=> .... tự tìm
Câu c dấu bằng chỗ nào ?
Lời giải:
a)
\((x-2)(x-3)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)(x-2-1)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)^2-(x-2)+2x=(x-2)^2-2\)
\(\Leftrightarrow x+4=0\Rightarrow x=-4\)
b)
\((x-1)^2+3x(x-1)+7=(2x-1)^2+5(x-3)\)
\(\Leftrightarrow (x-1)^2+3x(x-1)+7=x^2+(x-1)^2+2x(x-1)+5(x-3)\)
\(\Leftrightarrow x(x-1)+7=x^2+5(x-3)\)
\(\Leftrightarrow 6x=22\Rightarrow x=\frac{11}{3}\)
c)
\(5(x^2-2x-1)+2(3x-2)=5(x+1)^2=5(x^2-2x+1)\)
\(\Leftrightarrow -5+2(3x-2)=5\)
\(\Leftrightarrow 3x-2=5\Rightarrow x=\frac{7}{3}\)
d)
\((x-1)(x^2+x+1)-2x=x(x-1)(x+1)=x(x^2-1)\)
\(\Leftrightarrow x^3-1-2x=x^3-x\Leftrightarrow -1-x=0\Rightarrow x=-1\)
\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(4x+26=-12\)
\(4x=-38\)
\(x=\frac{-19}{2}\)
x(3x+2) + (x+1)2 - (2x-5)(2x+5)= -12
(3x2+2x) + (x2+2x+1) - (4x2 - 25) = -12
3x2 + 2x + x2 + 2x + 1 - 4x2 +25 = -12
(3x2 + x2 - 4x2) + ( 2x+2x) + (1+25) = -12
0 + 4x + 26 = -12
4x = -12 - 26
4x = -38
x = -9.5
<=>(2x2-x)(x+5)-(2x3+9x2+x+4,5)=-2,5
<=>(2x3+10x2-x2-5x)-2x3-9x2-x-4,5+2,5=0
<=>2x3+10x2-x2-5x-2x3-9x2-x-4,5+2,5=0
<=>-9x=2
<=>x=-2/9