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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 ?
a: \(=4x^2-25-4x^2+12x-9-12x=-34\)
b: \(=8y^3-12y^2+6y-1-2y\left(4y^2-12y+9\right)-12y^2+12y\)
\(=8y^3-24y^2+18y-1-8y^3+24y^2-18y=-1\)
c: \(=x^3+27-x^3-20=7\)
d: \(=3y\left(9y^2+12y+4\right)-27y^3+1-36y^2-12y-1\)
\(=27y^3+36y^2+12y-27y^3-36y^2-12y\)
=0
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
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: =>2x-5=4 hoặc 2x-5=-4
=>2x=9 hoặc 2x=1
=>x=9/2hoặc x=1/2
2: \(\Leftrightarrow\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}=\dfrac{-1}{8}\)(vô lý)
3: \(\Leftrightarrow\left|5x-3\right|=x+5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(5x-3-x-5\right)\left(5x-3+x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(4x-8\right)\left(6x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)
1. \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
2. \(\left(x-2\right)\left(6x+2\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right).7x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\7x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
3.
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4.
\(x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)