x(3x^2+1)=4x-x^2
tìm x
giup minh nha
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c: ta có: \(7x^2-2x-5=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{7}\end{matrix}\right.\)
\(\left(3x+1\right)^2=9\left(x-2\right)^2\)
\(\Leftrightarrow9x^2+6x+1=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2+6x+1=9x^2-36x+36\)
\(\Leftrightarrow9x^2+6x+1-9x^2+36x-36=0\)
\(\Leftrightarrow42x-35=0\)
\(\Leftrightarrow42x=35\)
\(\Leftrightarrow x=\dfrac{35}{42}=\dfrac{5}{6}\)
Vậy: \(S=\left\{\dfrac{5}{6}\right\}\)
\(x\inƯC\left(120;150\right)\)
mà x>12
nên \(x\in\left\{15;30\right\}\)
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
Để A là số nguyên thì \(2x-1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{1;0;3;-2\right\}\)
Cho \(M\left(x\right)=0\)
hay \(x^2-3x+2=0\)
⇒ \(x^2-2x-x+2=0\)
\(x.x-2x-x+2=0\)
\(x.\left(x-2\right)-\left(x+2\right)=0\)
⇒ \(\left(x-1\right).\left(x-2\right)=0\)
⇒ \(x-1=0\) hoặc \(x-2=0\)
* \(x-1=0\) * \(x-2=0\)
\(x\) \(=0+1\) \(x\) \(=0+2\)
\(x\) \(=1\) \(x\) \(=2\)
Vậy \(x=1\) hoặc \(x=2\) là nghiệm của \(M\left(x\right)\)
\(a,2x\left(x^3-3\right)-2x^4=18\\ 2x^4-6x-2x^4=18\\ -6x=18\\ x=-3\)
\(b,9x\left(4-x\right)+\left(3x+1\right)^2=2\\ 36x-9x^2+9x^2+6x+1=2\\ 42x=2-1\\ 42x=1\\ x=\dfrac{1}{42}\)
\(a,\Leftrightarrow2x^4-3x-2x^4=18\Leftrightarrow-3x=18\Leftrightarrow x=-6\\ b,\Leftrightarrow36x-9x^2+9x^2+6x+1=2\\ \Leftrightarrow42x=1\Leftrightarrow x=\dfrac{1}{42}\)
\(x^3+3x^2+x+a=x^2\left(x-2\right)+5x\left(x-2\right)+11\left(x-2\right)+22+a=\left(x-2\right)\left(x^2+5x+11\right)+22+a⋮\left(x-2\right)\)
\(\Rightarrow22+a=0\Rightarrow a=-22\)
\(x\left(3x^2+1\right)=4x-x^2\)
\(3x^3+x=4x-x^2\)
\(3x^3+x^2=4x-x\)
\(x^2.\left(3x+1\right)=3x\)
\(x.\left(3x+1\right)=3\)
\(\Rightarrow x;3x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
còn lại tự làm nôt
la sao