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Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
x2 ( 2x3 – 4x + 3)
a: \(=15x^5-25x^4+15x^3\)
b: \(=2x^3+10x^2-8x-x^2-5x+4\)
\(=2x^3+9x^2-13x+4\)
\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)
\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
Bài 1:
a) \(3x\left(5x^2-2x+1\right)\)
\(=15x^3-6x^2+3x\)
b) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^2\left(x^2-1\right)+2x\left(x^2-1\right)\)
\(=x^4-x^2+2x^3-2x\)
\(=x^4+2x^3-x^2-2x\)
Bài 2:
a) \(3x^2=2x\)
\(\Leftrightarrow3x^2-2x=0\)
\(\Leftrightarrow x\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)