tìm x:5x+2016=2006
2(x-4)-(x+5)=-13
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5x+2016 = 2006 => 5x = 2006 - 2016 = -10 => x = -10 : 5 = -2
2(x-4)-(x+5) = 2x-8-x-5 = (2x-x)-(8+5) = x-13 = -13 => x = -13+13 = 0
\(5x+2016=2006\)
\(\Rightarrow5x=2006-2016\)
\(\Rightarrow5x=-10\)
\(\Rightarrow x=-10:5\)
\(\Rightarrow x=-2\)
\(2\left(x-4\right)-\left(x+5\right)=-13\)
\(\Rightarrow2x-8-x-5=-13\)
\(\Rightarrow x=-13+8+5\)
\(\Rightarrow x=0\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+2016\)
\(=x\left(x^2+5x+4\right)\left(x^2+5x+6\right)+2016\)
\(=x\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)+2016\)
\(=x\left[\left(x^2+5x+5\right)-1\right]+2016\)
\(=x\left(x^2+5x+5\right)-x+2016\)
Đáp số : Dư \(-x+2016\)
\(5x-\dfrac{4}{5}=x+\dfrac{13}{15}\)
\(\Rightarrow5x-x=\dfrac{13}{15}+\dfrac{4}{5}\)
\(\Rightarrow4x=\dfrac{13}{15}+\dfrac{12}{15}\)
\(\Rightarrow4x=\dfrac{25}{15}\)
\(\Rightarrow4x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}:4\)
\(\Rightarrow x=\dfrac{5}{12}\)
5x-4/5=x+13/15
<=>5x-x=13/15+4/5
<=>4x=5/3
<=>x=5/12
vậy ...
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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
hình như câu dưới đề bài sai