Tìm x biết 5x-(-2x-0,3)=2,4
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Bài 2:
5x-(-2x-0,3)=2,4
=>5x+2x+0,3=2,4
=>7x=2,1
hay x=0,3
a: \(2,5:4x=0,5:0,2\)
=>\(2,5:4x=0,5\cdot5=2,5\)
=>4x=1
=>\(x=\dfrac{1}{4}\)
b: \(3,8:2x=\dfrac{1}{4}:2\dfrac{2}{3}\)
=>\(3,8:2x=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{1}{4}\cdot\dfrac{3}{8}=\dfrac{3}{32}\)
=>\(2x=3,8:\dfrac{3}{32}=\dfrac{19}{5}\cdot\dfrac{32}{3}=\dfrac{608}{15}\)
=>\(x=\dfrac{608}{15}:2=\dfrac{304}{15}\)
c: \(5,25:7x=3,6:2,4\)
=>\(5,25:7x=1,5\)
=>\(7x=5,25:1,5=3,5\)
=>\(x=\dfrac{3.5}{7}=0,5\)
d: \(1,8:1,3=-2,7:5x\)
=>\(5x=-2,7:\dfrac{18}{13}=-2,7\cdot\dfrac{13}{18}=-1,95\)
=>\(x=-1,95:5=-0,39\)
a)\(60x^2+35x-60x^2+15x=100\)
35x+15x=100
50x=100 =>x=2
b)\(10x^2-35x+16x-10x^2=5\)
-35x+16x=5
-19x=5 =>x=-5/19
b) \(\left(x+3\right)^2-5x-15=0\\ \Leftrightarrow\left(x+3\right)^2-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)
c) \(2x^5-4x^3+2x=0\\ \Leftrightarrow2x\left(x^4-2x^2+1\right)=0\\ \Leftrightarrow2x\left(x^2-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\\left(x^2-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của pt là : \(S=\left\{0;1;-1\right\}\)
b) \(x\left(2x+5\right)=0\)
TH1: \(x=0\)
TH2: \(2x+5=0\Leftrightarrow2x=-5\Leftrightarrow x=\dfrac{-5}{2}\)
\(a,\Rightarrow5x\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x-3\right)\left(5x-x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(4x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{4}\end{matrix}\right.\\ b,\Rightarrow x\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(\begin{array}{l}a){\rm{ }}3{x^2}-{\rm{ }}3x\left( {x{\rm{ }}-{\rm{ }}2} \right){\rm{ }} = {\rm{ }}36\\ \Leftrightarrow 3{x^2}-{\rm{ [}}3x.x + 3x.( - 2)] = 36\\ \Leftrightarrow 3{x^2} - (3{x^2} - 6x) = 36\\ \Leftrightarrow 3{x^2} - 3{x^2} + 6x = 36\\ \Leftrightarrow 6x = 36\\ \Leftrightarrow x = 36:6\\ \Leftrightarrow x = 6\end{array}\)
Vậy x = 6
\(\begin{array}{l}b){\rm{ }}5x\left( {4{x^2}-{\rm{ }}2x{\rm{ }} + {\rm{ }}1} \right){\rm{ }}-{\rm{ }}2x\left( {10{x^2}-{\rm{ }}5x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }} - 36\\ \Leftrightarrow 5x.4{x^2} + 5x.( - 2x) + 5x.1 - [2x.10{x^2} + 2x.( - 5x) + 2x.2] = - 36\\ \Leftrightarrow 20{x^3} - 10{x^2} + 5x - (20{x^3} - 10{x^2} + 4x) = - 36\\ \Leftrightarrow 20{x^3} - 10{x^2} + 5x - 20{x^3} + 10{x^2} - 4x = - 36\\ \Leftrightarrow (20{x^3} - 20{x^3}) + ( - 10{x^2} + 10{x^2}) + (5x - 4x) = - 36\\ \Leftrightarrow x = - 36\end{array}\)
Vậy x = -36
\(5x-\left(-2x-0,3\right)=2,4\)
\(\Leftrightarrow5x+2x+0,3=2,4\Leftrightarrow7x=2,4-0,3=2,1\Leftrightarrow x=0,3\)