Question

9x^2-5-(4-3x).(-3x)=1

Answer 1

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Answer 2

\(9x^2-5-\left(4-3x\right)\left(-3x\right)=1\)

\(\Rightarrow9x^2-5-\left[4\left(-3x\right)-3x\left(-3x\right)\right]=1\)

\(\Rightarrow9x^2-5-\left[-12x-\left(-9x^2\right)\right]=1\)

\(\Rightarrow9x^2-5-\left[-12x+9x^2\right]=1\)

\(\Rightarrow9x^2-5+12x-9x^2=1\)

\(\Rightarrow12x-5=1\)

\(\Rightarrow12x=6\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

Answer 3

\(9x^2-5-\left(4-3x\right)\left(-3x\right)=1\)

\(\Leftrightarrow9x^2-5-\left(-12x+9x^2\right)=1\)

\(\Leftrightarrow9x^2-5+12x-9x^2=1\)

\(\Leftrightarrow-5+12x=1\)

\(\Leftrightarrow12x=6\)

\(\Rightarrow x=\frac{6}{12}=\frac{1}{2}\)

Answer 4

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Answer 5

\(9x^2-5-\left(4-3x\right).\left(-3x\right)=1\)

\(=9x^2-5+3x\left(4-3x\right)=1\)

\(=9x^2+3x\left(4-3x\right)=6\)

\(\Rightarrow x=\frac{1}{2}\)