4x3 + 15 = 42
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=> 42 : x + 0 = 7
=> 42 : x = 7
=> x = 42 : 7
=> x = 6
Vậy x = 6
Tk mk nha
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\(a,x^4-4x^3-19x^2+106x-120=0\\ \Rightarrow\left(x-4\right)\left(x^3-19x+30\right)=0\Rightarrow\left(x-4\right)\left(x+5\right)\left(x^2-5x+6\right)=0\\ \Rightarrow\left(x-4\right)\left(x+5\right)\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-5\\x=2\\x=3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-5;2;3;4\right\}\)
\(b,4x^4+12x^3+5x^2-6x-15=0\\ \Rightarrow\left(x-1\right)\left(4x^3+16x^2+21x+15\right)=0\\ \Rightarrow\left(x-1\right)\left[\left(4x^3+10x^2\right)+\left(6x^2+15x\right)+\left(6x+15\right)\right]=0\\ \Rightarrow\left(x-1\right)\left[2x^2\left(2x+5\right)+3x\left(2x+5\right)+3\left(2x+5\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(2x+5\right)\left(2x^2+3x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\\2x^2+3x+3=0\left(vô.lí\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1;-\dfrac{5}{2}\right\}\)
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a: \(P\left(x\right)=-5x^3+3x^2+2x+5\)
\(Q\left(x\right)=-5x^3+6x^2+2x+5\)
b: Q(x)-P(x)=6
\(\Leftrightarrow-5x^3+6x^2+2x+5+5x^3-3x^2-2x-5=6\)
=>3x2=6
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
\(x=\sqrt[3]{\frac{27}{4}}\)