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\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=0\\ \Rightarrow\left(x^3-27\right)+x\left(4-x^2\right)=0\\ \Rightarrow x^3-27+4x-x^3=0\\ \Rightarrow4x-27=0\\ \Rightarrow4x=27\\ \Rightarrow x=\dfrac{27}{4}\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Rightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\\ \Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Rightarrow12x+6=0\\ \Rightarrow12x=-6\\ \Rightarrow x=-\dfrac{1}{2}\)
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a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
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\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
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\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
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\(\text{a , (x-3).(x^2+3x+9)+x(x+2).(2-x)=1 }\)
=(x3-33)+x(4-x2)=1
=x3-27+4x-x3=1
4x-27=1
4x=28
x=7
\(\text{b, (x+1)^3-(x-1)^3-6.(x-1)^2=-10}\)
=-0,5
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a)\(2x\left(x+1\right)-3-2x=5\)
\(\Leftrightarrow2x^2+2x-3-2x=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)
\(\Rightarrow x=2;-2\)
b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)
\(\Leftrightarrow6x^2+2x+4-2x=7\)
\(\Leftrightarrow6x^2+4=7\)
\(\Leftrightarrow6x^2=3\)
\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow3x^2+15x=0\)
\(\Leftrightarrow3x\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
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x^3 -9x^2 +27x -27 -(x^3 -27) +6(x^2 +2x+1) +3x^2 =-33
x^3 -9x^2 +27x -27 -x^3 +27 +6x^2 + 12x+ 6 +3x^2 =-33
39x+6=-33
39x=-39
x=-1
Vậy x=-1
a) \(=>x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=0\)
\(24x+25=0=>x=-\frac{25}{24}\)
Mình nghĩ đề thiếu =0
Hok tốt !