Tìm x thỏa mãn giá trị cho trước
a) (x-2)^3-(x-3).(x^2+3x+9)+6+(x+1)^2=15
b) 6x^2-6x.(-2+x)=36
c) (x+2)^2+(x-3)^2-2.(x-1).(x+1)=9
d) (x+5)^2-9=0
e) (x-2)^3=x^3+6x^2=7
Tìm x thỏa mãn giá trị cho trước
a) (x-2)^3-(x-3).(x^2+3x+9)+6+(x+1)^2=15
b) 6x^2-6x.(-2+x)=36
c) (x+2)^2+(x-3)^2-2.(x-1).(x+1)=9
d) (x+5)^2-9=0
e) (x-2)^3=x^3+6x^2=7
Đáp án:
$\begin{array}{l}
a){\left( {x – 2} \right)^3} – \left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right)\\
+ 6 + {\left( {x + 1} \right)^2} = 15\\
\Rightarrow {x^3} – 3.{x^2}.2 + 3.x.4 – 8 – \left( {{x^3} – {3^3}} \right)\\
+ 6 + {x^2} + 2x + 1 = 15\\
\Rightarrow {x^3} – 6{x^2} + 12x – 8 – {x^3}\\
+ 27 + {x^2} + 2x – 8 = 0\\
\Rightarrow – 5{x^2} + 14x + 9 = 0\\
\Rightarrow 5{x^2} – 14x – 9 = 0\\
\Rightarrow x = \frac{{12 \pm \sqrt {94} }}{5}\\
b)6{x^2} – 6x\left( { – 2 + x} \right) = 36\\
\Rightarrow 6{x^2} + 12x – 6{x^2} = 36\\
\Rightarrow 12x = 36\\
\Rightarrow x = 3\\
c){\left( {x + 2} \right)^2} + {\left( {x – 3} \right)^2} – 2\left( {x – 1} \right)\left( {x + 1} \right) = 9\\
\Rightarrow {x^2} + 4x + 4 + {x^2} – 6x + 9 – 2\left( {{x^2} – 1} \right) = 9\\
\Rightarrow 2{x^2} – 2x + 13 – 2{x^2} + 2 – 9 = 0\\
\Rightarrow – 2x + 6 = 0\\
\Rightarrow x = 3\\
d){\left( {x + 5} \right)^2} – 9 = 0\\
\Rightarrow {\left( {x + 5} \right)^2} = 9\\
\Rightarrow \left[ \begin{array}{l}
x + 5 = 3\\
x + 5 = – 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – 2\\
x = – 8
\end{array} \right.\\
e){\left( {x – 2} \right)^3} = {x^3} + 6{x^2} + 7\\
\Rightarrow {x^3} – 6{x^2} + 12x – 8 = {x^3} + 6{x^2} + 7\\
\Rightarrow 12{x^2} – 12x + 15 = 0\\
\Rightarrow 4{x^2} – 4x + 5 = 0\\
\Rightarrow \left( {4{x^2} – 4x + 1} \right) + 4 = 0\\
\Rightarrow {\left( {2x – 1} \right)^2} + 4 = 0\left( {\text{vô}\,\text{nghiệm}} \right)
\end{array}$
Vậy pt vô nghiệm.