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\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
a) => x - 8 = 0 hoặc x3 + 8 = 0
+) x - 8 = 0 => x = 8
+) x3 + 8 = 0 => x3 = - 8 = (-2)3 => x = -2
Vậy x = 8; -2
b) => 4x - 3 - x - 5 = 30 - 3x
=> 3x - 8 = 30 - 3x
=> 3x + 3x = 30 + 8
=> 6x = 38 => x = 38/6 = 19/3
Vậy x = 19/3
a) => x - 8 = 0 hoặc x3 + 8 = 0
+) x - 8 = 0 => x = 8
+) x3 + 8 = 0 => x3 = - 8 = (-2)3 => x = -2
Vậy x = 8; -2
b) => 4x - 3 - x - 5 = 30 - 3x
=> 3x - 8 = 30 - 3x
=> 3x + 3x = 30 + 8
=> 6x = 38 => x = 38/6 = 19/3
Vậy x = 19/3
Do \(\left|10-x\right|,\left|8-x\right|\ge0\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}10-x=0\\8-x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=10\\x=8\end{matrix}\right.\)(Vô lý)
Vậy \(S=\varnothing\)
a)(x-8)(x3+8)=0
<=>x-8=0 hoặc x3+8=0
<=>x=8 hoặc x3=-8
<=>x=8 hoặc x=-2
b)(4x-3)-(x+5)=3(10-x)
<=>4x-3-x-5=30-3x
<=>(4x-x)+(-3-5)=30-3x
<=>3x-8=30-3x
<=>6x=38
<=>x=\(\frac{38}{6}=\frac{19}{3}\)
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
\(\frac{x+5}{7}-\frac{x+18}{8}+\frac{x+8}{9}=0\)
\(72\left(x+5\right)-63\left(x+18\right)+56\left(x+8\right)=0\)
\(72x+360-63x-1134+56x+448=0\)
\(65x-326=0\)
\(65x=326\)
\(x=\frac{326}{65}\)
a) \(\left(x-8\right)\left(x^2+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\x^2+8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\\x^2=-8\left(loai\right)\end{cases}}\)
Vậy x=8
b) \(\left(4x-3\right)-\left(x+5\right)=3.\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow4x-x+3x=30+5+3\)
\(\Leftrightarrow6x=38\)
\(\Leftrightarrow x=\frac{19}{3}\)
Vaayjh\ ...
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\x^3+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}}\)
Vậy ....
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\x^3+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x^3=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=-2\end{cases}}}\)