\(A=x+\left\{\left(x+5\right)-\left[\left(5-x\right)-\left(-x-3\right)\right]\right\}\)
\(=x+\left\{\left(x+5\right)-\left[5-x+x+3\right]\right\}\)
\(=x+\left\{\left(x+5\right)-\left(5+3\right)\right\}\)
\(=x+\left\{\left(x+5\right)-8\right\}\)
\(=x+\left\{x+5-8\right\}=x+\left\{x-3\right\}\)
\(=x+x-3=2x-3\)
\(B=x.\left\{\left[-x-2-\left[x+\left(3-x\right)-\left(x+3\right)\right]\right]\right\}\)
\(=x.\left\{\left[-x-2-\left[x+3x-x-x-3\right]\right]\right\}\)
\(=x\left\{\left[-x-2-\left(4x-2x-3\right)\right]\right\}\)
\(=x\left\{\left[-x-2-\left(2x-3\right)\right]\right\}\)
\(=x\left\{-x-2-2x+3\right\}\)
\(=x\left(1-3x\right)=x-3x^2\)
Ta có : \(A+B=0\)
\(< =>2x-3+x-3x^2=0\)
\(< =>-3x^2+3x-3=0\)
\(< =>3\left(-x^2-x-1\right)=0\)
\(< =>-x^2-x-1=0\)
\(< =>-\left[x\left(x+1\right)\right]=1\)
\(< =>x\left(x+1\right)=-1.1=1.\left(-1\right)\)
\(th1\hept{\begin{cases}x=-1\\x+1=1\end{cases}< =>\hept{\begin{cases}x=-1\\x=0\end{cases}}\left(loai\right)}\)
\(th2\hept{\begin{cases}x=1\\x+1=-1\end{cases}< =>\hept{\begin{cases}x=1\\x=-2\end{cases}\left(loai\right)}}\)
Vậy không có giá trị x thỏa mãn A+B=0
dcv_new làm sai từ dọng 2 của biểu thức B nha!!!
trong ngoặc là 3 - x sao phá hoặc thành 3x - x được???
a ) \(A=x+\left\{\left(x+5\right)-\left[\left(5-x\right)-\left(-3-x\right)\right]\right\}\)
\(A=x+\left(x+5\right)-\left[5-x+x+3\right]\)
\(A=x+x+5-8\)
\(A=2x-3\)
\(B=x.\left[-x-2-\left[x+\left(3+x\right)-\left(x+3\right)\right]\right]\)
\(B=x.\left[-x-2-\left[x+3-x-x-3\right]\right]\)
\(B=x.\left\{-x-2-x-3+x+x+3\right\}\)
\(B=-2x\)
b ) \(A+B=2x-3-2x=-3\)
Vậy không có giá trị \(x\) nào để \(A+B=0\)
\(A=x+\left\{\left(x+5\right)-\left[\left(5-x\right)-\left(-x-3\right)\right]\right\}\)
\(A=x+x+5-\left(5-x+x+3\right)\)
\(A=2x+5-8\)
\(A=2x-3\)
\(B=x.\left\{\left[-x-2-\left[x+\left(3-x\right)-\left(x+3\right)\right]\right]\right\}\)
\(B=x.\left[-x-2-\left(x+3-x-x+3\right)\right]\)
\(B=x.\left(-x-2-6+x\right)\)
\(B=x.\left(-8\right)\)
Thay \(A=2x-3;B=x.\left(-8\right)\)vào A + B = 0, ta được:
2x - 3 + x. (-8) = 0
x. (2 - 8) = 3
x. (-6) = 3
x = \(\frac{-1}{2}\)
\(A=x+\left\{\left(x+5\right)-\left[\left(5-x\right)-\left(-x-3\right)\right]\right\}\)
\(=x+\left\{\left(x+5\right)-\left[5-x+x+3\right]\right\}\)
\(=x+\left\{\left(x+5\right)-8\right\}=x+\left\{x-3\right\}=x+x-3\)
\(=2x-3\)
\(B=x\left\{\left[-x-2-\left[x+\left(3-x\right)-\left(x+3\right)\right]\right]\right\}\)
\(=x\left\{\left[-x-2-\left[x+3-x-x-3\right]\right]\right\}\)
\(=x\left\{\left[-x-2-\left[-2x\right]\right]\right\}=x\left\{-x-2+2x\right\}=x\left[x-2\right]\)
\(=x^2-2x\)
Ta có : \(A+B=0\)
\(2x-3+x^2-2x=0\Leftrightarrow x^2-3=0\)
\(\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)
a) A=x+(x+5)−[(5−x)−(−x−3)]
A=x+(x+5)−[5−x+x+3]
A=x+x+5−8
A=2x−3
B=x.[−x−2−[x+(3−x)−(x+3)]
B=x.[−x−2−[x+3−x−x−3]
B=x.{−x−2−x−3+x+x+3}
B= -2x
b, Ta có :
A + B = 2x - 3 + ( -2x ) = -3
Vậy A + B = -3 không thể = 0 => không có giá trị x nào thỏa mãn
