Nhanh nhé các bn, mai mik phải nộp r...huhu
a) | x2 + 2 | + | x2 + 1 | = x2 + 2 + x2 + 1 = 2x2 + 3
b) | 2x - 3 | + | 3x - 2 | = 2x - 3 + 3x - 2 = 5x - 5 = 5( x - 1 ) với x > 2
c) | x - 4 | + | 5 - x | = -( x - 4 ) + 5 - x = -x + 4 + 5 - x = -2x + 9 ( với 4 > x )
d) | 1 - x2 | - | 1 + x2 | = -( 1 - x2 ) - ( 1 + x2 ) = -1 + x2 - 1 - x2 = -2 ( với x > 1 )
a. Vì \(x^2\ge0\forall x\)
\(\Rightarrow\left|x^2+2\right|+\left|x^2+1\right|=x^2+2+x^2+1=2x^2+3\)
b. Vì \(x>2\)\(\Rightarrow2x-3>1;3x-2>4\)
\(\Rightarrow\left|2x-3\right|+\left|3x-2\right|=2x-3+3x-2=5x-5\)
c. Vì \(4>x\)\(\Rightarrow x-4< 0;5-x>1\)
\(\Rightarrow\left|x-4\right|+\left|5-x\right|=-x+4+5-x=-2x+9\)
d. Vì \(x>1\)\(\Rightarrow1-x^2< 0;1+x^2>2\)
\(\Rightarrow\left|1-x^2\right|+\left|1+x^2\right|=-x+x^2+1+x^2=2x^2-x+1\)
Cảm ơn nhiều nhé các bn
Xin phép sửa câu d
\(\left|1-x^2\right|+\left|1+x^2\right|=-1+x^2+1+x^2=2x^2\)
a) Vì \(x^2\ge0\forall x\)\(\Rightarrow\)\(\hept{\begin{cases}x^2+2\ge2>0\forall x\\x^2+1\ge1>0\forall x\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}\left|x^2+2\right|=x^2+2\\\left|x^2+1\right|=x^2+1\end{cases}}\)
\(\Rightarrow\left|x^2+2\right|+\left|x^2+1\right|=x^2+2+x^2+1=2x^2+3\)
b) Vì \(x>2\)\(\Rightarrow\)\(\hept{\begin{cases}\left|2x-3\right|=2x-3\\\left|3x-2\right|=3x-2\end{cases}}\)
\(\Rightarrow\left|2x-3\right|+\left|3x-2\right|=2x-3+3x-2=5x-5\)
c) Vì \(4>x\)\(\Rightarrow\)\(\hept{\begin{cases}\left|x-4\right|=4-x\\\left|5-x\right|=5-x\end{cases}}\)
\(\Rightarrow\left|x-4\right|+ \left|5-x\right|=4-x+5-x=9-2x\)
d) Vì \(x>1\)\(\Rightarrow\)\(\hept{\begin{cases}\left|1-x^2\right|=x^2-1\\\left|x^2+1\right|=x^2+1\end{cases}}\)
\(\Rightarrow\left|1-x^2\right|-\left|x^2+1\right|=x^2-1-x^2-1=-2\)
chúc bn hok tốt
