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\(\left|x+3\right|+\left|x-2\right|=5\left(1\right)\)
Xét \(x\le-3\), \(\left(1\right)\Leftrightarrow-x-3+2-x=5\)
\(\Leftrightarrow-2x=6\Leftrightarrow x=-3\) (thỏa mãn)
Xét \(-3< x\le2\), \(\left(1\right)\Leftrightarrow x+3+2-x=5\)
\(\Leftrightarrow5=5\Leftrightarrow x\in Z\)
Xét \(x>5\), \(\left(1\right)\Leftrightarrow x+3+x-2=5\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\) (loại)
\(5A=5^2+5^3+5^4+...+5^{2017}\)
\(A=\frac{5A-A}{4}=\frac{5^{2017}-5}{4}=\frac{5\left(5^{2016}-1\right)}{4}\)
\(\Rightarrow4.A+5=\frac{4.5\left(5^{2016}-1\right)}{4}+5=5\left(5^{2016}-1\right)+5=5^x\)
\(\Rightarrow\frac{5\left(5^{2016}-1\right)}{5}+\frac{5}{5}=\frac{5^x}{5}\Rightarrow5^{2016}-1+1=5^{x-1}\)
\(\Rightarrow5^{2016}=5^{x-1}\Rightarrow x-1=2016\Rightarrow x=2017\)
( 1000 + x ) . 5 = 10 . x + 5
1000 . 5 + 5 . x = 10 . x + 5
10 . x - 5 . x = 1000 . 5 - 5
5 . x = 5 . ( 1000 - 1 )
5 . x = 5 . 999
x = 999
( 1000 + x ) * 5 = 10 * x + 5
=> 1000 * 5 + 5 = 10 * x + 5
=> 10* x - 5 = 1000 * 5 - 5 ( Vì chuyển vế phải đổi dấu )
=> 5 * x = 5 * ( 1000 - 1 )
=> 5 * x = 5 * 999
=> 5 * x = 4995
=> x = 4995 : 5
=> x = 999
Chúc bn học vui^^
\(\left(15-5x\right)^3=-125=\left(-5\right)^3\)
\(\Leftrightarrow15-5x=-5\)
\(\Leftrightarrow5x=20\)
\(\Leftrightarrow x=4\)
Vậy : \(x=4\)
Ta có: (15-5x)3=-125
<=>(15-5x)3=-53
<=>15-5x=-5
<=>5x=15-(-5)
<=>5x=15+5
<=>5x=20
<=>x=4
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)
\(\left(2x-1\right)=-\frac{4}{63}\)
2x= -4/63 + 1
2x = 59/63
x = 59/63 : 2
x = 59/126
1/3:(2.x-1)=-5-1/4
1/3:(2.x-1)=-21/4
2.x-1=1/3:-21/4
2.x-1=-4/63
2.x=-4/63+1
2.x=\(3\frac{59}{63}\)
x=\(3\frac{59}{63}\):2
x=\(1\frac{61}{63}\)
\(\left(x^2+5\right)\left(x-3\right)>0\)
\(=>x^3-3x^2+5x-15>0\)
\(=>x^2\left(x-3\right)+5\left(x-3\right)>0\)
\(=>\left(x-3\right)\left(x^2+5\right)>0\)
TH1 : \(\orbr{\begin{cases}x-3>0\\x^2+5>0\end{cases}}=>\orbr{\begin{cases}x>3\\x^2>-5\end{cases}}=>\orbr{\begin{cases}x>3\\x>\sqrt{\left(-5\right)}\end{cases}}\)
TH2 : \(\orbr{\begin{cases}x-3< 0\\x^2+5< 0\end{cases}=>\orbr{\begin{cases}x< 3\\x^2< -5\end{cases}=>\orbr{\begin{cases}x< 3\\x< \sqrt{\left(-5\right)}\end{cases}}}}\)