đố biết :201-20
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\(x^2\ge0\Rightarrow x^2+1\ge1>0\Rightarrow\left|x^2+1\right|=x^2+1\)
<=>\(x^2+1-\left|x^2-4\right|=1\Leftrightarrow x^2-\left|x^2-4\right|=0\Leftrightarrow x^2=\left|x^2-4\right|\)
+)\(x^2-4>0\Leftrightarrow x^2>4\Leftrightarrow x< -2;x>2\)
<=>\(x^2-4=x^2\Leftrightarrow0=4\) vô lý
+)\(x^2-4\le0\Leftrightarrow x^2\le4\Leftrightarrow-2\le x\le2\)
<=>\(4-x^2=x^2\Leftrightarrow4=2x^2\Leftrightarrow x^2=2\Leftrightarrow\orbr{\begin{cases}x=-\sqrt{2}\\x=\sqrt{2}\end{cases}}\)(nhận)
Vậy ...
1984= 1x1000+9x........+8x10+4x1=
1984= 1000+ 9x.........+80+4=
1984= 1084 + 9x..............=
9x..............= 1984 - 1084
9x..............=900
...... =900: 9
..... =100
a, \(\frac{23+x}{201-x}=\frac{3}{5}\)
\(\Rightarrow\left(23+x\right)5=3\left(201-x\right)\)
\(\Rightarrow115+5x=603-3x\)
\(\Rightarrow5x+3x=603-115\)
\(\Rightarrow8x=448\Rightarrow x=61\)
Vậy x = 81
a) 126+(−20)+2004+(−106)126+(−20)+2004+(−106)
=2004+126+[(−20)+(−106)]=2004+126+[(−20)+(−106)]
=2004+126+[−(20+106)]=2004+126+[−(20+106)]
=2004+126+(−126)=2004+126+(−126)
=2004+[126+(−126)]=2004+0=2004=2004+[126+(−126)]=2004+0=2004.
b) (−199)+(−200)+(−201)=−(199+200+201)(−199)+(−200)+(−201)=−(199+200+201)
=−(199+201+200)=−[(199+201)+200]=−(199+201+200)=−[(199+201)+200]
=−(400+200)=−600=−(400+200)=−600.
\(A=1+3+3^2+...+3^{2000}\)
\(3A=3\left(1+3+3^2+...+3^{2000}\right)\)
\(3A=3+3^2+...+3^{2001}\)
\(3A-A=\left(3+3^2+...+3^{2001}\right)-\left(1+3+...+3^{2000}\right)\)
\(2A=3^{2001}-1\). Suy ra
\(3^{2001}-1=3^n-1\Leftrightarrow3^{2001}=3^n\Leftrightarrow n=2001\)
201-20=181
k mình nhé