32 x ( 200 + 3 ) =
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a) x = 2.
b) x = 7.
c) x= 12.
d) x= 45.
e) x = 18.
f) x = 10.
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`a, (4x+5) : 3 -121:11=14`
`=> (4x+5) : 3 - 11=14`
`=> (4x+5) : 3 =14+11`
`=> (4x+5) : 3 = 25`
`=> 4x+5=25xx3`
`=>4x+5= 75`
`=> 4x=75-5`
`=>4x=70`
`=>x= 70/4`
`=>x= 35/2`
`b, 2+4+6+...+x=2450`
Số lượng của dãy là :
`(x-2)/2 + 1= (x-2)/2 +2/2= x/2`
Tổng số lượng là :
\(\dfrac{\left(x+2\right)\cdot\dfrac{x}{2}}{2}=\dfrac{\dfrac{x^2}{2}+\dfrac{2x}{2}}{2}=\dfrac{x\left(x+2\right)}{4}\)
\(\dfrac{x\left(x+2\right)}{4}=2450\)
\(\Rightarrow x\left(x+2\right)=2450\cdot4\\ \Rightarrow x\left(x+2\right)=9800\)
`=>x=98`
`c,` `1` nhân `32` sao?
`d, 2x+3x=1505`
`=> (2+3)x=1505`
`=> 5x=1505`
`=> x= 1505:5`
`=>x= 301`
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1)\(\left(x-280\right):35=56.5\)
\(\left(x-280\right):35=280\)
\(x-280=280.35\)
\(x-280=9800\)
\(x=9800+280\)
\(x=10080\)
2) \(\left(x-128+20\right):192=0\)
\(\Rightarrow\left(x-148\right):192=0\)
\(\Rightarrow x-148=0\)
\(\Rightarrow x=0+148\)
\(\Rightarrow x=148\)
3) \(460+85.4=\left(x+200\right).4\)
\(\Rightarrow\left(x+200\right).4=460+340\)
\(\Rightarrow\left(x+200\right).4=800\)
\(\Rightarrow x+200=800:4=200\)
\(\Rightarrow x+200=200\)
\(\Rightarrow x=200-200=0\)
4) \(x+5.2-\left(32+16.3:16-15\right)=0\)
\(\Rightarrow x+10-\left(32+3-15\right)=0\)
\(\Rightarrow x+10-20=0\)
\(\Rightarrow x+10=20\)
\(\Rightarrow x=20-10=10\)
nha m.n
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\(a,\Rightarrow x+35=103\\ \Rightarrow x=68\\ b,\Rightarrow x+7=42\\ \Rightarrow x=35\\ c,\Rightarrow x-15=9\\ \Rightarrow x=24\\ d,\Rightarrow x+23=31\\ \Rightarrow x=8\\ e,\Rightarrow18-x=6\\ \Rightarrow x=12\\ f,\Rightarrow x+14=34\\ \Rightarrow x=20\\ g,\Rightarrow2x+1=21\\ \Rightarrow x=11\\ h,\Rightarrow3\left(x+1\right)=99\\ \Rightarrow x+1=33\Rightarrow x=32\\ i,5\left(x-3\right)=25\\ \Rightarrow x-3=5\\ \Rightarrow x=8\\ j,\Rightarrow8\left(2x+7\right)=88\\ \Rightarrow2x+7=11\Rightarrow x=2\\ k,\Rightarrow5\left(x+4\right)=85\\ \Rightarrow x+4=17\\ \Rightarrow x=13\)
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a) Ta có \(0,625^{200}=\left(\dfrac{5}{8}\right)^{200}\) và \(0,5^{1000}=\left(\dfrac{1}{2}\right)^{1000}=\left(\dfrac{1}{2}\right)^{5.200}\) \(=\left[\left(\dfrac{1}{2}\right)^5\right]^{200}\) \(=\left(\dfrac{1}{32}\right)^{200}\). Mà hiển nhiên \(\left(\dfrac{5}{8}\right)^{200}>\left(\dfrac{1}{32}\right)^{200}\) nên suy ra \(0,625^{200}>0,5^{1000}\)
b) Ta thấy \(\left(-32\right)^{27}< 0\) trong khi \(\left(-27\right)^{32}>0\) nên đương nhiên \(\left(-32\right)^{27}< \left(-27\right)^{32}\)
c) Ta thấy \(-\dfrac{3}{2}>-2\) nên \(\left(-\dfrac{3}{2}\right)^5>\left(-2\right)^5\)
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a) \(\frac{47}{32}>\frac{36}{46}\)Vì phân số \(\frac{47}{32}>1;\frac{36}{46}< 1\)nên \(\frac{47}{32}>\frac{36}{46}\)
b) \(\frac{199}{200}< \frac{200}{201}\)Vì \(\frac{199}{200}=\frac{199.201}{200.201};\frac{200}{201}=\frac{200.200}{201.200}\)
ta so sánh 199.201 và 200.200
199.201 = 199.(200+1) = 199.200+199
200.200 = 200.(199+1) = 200.199+200
Vì 199.200 + 199 < 200.199+200 nên \(\frac{199}{200}< \frac{200}{201}\)
c) \(\frac{2012.2010}{2011.2011}=\frac{\left(2011+1\right).2010}{2011.\left(2010+1\right)}=\frac{2011.2010+2010}{2011.2010+2011}\)
\(\frac{2013.2009}{2014.2008}=\frac{2013.\left(2008+1\right)}{\left(2013+1\right).2008}=\frac{2013.2008+2013}{2013.2008+2008}\)
ta so sánh : \(\frac{2010}{2011}< \frac{2013}{2008}\) vì \(\frac{2010}{2011}< 1;\frac{2013}{2008}>1\)
\(32\left(200+3\right)=32\cdot200+32\cdot3=6400+96=6496\)