(x-5)^5=(x-5)^15
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1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
a, \(\dfrac{5\times11\times15\times18}{6\times15\times22\times10}\)
= \(\dfrac{5\times11\times15\times6\times3}{6\times15\times11\times2\times2\times5}\)
= \(\dfrac{5\times11\times15\times6}{5\times11\times15\times6}\) \(\times\) \(\dfrac{3}{2\times2}\)
= 1 \(\times\) \(\dfrac{3}{4}\)
= \(\dfrac{3}{4}\)
b, \(\dfrac{7}{12}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{5}{12}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{7}{12}+\dfrac{5}{12}\))
= \(\dfrac{3}{5}\) \(\times\) \(\dfrac{12}{12}\)
= \(\dfrac{3}{5}\) \(\times\) 1
= \(\dfrac{3}{5}\)
a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)
\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)
=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)
b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)
Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)
c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)
\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)
=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)
d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)
\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6
=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)
`1, -2/9 xx 15/17 + (-2/9) xx 2/17`
`= -2/9 xx (15/17 + 2/17)`
`= -2/9 xx 17/17`
`=-2/9xx1`
`=-2/9`
__
`-5/3 xx 6/5 + (-7/9) xx 3/10`
`= -30/15 + (-21/90)`
`= -2 + (-7/30)`
`=-60/30 +(-7/30)`
`=-67/30`
__
`15/20 xx 7/5 + (-9/7) xx (-6/4)`
`=3/4 xx7/5 + (-9/7) xx(-6/4)`
`= 21/20 + 54/28`
`= 21/20 + 27/14`
`=417/140`
__
`-25/13 xx 5/19 + (-25/13) xx 14/19`
`=-25/13 xx (5/19 +14/19)`
`=-25/13 xx 19/19`
`= -25/13 xx 1`
`=-25/13`
__
`-7/13 xx 13/5 + (-9/7) xx 5/3`
`=-7/5 +(-15/7)`
`=-124/35`
1) -5 . x - ( - 3 ) = 13
-5 . x + 3 = 13
-5 . x = 13 - 3
-5 . x =10
x= 10 : ( -5 )
x = -2
2) 3 . x + 17 = 2
3 . x = 2 - 17
3 . x = -15
x = -15 : 3
x = -5
3) (-5) . x =15
x = -3
4) 2x - (-17) = 15
2x + 17 = 15
2x = -2
x = -2 : 2
x= -1
5) -12 . ( x - 5 ) + 7 .(3-x) = 5
a, thiếu đề
b, -5 ( 2 - x ) + 4 ( x - 3 ) = 10x - 15
<=> -10 + 5x + 4x - 12 = 10x - 15
<=> -x = 7 <=> x = 7
c,-7(5-x)-2(x-10)=15
<=> -35 + 7x - 2x + 20 = 15 <=> 5x = 30 <=> x = 6
d, -4(x+1) + 89x - 3 = 24
<=> 85x = 31 <=> x = 31/85
e, 5(x-30-2(x+6)) = 9
<=> 5 (-x-49) = 9 <=> -5x = 254 <=> x = -254/5
b: =>-10+5x+4x-12=10x-15
=>9x-10x=-15+22
=>-x=7
hay x=-7
c: =>-35+7x-2x+20=15
=>5x-15=15
=>x=6
d: =>-4x-4+72x-24=24
=>68x-32=24
hay x=14/17
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
\(\left|x\right|=\left|-5\right|-\left|-10\right|+15\)
\(\Rightarrow\left|x\right|=5-10+15\)
\(\Rightarrow\left|x\right|=10\)
\(\Rightarrow x=\pm10\)
\(\left|x\right|=-(-12)+(-15)+8\)
\(\Rightarrow\left|x\right|=+12+(-15)+8\)
\(\Rightarrow\left|x\right|=12+(-15)+8\)
\(\Rightarrow\left|x\right|=(-3)+8\)
\(\Rightarrow\left|x\right|=5\)
\(\Rightarrow x=\pm5\)
Bài dưới tương tự
|x|=|−5|−|−10|+15
⇒|x|=5−10+15
⇒|x|=10
⇒x=±10
|x|=−(−12)+(−15)+8
⇒|x|=+12+(−15)+8
⇒|x|=12+(−15)+8
⇒|x|=(−3)+8
⇒|x|=5
⇒x=±5
\(x-5+x-5+x-5=15\)
\(\left(x-5\right)\times3=15\)
\(x-5=15\div3\)
\(x-5=5\)
\(x=5+5\)
\(x=10\)
Ta có: x-5+x-5+x-5=15
=>(x-5).3=15
=>x-5=15:3
=>x-5=5
=>x=5+5
=>x=10
\(\left(x-5\right)^5=\left(x-5\right)^{15}\)
\(\left(x-5\right)^{15}-\left(x-5\right)^5=0\)
\(\left(x-5\right)^5\left[\left(x-5\right)^{10}-1\right]=0\)
\(\left(x-5\right)^5=0\) hoặc \(\left(x-5\right)^{10}-1=0\)
*) \(\left(x-5\right)^5=0\)
\(x-5=0\)
\(x=5\)
*) \(\left(x-5\right)^{10}-1=0\)
\(\left(x-5\right)^{10}=1\)
\(x-5=1\) hoặc \(x-5=-1\)
+) \(x-5=1\)
\(x=1+5\)
\(x=6\)
+) \(x-5=-1\)
\(x=-1+5\)
\(x=4\)
Vậy \(x=4;x=5;x=6\)