(1-x)^3=96
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\(x+x\times\frac{1}{3}=\frac{96}{108}\)
\(x\times\left(1+\frac{1}{3}\right)=\frac{8}{9}\)
\(x\times\frac{4}{3}=\frac{8}{9}\)
\(x=\frac{8}{9}\div\frac{4}{3}\)
\(x=\frac{2}{3}\)
Vậy .............
Ta có : \(x\)+ \(x\)x \(\frac{1}{3}\)= \(\frac{96}{108}\)
\(x\)x \(\left(1+\frac{1}{3}\right)\)= \(\frac{96}{108}\)
\(x\)x \(\frac{4}{3}\)= \(\frac{96}{108}\)
\(x\) = \(\frac{96}{108}\): \(\frac{4}{3}\)
\(x\) = \(\frac{96}{108}\)x \(\frac{3}{4}\)
\(x\) = \(\frac{24}{27}\)x \(\frac{3}{4}\)
\(x\) = \(\frac{6}{9}\)
96 - 3(x+1)= 42
3(x+1) = 96 - 42 = 54
x + 1 = 54 : 3 = 18
x = 18 - 1 = 17
c)96 - 3.(x + 1) = 42
3.(x + 1)= 96 - 42
3.(x + 1) = 54
x + 1= 54:3
x + 1= 18
x = 18 - 1
x = 17
\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\)
<=> x=-100
ko chép đề nhé
\(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95} \)
=> \((x+100)(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95})=0\)
vì \((\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}) khác 0\)
=>\(x+100=0\)
=>x=-100
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}=\dfrac{x+4+96}{96}+\dfrac{x+5+95}{95}\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
=>x+100=0
hay x=-100
\(96+98\times96+1\times96\)
\(=1\times96+98\times96+1\times96\)
\(=\left(1+1+98\right)\times96\)
\(=100\times96\)
\(=9600\)
96 + 98 x 96 + 1 x 96
= 96 x ( 1 + 98 + 1 )
= 96 x 100
= 9600
Ủng hộ nha!
`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
`=>(x+1)/99+1+(x+2)/98+1+(x+3)/97+1+(x+4)/96+1=-4+4`
`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`
`=>(x+100)(1/99+1/98+1/97+1/96)=0`
`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)
`=>x=-100`
Vậy ...
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)
mà `1/99+1/98+1/97+1/96 \ne 0`
nên `x+100=0`
`x=-100`
\((1-x)^3=96\)
\((1-x)^3=\sqrt[3]{96}\)
\(1-x=\sqrt[3]{96}\)
\(x=1-\sqrt[3]{96}\)
Lời giải:
$(1-x)^3=96$
$\Rightarrow 1-x=\sqrt[3]{96}$
$\Rightarrow x=1-\sqrt[3]{96}$