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a, 3.(37 - \(x\)) - 56 = 1
3.(37 - \(x\)) = 1 + 56
3.(37 - \(x\)) = 57
37 - \(x\) = 57 : 3
37 - \(x\) = 19
\(x\) = 37 - 19
\(x\) = 18
b, 95 - (2\(x\) - 1).4 = 14
(2\(x\) - 1).4 = 95 - 14
(2\(x\) - 1).4 = 81
2\(x\) - 1 = 81 : 4
2\(x\) = \(\dfrac{81}{4}\) - 1
2\(x\) = \(\dfrac{85}{4}\)
\(x\) = \(\dfrac{85}{4}\) : 2
\(x\) = \(\dfrac{85}{8}\)
95-3.(x+7)=23
3.(x+7)=95-23
3.(x+7)=72
x+7=72:3
x+7=24
x=24-7
x=17
k mình với
3.(x+7) = 95 - 23
3. ( x+ 7) = 72
x + 7 = 72 : 3
x + 7 = 24
x = 17
95-3*(x+7)=23
3*(x+7)=95-23
3*(x+7)=72
x+7=72/3
x+7=24
x=24-7
x=17
a)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)
$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$
Do đó $x-23=0\Rightarrow x=23$
b)
PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$
$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$
$\Rightarrow x+100=0\Rightarrow x=-100$
c)
PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$
Do đó $x+2005=0\Rightarrow x=-2005$
d)
PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)
\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)
\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$
Do đó $300-x=0\Rightarrow x=300$
\(a.\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\\\Leftrightarrow \left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\\\Leftrightarrow x-23=0\left(vi\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\right)\\ \Leftrightarrow x=23\)
Này tớ làm tắt có gì cậu không hiểu nói tớ nhé
\(b.\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\\ \Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1-\left(\frac{x+4}{96}+1+\frac{x+5}{95}+1\right)=0\\\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\\\Leftrightarrow \left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\\ \Leftrightarrow x+100=0\left(Vi\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\\\Leftrightarrow x=-100\)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\right)=0\)
\(\Leftrightarrow x-23=0\left(vì\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\ne0\right)\)
\(\Leftrightarrow x=23\)
vậy................
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\left(vì\frac{1}{99}+\frac{1}{97}+\frac{1}{95}>0\right)\)
\(\Leftrightarrow x=300\)
vậy..........
a. 2 3 − 3 8 = 16 24 − 9 24 = 7 24
b. 2 3 × 9 5 = 2 × 9 3 × 5 = 18 15 = 6 5
=121 nhé
bằng 121