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92 + 71 + 108 = (92 + 108) + 71 = 200 + 71 = 271
32 x 47 + 53 x 47 + 47 x 15 = 47(32 + 53 + 15)
= 47.100
= 4700
90 x 143 - 43 x 90 = 90.(143 - 43) = 90.100 = 9000
(1200 + 60) : 12 = 1800 : 12 = 150
2 x 18 x 24 + 3 x 50 x 16 + 12 x 32 x 4 = 18.(2.24) + 50.(3.16) + 32.(12.4) = 18.48 + 50.48 + 32.48 = 48(18 + 50 + 32) = 48.100 = 4800
43 x 53 + 47 x 196 - 47 x 52
= 43 x 53 + 47 x ( 196 - 52 )
= 43 x 53 + 47 x 144 -> Hết thuận tiện rồi bạn ơi ( kiểm tra lại bài thử nha ).
- (-15) – (-3 +15 – 8)
= 15 + 3 - 15 + 8
= 11
b) + ( - 143 + 17 – 35) – ( -143 + 37 – 35)
= -143 + 17 - 35 + 143 - 37 + 35
= -20
c) 50 - ( 47 + 50 - 18 ) - ( 47 + 18 )
= 50 - 47 - 50 + 18 - 47 - 18
= -47 - 47
= -94
d) 16 - ( -15 ) + ( -20 ) - | -24 |
= 16 + 15 - 20 - 24
= -13
130-[5.(9-x)+43]=47
5.(9-x)+43=130-47
5.(9-x)+43=83
5.(9-x)=83-43
5.(9-x)=40
9-x=40:5
9-x=8
x=9-8
x=1
\(130-\left[5\times\left(9-x\right)+43\right]=47\)
\(\Rightarrow\left[5\times\left(9-x\right)+43\right]=130-47\)
\(\Rightarrow\left[5\times\left(9-x\right)\right]=83\)
\(\Rightarrow\left(9-x\right)=83\div5\)
\(\Rightarrow\left(9-x\right)=\frac{83}{5}\)
\(\Rightarrow x=9-\frac{83}{5}\)
\(\Rightarrow x=-\frac{38}{5}\)
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
143 - 43.(x+1)=47
=> 43.(x+1) = 96
x+1=\(\frac{96}{43}\)
\(\Rightarrow x=\frac{53}{43}\)
143 - 43. (x+1) = 47
43.(x+1) = 143 - 47
43.(x+1) = 96
x+1 = 96 : 43
x+1 = 96/43
Học tốt!!!