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\(a,\left(13+150+47\right)-\left(13+47\right)\)
\(=13+150+47-13-47\)
\(=\left[13+\left(-13\right)\right]+\left[47+\left(-47\right)\right]+150\)
\(=0+0+150\)
\(=150\)
\(b,\left(-237\right)-\left[\left(-237+155\right)-55\right]\)
\(=\left(-237\right)+237+155-55\)
\(=0+100\)
\(=100\)
\(c,57.\left(62-43\right)-62.\left(57-43\right)\)
\(=57.62-57.43-62.57+62.43\)
\(=-57.43+62.43\)
\(=43.\left(-57+62\right)\)
\(=43.5\)
\(=215\)
a , ( 13 + 150 + 47 ) - ( 13 + 47 )
= 12 + 150 + 47 - 13 - 47
= [ 12 + ( - 13 ) ] + [ 47 + ( - 47 ) ] + 150
= - 1 + 0 + 150
= 149
b , ( - 237 ) - [ ( - 237 + 155 ) - 55 ]
= - 237 - [ - 237 + 155 - 55 ]
= - 237 + 237 - 155 + 55
= [ ( - 237 ) + 237 ] + [ ( - 155 ) + 55 ]
= 0 + ( - 100 )
= - 100
c , 57 . ( 62 - 43 ) - 62 . ( 57 - 43 )
= 57 . 62 - 57 . 43 - 62 . 57 - 62 . 43
= [ 57 . 62 - 62 . 57 ] . [ 62 . 43 - 62 . 43 ]
= 0 . 0
= 0
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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 \(100-x=0\Leftrightarrow x=100\)
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42 + 43 + 44 + 45 - 32 - 33 - 34 - 35
= (42 - 32)+(43 - 33)+(44 - 34)+(45 - 35)
= 10+10+10+10
= 40
(64+65+66+67+68) - (54+55+56+57+58)
= 64+65+66+67+68 - 54-55-56-57-58
= (64 - 54)+(65 - 55) +(66 - 56) + (67- 57) + (68 -58)
= 10+10+10+10+10
= 50
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\(42\cdot47+43\cdot42-600\)
\(=42\cdot90-600\)
\(=3180\)
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\(\left(4^2+4^3+4^7\right)\left[55+\left(-55\right)\right]\\ =\left(16+64+16384\right).0\\ =0\)
\(\left(4^2+4^3+4^7\right).\left(55+-55\right)\)
\(=\left(4^2+4^3+4^7\right).0=0\)