Question

Calculați folosind metoda factorului comun a 12 102 plus 12 b 144 ori 39 minus 75 ori 39 plus 39 ori 31 e 3798×26+3799×75-3798​

Answer 1

Răspuns:

• $\bf a)~~\red{1236}$
• $\bf b)~~\blue{3900}$
• $\bf c)~~\green{~0~}$
• $\bf d)~~\orange{1050}$
• $\bf e)~~\purple{379~800}$
• $\bf f)~~\pink{0}$
• $\bf g)~~\red{\underline{1996}}$
• $\bf h)~~\blue{0}$
• $\bf i)~~\green{1}$

Explicație pas cu pas:

$\bf a) ~12\cdot 102 + 12=12\cdot\big(102 + 1\big)=12 \cdot 103=\red{\underline{1236}}$

$\it~$

$\bf b) ~144\cdot 39 -75\cdot 39+ 39\cdot 31=39\cdot\big(144-75+31\big)=$

$\bf 39 \cdot 100=\blue{\underline{3900}}$

$\bf ~$

$\bf c) ~495+99\cdot 3 -99\cdot 8=495+99\cdot\big(3-8\big)=$

$\bf 495+99\cdot\big(-5\big)=495 - 495=\green{\underline{~0~}}$

$\bf~$

$\bf d) ~5+10+15+20+25+30+...+100=5\cdot\big(1+2+3+4+5+6+...+20\big)=$

$\bf5\cdot\big[\big(20\cdot21\big):2\big]=5\cdot\big(420:2\big)=5\cdot210=\orange{\underline{1050}}$

$\it~$

$\bf e) ~3798\cdot 26+3798\cdot 75-3798=3798\cdot\big(26+75-1\big)=$

$\bf 3798\cdot 100=\purple{\underline{379~800}}$

$\bf~$

$\bf f) ~8124\cdot 18-8124\cdot 17-8124=8124\cdot\big(18-17-1\big)=$

$\bf 8124\cdot 0=\pink{\underline{0}}$

$\bf ~$

$\bf g) ~1994+1994\cdot 1995-1996\cdot 1993=$

$\bf 1994\cdot\big(1+1995\big)-1996\cdot 1993=1994\cdot1996-1996\cdot 1993=$

$\bf 1996\cdot\big(1994-1993\big)=1996\cdot 1=\red{\underline{1996}}$

$\it~~$$\it~$

$\bf h) ~\big(2022\cdot 35 - 35\cdot2021\big)\cdot 84-35\cdot\big(1964\cdot 84 - 84\cdot1963\big)=$

$\bf 35\cdot\big(2022 - 2021\big)\cdot 84-35\cdot84\cdot\big(1964 -1963\big)=$

$\bf 35\cdot1\cdot 84-35\cdot84\cdot1=2940-2940=\blue{\underline{~0~}}$

$\it~$

$\bf i) ~\big(19\cdot 21+21\big):\big(35 + 35\cdot11\big)=$

$\bf \big[21\cdot\big(19+1\big)\big]:\big[35\cdot\big(1 + 11\big)\big]=\big(21\cdot20\big):\big(35\cdot12\big)=$

$\bf 420 : 420 = \green{\underline{~1~}}$

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==pav38==

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