a: \(20-\left[30-\left(5-1\right)^2\right]\)

\(=20-\left[30-4^2\right]\)

\(=20-14=6\)

b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)

\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)

\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)

c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)

\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)

\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)

d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)

\(=411-\left[\dfrac{110}{5}-4\right]\)

=410-22+4

=410-18

=392

e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)

\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)

\(=450-5\cdot\left[9\cdot8-12\right]+18\)

=468-5*60

=468-300

=168

f:

\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)

\(=102-150:\left[18-2\cdot4\right]+1\)

\(=103-\dfrac{150}{18-8}=103-15=88\)

ta có : \(A=6.7+6.7^2+6.7^3+...+6.7^{100}\)

\(\Rightarrow7A=7.\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)

\(7A=6.7^2+6.7^3+6.7^4+...+6.7^{101}\)

\(\Rightarrow7A-A=6A=\left(6.7^2+6.7^3+6.7^4+...+6.7^{101}\right)-\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)

\(6A=6.7^{101}-6.7=6\left(7^{101}-7\right)\Leftrightarrow A=7^{101}-7\)

vậy \(A=7^{101}-7\)

ta có : \(B=6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)

\(\Rightarrow5B=5\left(6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\right)\)

\(5B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}\)

\(\Rightarrow5B+B=6B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}+6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)

\(6B=6.5-6.5^{101}=6.\left(5-5^{101}\right)\Leftrightarrow B=5-5^{101}\)

vậy \(B=5-5^{101}\)

\(A=6\cdot7+6\cdot7^2+6\cdot7^3+...+6\cdot7^{100}\\ =6\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot7+\left(7-1\right)\cdot7^2+\left(7-1\right)\cdot7^3+...+\left(7-1\right)\cdot7^{100}\\ =7^2-7+7^3-7^2+7^4-7^3+...+7^{101}-7^{100}\\ =7^{101}-7=7\cdot\left(7^{100}-1\right)\)

\(B=6\cdot5-6\cdot5^2+6\cdot5^3-...+6\cdot5^{99}-6\cdot5^{100}\\ =6\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\ =\left(5+1\right)\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\=\left(5+1\right)\cdot5-\left(5+1\right)\cdot5^2+\left(5+1\right)\cdot5^3-...+\left(5+1\right)\cdot5^{99}-5^{100}\\ =5^2+5-5^3-5^2+5^4+5^3+...+5^{100}+5^{99}-5^{101}-5^{100}\\ =5-5^{101}\\ =5\cdot\left(1-5^{100}\right)\)

Tính thuận tiện 

5.3 × 6.7 + 6.7 × 4.7

= (5,3 + 4,7) x 6,7

= 67

Tính thuận tiện 

5.3 × 6.7 + 6.7 × 4.7

\(=6,7x\left(5,3+4,7\right)\)

\(=6,7x10\)

\(=67\)

Công thức :a x b + a x c = a x (b+c)

5,3 x 6,7 + 6,7 x 4,7

= (5,3 + 4,7) x 4,7

= 10 x 4,7

= 47

~Học tốt~

a) 2,5x9,3x4=(2,5x4)x9,3=10x9,3=93

b) 0,5x3,8x2=(0,5x2)x3,8=1x3,8=3,8

c)7,61x5x0,2=7,61x(5x0,2)=7,61x1=7,61

d) 5,3x6,7+6,7x4,7=6,7x(5,3+4,7)=6,7x10=67

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)

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