\(=\left(40^2-31^2\right)-\left(39^2-32^2\right)+\left(38^2-33^2\right)-\left(37^2-34^2\right)+\left(36^2-35^2\right)\)

\(=\left(40-31\right)\left(40+31\right)-\left(39-32\right)\left(39+32\right)+\left(38-33\right)\left(38+33\right)-\)

\(\left(37-34\right)\left(37+34\right)+\left(36-35\right)\left(36+35\right)\)

\(=9.71-7.71+5.71-3.71+1.71\)

\(=\left(9-7+5-3+1\right).71=5.71=355\)

\(40^2-39^2+38^2-37^2+....+32^2-31^2\)

\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(32-31\right)\left(32+31\right)\)

\(=40+39+38+37+....+32+31\)

Số số hạng của dãy trên là: (40-31):1+1= 10 (số)

Tổng trên là: (40+31) x 10 : 2 = 355

Vậy ....

132^2 - 112^2 = (132 - 112) (132 + 112) = 20 . 244 = 4880

39 . 41 = (40 - 1) (40 + 1) = 40² - 1 = 1599

40² - 39² + 38² - 37² + ... + 2² - 1 = (40² - 39²) + (38² - 37²) + ... + (2² - 1)

= (40 - 39) (40 + 39) + (38 - 37) (38 + 37) + ... + (2 - 1) (2 + 1)

= 40 + 39 + 38 + 37 + 36 + ... + 2 + 1

= \(\frac{\left(40-1+1\right)\left(40+1\right)}{2}=\frac{40.41}{2}=820\)

tìm 2 số lẻ liên tiếp biết hiệu b của phương của chúng bằng 56

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a)2011²-2010.2012

=(2010+1)²-2010.(2010+2)

=2010²+2.2010+1-2010²-2.2010

=1

bằng 40 nha

\(40^2-39^2+38^2-37 ^2+...+2^2-1^2\)

= \(\left(40+39\right)\left(40-39\right)+\left(38+37\right)\left(38-37\right)+....+\left(2+1\right)\left(2-1\right)\)

= \(79.1+75.1+....+3.1\)

= \(79+75+....+3\)

= \(\left(79+3\right)\left[\left(79-3\right):4+1\right]:2\)

= \(82.20:2\)

= \(820\)

\(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)

=> \(9x^2-6x+1+2x^2+12x+18-11x^2+11=6\)

=> \(6x+30=6\)

=> \(6x=6-30\)

=> \(6x=-24\)

=> \(x=-24:6=-4\)

  \(\text{a) }40^2-39^2+38^2-37^2+...+2^2-1^2\)

\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+...+\left(2^2-1^2\right)\)

\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=1.79+1.75+...+1.3\)

\(=79+75+...+3\)
\(\text{Từ 3 đến 79 có: (79 - 3) : 2 + 1 = 39 (số hạng)}\)

\(\text{Tổng là: }\frac{\left(79+3\right)\times39}{2}=1599\)

\(\text{b) }\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)

\(\Leftrightarrow\left(9x^2-6x+1\right)+2\left(x^2+6x+9\right)+11\left(1-x^2\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18+11-11x^2=6\)

\(\Leftrightarrow\left(9x^2+2x^2-11x^2\right)+\left(-6x+12x\right)+\left(1+18+11\right)=6\)

\(\Leftrightarrow6x+30=6\)

\(\Leftrightarrow6x=6-30\)

\(\Leftrightarrow6x=-24\)

\(\Leftrightarrow x=-4\)

+) KHÔNG DÙNG MÁY TÍNH.HÃY TÍNH NHANH GIÁ TRỊ CỦA BIỂU THỨC SAU:

40²−39²+38²−37²+...+2²−1²

\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+....+\left(2^2-1^2\right)\)

\(=\left(40-39\right).\left(40+39\right)+\left(38-37\right).\left(38+37\right)+...+\left(2-1\right).\left(2+1\right)\)

\(=40+39+38+37+....+2+1\)

\(=\frac{40.41}{2}=820\)
 

\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{37\cdot38\cdot39}\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{38\cdot39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}\cdot\frac{370}{741}\)

\(=\frac{185}{741}\)