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

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

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

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

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

= \(79.1+75.1+....+3.1\)

= \(79+75+....+3\)

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

= \(82.20:2\)

= \(820\)

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

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

=> \(6x+30=6\)

=> \(6x=6-30\)

=> \(6x=-24\)

=> \(x=-24:6=-4\)

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow6x+30=6\)

\(\Leftrightarrow6x=6-30\)

\(\Leftrightarrow6x=-24\)

\(\Leftrightarrow x=-4\)

Lên mạng tra nha cou có đó

Lên mạng tra cho nó nhanh

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

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

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

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

\(=\dfrac{40.41}{2}=820\)

đặt biểu thức \(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\) là A

Ta có:\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)

\(\Rightarrow15.A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)

\(\Rightarrow15.A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)

\(\Rightarrow15.A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)

\(\Rightarrow15.A=\left(4^{16}-1\right)\left(4^{16}+1\right)\)

\(\Rightarrow15.A=4^{32}-1\)

\(\Rightarrow A=\dfrac{4^{32}-1}{15}\)

Vậy giá trị biểu thức trên là \(\dfrac{4^{32}-1}{15}\)

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

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

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

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

\(=\dfrac{\left(40+1\right).40}{2}=\dfrac{41.40}{2}=820\)

\(a)=\left(27+73\right)^2=100^2=10000\)

\(b)=\left(63-13\right)^2=50^2=2500\)