\(E=1\cdot1+2\cdot2+3\cdot3+...+15\cdot15\)

\(=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+3\cdot\left(4-1\right)+...+15\cdot\left(16-1\right)\)

\(=1\cdot2-1+2\cdot3-2+3\cdot4-3+...+15\cdot16-15\)

\(=\left(1\cdot2+2\cdot3+3\cdot4+...+15\cdot16\right)-\left(1+2+3+...+15\right)\)

\(=1360-120=1240\)

Phương trình −2 x 2 − 6x − 1 = 0 có  = ( − 6 ) 2 – 4.(− 2).(−1) = 28 > 0 nên phương trình có hai nghiệm x 1 ;   x 2

Theo hệ thức Vi-ét ta có  x 1 + x 2 = − b a x 1 . x 2 = c a ⇔ x 1 + x 2 = − 3 x 1 . x 2 = 1 2

Ta có

N = 1 x 1 + 3 + 1 x 2 + 3 = x 1 + x 2 + 6 x 1 . x 2 + 3 x 1 + x 2 + 9 = − 3 + 6 1 2 + 3. − 3 + 9 = 6

Đáp án: A

Sửa đề: \(a=1\cdot2^2+2\cdot3^2+3\cdot4^2+\cdots+98\cdot99^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+4^2\left(4-1\right)+\cdots+99^2\left(99-1\right)\)

\(=\left(2_{}^3+3^3+\cdots+99^3\right)-\left(2^2+3^2+\cdots+99^2\right)\)

\(=\left(1^3+2^3+3^3+\cdots+99^3\right)-\left(1^2+2^2+3^2+\cdots+99^2\right)\)

\(=\left(1+2+\cdots+99\right)^2-\frac{99\cdot\left(99+1\right)\left(2\cdot99+1\right)}{6}\)

\(=\left(99\cdot\frac{100}{2}\right)^2-\frac{99\cdot100\cdot199}{6}=\left(99\cdot50\right)^2-33\cdot50\cdot199\)

\(=4950^2-328350=24174150\)

S5=5x5-(4x4-(3x3-2x2)

S5=5x5-(4x4-(3x3-(2x2-1x1)))

S2011=2001x2001-(2000x2000-(1999x1999-(....)))

Bạn ơi tính như vậy thì phần b tính kiểu jr bao nhiêu dấu ngoạc làm sao tính được

Bài 1:

123.4=492

145.2=290

420.2=840

Bài 2:

23-14+17+24-13+73=110

2+3+4+5+...+26=350

1.1+2.2+3.3+...+9.9=285

Bài 1:

123.4=492

145.2=290

420.2=840

Bài 2:

23-14+17+24-13+73=110

2+3+4+5+...+26=350

1.1+2.2+3.3+...+9.9=285

a: \(=60\cdot\dfrac{17}{20}=51\)

b: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{1}{5}\)

a) 1+1x1−1x1+1x1−1x =(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1=(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1

b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1 =(1−2x+1):(1−x2−2x2−1)=(1−2x+1):(1−x2−2x2−1)

                       =x+1−2x+1:x2−1−(x2−2)x2−1=x+1−2x+1:x2−1−(x2−2)x2−1

                       =x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)

                        =x−1x+1.(x−1)(x+1)1=(x−1)2=x−1x+1.(x−1)(x+1)1=(x−1)2.

a) 1+1x1−1x1+1x1−1x =(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1=(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1 b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1 =(1−2x+1):(1−x2−2x2−1)=(1−2x+1):(1−x2−2x2−1)                        =x+1−2x+1:x2−1−(x2−2)x2−1=x+1−2x+1:x2−1−(x2−2)x2−1                        =x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)                         =x−1x+1.(x−1)(x+1)1=(x−1)2=x−1x+1.(x−1)(x+1)1=(x−1)2.