Tính nhanh
E = 1x1+2x2+3×3+...+15×15
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\(E=1\cdot1+2\cdot2+3\cdot3+...+15\cdot15\)
\(=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+3\cdot\left(4-1\right)+...+15\cdot\left(16-1\right)\)
\(=1\cdot2-1+2\cdot3-2+3\cdot4-3+...+15\cdot16-15\)
\(=\left(1\cdot2+2\cdot3+3\cdot4+...+15\cdot16\right)-\left(1+2+3+...+15\right)\)
\(=1360-120=1240\)
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
S5=5x5-(4x4-(3x3-(2x2-1x1)))
S2011=2001x2001-(2000x2000-(1999x1999-(....)))
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.
E = 1 x 1 + 2 x2 + 3 x 3 + . . . + 15 x 15
E = 1( 2 -1 ) + 2 ( 3 -1 ) + 3 ( 4-1 ) + . . . + 15 ( 16 -1 )
E = 1.2 -1 + 2.3 - 2 + 3.4 - 3 + . . . + 15 . 16 - 15
E = ( 1.2 + 2.3 + 3.4+ . . . + 15 .16 ) - ( 1 + 2 + 3 + . . . + 15 )
E = 1360- 120
E = 1240
Tk mk nha