Cho A = 1 x 2 x 3 x 4 + ... + 19 x 20.
Tính A x 3 = ?
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Nhân cả 2 vế với 4 ta có:
A × 4 = 1 × 2 × 3 ×4 + 2 × 3 × 4×(5 - 1) + 3 × 4 × 5 × (6 - 2) + ... + 18 × 19 × 20 × (21 - 17).
Khi nhân vào và làm phép trừ để triệt tiêu các số giống nhau ta còn lại là:
A × 4 = 18 × 19 × 20 × 21
=> A = 18 ×19 × 20 × 21 : 4 = 35910
Ax3=1x2x3+2x3x3+...+19x20x3
=1x2x3+2x3x(4-1)+...+19x20x(21-18)
=1x2x3+2x3x4-1x2x3+....+19x20x21-18x19x20
=19x20x21=7980
A = 1 x 2 + 2 x 3 + 3 x 4 ... 19 x 20
=>A x 3=1x2x3+2x3x3+3x3x4+...+3x19x20
=1x2x3 - 0x1x2 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4+....+19x20x21 - 18x19x20
=19x20x21
=7980
a/ A = 12 + 22 + 32 +...+ 202
=> A = 1 + 2.( 1 + 1 ) + 3( 1 + 2 ) +...+ 20.( 1 + 19 )
=> A = 1 + 2 + 1.2 + 3 + 2.3 +...+ 20 + 19.20
=> A = ( 1 + 2 + 3 +...+ 20 ) + ( 1.2 + 2.3 +...+ 19.20 )
=> A = 210 + ( 1.2 + 2.3 +...+ 19.20)
GỌI ( 1.2 + 2.3 + ... + 19.20 ) là B. Ta có:
B = 1.2 + 2.3 +...+ 19.20
3B = 1.2.3 + 2.3.( 4 - 1 ) +...+ 19.20.( 21 - 18 )
3B = 1.2.3 + 2.3.4 - 1.2.3 + ...+ 19.20.21 - 18.19.20
3B = 19.20.21
B = 19.20.7
B = 2660
Thay B vào A ta có:
A= 210 + 2660
A = 2870
Vậy A = 2870
b/ B= 1.2.3 + 2.3.4 + 3.4.5 +...+ 18.19.20
4B = 1.2.3.4 + 2.3.4.( 5-1 ) + 3.4.5.( 6-2 ) +...+ 18.19.20.( 21-17 )
4B= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 +...+ 18.19.20.21 - 17.18.19.20
4B = 18.19.20.21
B = \(\frac{18.19.20.21}{4}\)
B= 18.19.5.21
B = 34200
Bài 4:
a: xy=-2
=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)
=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)
b: \(\left(x-1\right)\left(y+2\right)=-3\)
=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)
=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)
Bài 3:
a: \(x\left(x+9\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b: \(\left(x-5\right)^2=9\)
=>\(\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)
c: \(\left(7-x\right)^2=-64\)
mà \(\left(7-x\right)^2>=0\forall x\)
nên \(x\in\varnothing\)
Bài 2:
a: \(\left(-31\right)\cdot x=-93\)
=>\(31\cdot x=93\)
=>\(x=\dfrac{93}{31}=3\)
b: \(\left(-4\right)\cdot x=-20\)
=>\(4\cdot x=20\)
=>\(x=\dfrac{20}{4}=5\)
c: \(5x+1=-4\)
=>\(5x=-4-1=-5\)
=>\(x=-\dfrac{5}{5}=-1\)
d: \(-12x+1=-4\)
=>\(-12x=-4-1=-5\)
=>\(12x=5\)
=>\(x=\dfrac{5}{12}\)
A= 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ... + 1/19 x 20
--> A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 +...+ 1/19 - 1/20
= 1/2 - 1/20 = 10/20 - 1/20 = 9/20
\(A=1\cdot2\cdot3\cdot...\cdot19\cdot20\)
\(\Leftrightarrow A=1\cdot2\cdot3+2\cdot3\cdot3\cdot+3\cdot3\cdot4+...+3\cdot19\cdot20\)
\(=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+19\cdot20\cdot21-18\cdot19\cdot20\)\(=19\cdot20\cdot21\)
\(=7980\)
tICH NHA BAN HIEN
A x3 = 7980