Giúp mình câu 98-100
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\(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+...+\left(-4\right)\) Vì dãy số trên có 100 số hạng => Dãy số trên có 25 cặp -4.
\(=\left(-4\right).25\)
\(=-100\)
Tk mk nhé!Thank you!
Có: 1+2-3-4+......+97+98-99-100
= 1+(2-3-4+5)+......+(94-95-96+97)+98-99-100
=1+98-99-100
=99-99-100
= -100
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
A=1-2-3+4+5-6-7+8+...+97-98-99+100
=>A=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
=>A=0+0+....+0=0
vậy A=0
B=1-2+2^2-2^3+...+2^100
=>2B=2-2^2+2^3-2^4+....+2^101
=>2B+B=1-2^101=3B
=>B=1-2^101/3
C= 2^100-2^99-2^98-...-2^2-2-1
=>C=2^100-(2^99+2^98+.....+2^2+2+1)
Đặt D=2^99+2^98+.....+2^2+2+1
=>2D=2^100+2^99+.....+2^3+2^2+2
=>2D-D=2^100-1=D
=>C=2^100-(2^100-1)=1
tick nha
hic!ngày kia phải nộp rồi ! mọi người giúp mình nhanh nha!
18576: {\(105^0\)+[2.(102+101−100−99+98+97−96−95+.........+6+5−4−3+2+1)−201]}^3
Đặt A=102+101−100−99+98+97−96−95+...............+6+5−4−3+2+1
A=(102+101−100−99)+(98+97−96−95)+........+(6+5−4−3)+2+1
A=4+4+...........+4+3
A=4.25+3
A=103
⇒18576:{1050+[2.(102+101−100−99+98+97−96−95+.........+6+5−4−3+2+1)−201]}^3
=18576:[1+(2.103)−201]^3
=18576:63
=18576:216
=86
Đặt S = 2.4 + 4.6 + 6.8 + .... + 98.100 + 100.102
<=> S = 2.( 2 + 2 ) + 4.( 4 + 2 ) + 6.( 6 + 2 ) + ...... + 98.( 98 + 2 ) + 100.( 100 + 2 )
<=> S = 2.2 + 22 + 2.4 + 42 + 2.6 + 62 + .... + 2.98 + 982 + 2.100 + 1002
<=> S = ( 22 + 42 + ... + 982 + 1002 ) + ( 2.2 + 2.4 + 2.6 + .... + 2.98 + 2.100 )
<=> S = 22.( 12 + 22 + ... +492 + 502 ) + 4.( 1 + 2 + 3 + .... + 49 + 50 )
Đặt A = 12 + 22 + 32 + .... + 492 + 502
B = 1 + 2 + 3 + .... + 49 + 50
=> S = 4A + 4B
A = 12 + 22 + 32 + .... + 492 + 502
<=> A = 1.1 + 2.2 + 3.3 + .... + 49.49 + 50.50
<=> A = 1.( 2 - 1 ) + 2.( 3 - 1 ) + 3.( 4 - 1 ) + .... + 49.(50 - 1 ) + 50.( 51 - 1 )
<=> A = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + .... + 49.50 - 49 + 50.51 - 50
<=> A = ( 1.2 + 2.3 + 3.4 + .... + 49.50 + 50.51 ) - ( 1 + 2 + 3 + ... + 49 + 50 )
Đặt C = 1.2 + 2.3 + 3.4 + .... + 49.50 + 50.51
D = 1 + 2 + 3 + .... + 49 + 50
=> A = C - D
C = 1.2 + 2.3 + 3.4 + ... + 49.50 + 50.51
<=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ..... + 49.50.3 + 50.51.3
<=> 3C = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + 4.5.( 6 - 3 ) + ..... + 49.50.( 51 - 48 ) + 50.51.( 52 - 49 )
<=> 3C = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 49.50.51 - 48.49.50 + 50.51.52 - 49.50.51
<=> 3C = 50.51.52
=> C = ( 50.51.52 ) : 3 = 44200
D = 1 + 2 + 3 + .... + 50
SSH : ( 50 - 1 ): 1 + 1 = 50 ( SH )
=> D = ( 50 + 1 ) . 50 : 2 = 1275
=> A = 44200 - 1275 = 42925
B = 1 + 2 + 3 + ... + 49 + 50
SSH : ( 50 - 1 ) : 1 + 1 = 50 ( SH )
=> B = ( 50 +1 ) . 50 : 2 = 1275
=> S = ( 42925 + 1275 ) . 4 = 176800
Vậy S = 176800
98/ \(z_2=3-5i\Rightarrow\overline{z}^2=\left(3+5i\right)^2=9+30i-25=30i-16\)
\(\Rightarrow w=\left(2+4i\right)\left(30i-16\right)=60i-32-120-64i=-4i-152\)
=> D
99/ Voi so phuc \(z=x+yi\left(x,y\in R\right)\) duoc bieu dien boi diem \(M\left(x;y\right)\)
\(pt\Leftrightarrow\left(3+2i\right)\left(x+yi\right)+\left(2-i\right)^2-4-i=0\)
\(\Leftrightarrow3x+3yi+2xi-2y+4-4i-1-4-i=0\)
\(\Leftrightarrow3x-2y-1+\left(3y+2x-5\right)i=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2y-1=0\\3y+2x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\Rightarrow M\left(1;1\right)\) =>C
100/ \(pt\Leftrightarrow\left(1-2\sqrt{3}i-3\right)\left(x+yi\right)-4+3i=0\)
\(\Leftrightarrow x+yi-2\sqrt{3}xi+2\sqrt{3}y-3x-3yi-4+3i=0\)
\(\Leftrightarrow-2x+2\sqrt{3}y-4+\left(y-2\sqrt{3}x-3y+3\right)i=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x+2\sqrt{3}y-4=0\\-2y-2\sqrt{3}x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x+2\sqrt{3}y=4\\-2\sqrt{3}y-6x=-3\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3\sqrt{3}-4}{8}\\y=\dfrac{4+\dfrac{3\sqrt{3}-4}{4}}{2\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow Modun=\sqrt{x^2+y^2}=\dfrac{5}{4}\) =>A
P/s: Nhìn chữ Chuyên Hạ Long lại nhớ Quảng Ninh zl :(
Tks bn nha