100 – 7 = 93
Đáp án cần chọn là B
A. 107
B. 93
C. 92
D. 94
a/ Chứng tỏ \(A⋮50\) biết:
A = (1+2+3+4+5+6+7+8+9+...+91+92+93+94+95+96+97+98+99+100)
b/ Nếu B=199000
Hỏi B có chia hết cho C không?
Biết C=[(10+20+30+40+50+60+70+80+90+...+950+960+970+980+990+100)+148500]
a: 10100
b: có, đáp số là: 0,8037156704
A có 100 số hạng
Tổng A :
( 100 + 1 ) x 100 : 2 = 5050 \(⋮\)5
=> A \(⋮\)5
A=100+98+96+94+92+...+2-99-97-95-93-91-...-1 = ?
92+93+94+95+96+97+98+99+100
92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 + 100
= (92 + 100) + (93 + 99) + (94 + 98) + (95 + 97) + 96
= 192 + 192 + 192 + 192 + 96
= 768 + 96
= 864
92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 + 100 = 864
92+93+94+95+96+97+98+99+100=864
Tk cho mk nhé!
100-99+98-97+96-95+94-93+92-91+90
= ?
TÍNH NHANH
\(100-99+98-97+96-95+94-93+92-91+90\)
\(=\left(100-99\right)+\left(98-97\right)+\left(96-95\right)+\left(94-93\right)+\left(92-91\right)+90\)
\(=1+1+1+1+1+90\)
\(=95\)
(23 x 94+93 x45): (92x10-92)
mik cần gấp
\(=\dfrac{9^3\left(2^3\cdot9+45\right)}{9^2\cdot9}=8\cdot9+45=117\)
100+99-92-97+96+96-94-93+...+4+3-2-1
\(\frac{108-x}{+92}+\frac{107-x}{93}+\frac{106-x}{94}+\frac{105-x}{95}+4=0\)
giải pt hộ mk nhé
\(\Leftrightarrow\frac{108-x}{92}+1+\frac{107-x}{93}+1+\frac{106-x}{94}+1+\frac{105-x}{95}=0.\)
\(\Leftrightarrow\frac{108+92-x}{92}+\frac{107+93-x}{93}+\frac{106+94-x}{94}+\frac{105+95-x}{95}=0\)
\(\Leftrightarrow\frac{200-x}{92}+\frac{200-x}{93}+\frac{200-x}{94}+\frac{200-x}{95}=0\)
\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}200-x=0\Leftrightarrow x=200\\\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\ne0\end{cases}}\)
Vậy nghiệm của phương trình là 200
Ta có :
\(\frac{108-x}{92}+\frac{107-x}{93}+\frac{106-x}{94}+\frac{105-x}{95}+4=0\)
\(\Leftrightarrow\)\(\left(\frac{108-x}{92}+1\right)+\left(\frac{107-x}{93}+1\right)+\left(\frac{106-x}{94}+1\right)+\left(\frac{105-x}{95}+1\right)+\left(4-4\right)=0\)
\(\Leftrightarrow\)\(\frac{200-x}{92}+\frac{200-x}{93}+\frac{200-x}{94}+\frac{200-x}{95}=0\)
\(\Leftrightarrow\)\(\left(200-x\right)\left(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\ne0\)
\(\Rightarrow\)\(200-x=0\)
\(\Rightarrow\)\(x=200\)
Vậy \(x=200\)
Chúc bạn học tốt ~
100 + 98 + 96 + 94...... + 2 - 97 - 95 - 93 - 92 - 91...- 1
( 100 + 98 + 96 + .......... + 4 + 2 ) - ( 97 + 95 + 93 +...... + 1 )
= [ ( 100+ 2 ) x 50 : 2 ] - [ ( 97 + 1 ) x 49 : 2 ]
= 2550 - 2401
= 149
=100+98-97+96-95+94-93+...+2-1
=100+1+1+1+..+1=100+49.1
=149
hok tốt