A=(1+2/1.4)x (1+2.5)x(1+3/6)x...x(1+2/100/103)
K
Khách
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31 tháng 12 2017
1
b;
B=1+ (7-5) + (11-9) + ...+(101-99)
B=1+2+2+..+2
B=1+25.2=51
NT
31 tháng 12 2017
2.
a.
ĐK : x+2 >=0 => x>=-2
\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)
Vậy x=-2
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 12 2018
\(A=\left(\dfrac{6}{1.4}\right)\left(\dfrac{12}{2.5}\right)\left(\dfrac{20}{3.6}\right)\left(\dfrac{x^2+3x+2}{x\left(x+3\right)}\right)\)
\(A=\dfrac{2.3}{1.4}.\dfrac{3.4}{2.5}.\dfrac{4.5}{3.6}...\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+3\right)}\)
\(A=\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{3.4.5...\left(x+2\right)}{4.5.6...\left(x+3\right)}=\left(x+1\right)\dfrac{3}{x+3}=\dfrac{3\left(x+1\right)}{x+3}\)
LM
Lê Minh Vũ
VIP
13 tháng 10 2021
\(a)\)\(\left(50-6.x\right).18=2^3.3^2.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=8.9.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=360\)
\(\Leftrightarrow\)\(\left(50-6.x\right)=360\div18\)
\(\Leftrightarrow\)\(50-6.x=20\)
\(\Leftrightarrow\)\(6.x=50-20\)
\(\Leftrightarrow\)\(6.x=30\)
\(\Leftrightarrow\)\(x=5\)
\(b)\)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7450\)
\(\Leftrightarrow\)\(100x+\left(1+2+3+...+100\right)=7450\)
\(\Leftrightarrow\)\(100x+5050=7450\)
\(\Leftrightarrow\)\(100x=7450-5050\)
\(\Leftrightarrow\)\(100x=2400\)
\(\Leftrightarrow\)\(x=24\)
13 tháng 10 2021
b.
(x+1)+(x+2)+...+(x+100)=7450
=> 100x + (1+2+3+...+100)=7450
=>100x + (100+1).50=7450
=>100x=2400
=>x=24
Ta có : (12+13+14+....+110)×x=19+28+....+82+91(12+13+14+....+110)×x=19+28+....+82+91
Đặt A=19+28+....+82+91A=19+28+....+82+91
=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)
=1+102+103+...+109=102+103+....+109+1010=1+102+103+...+109=102+103+....+109+1010
=(12+13+14+....+110)×10=(12+13+14+....+110)×10 (1)
Từ (1), suy ra:
(12+13+14+...+110)×x=(12+13+14+....+110)×10(12+13+14+...+110)×x=(12+13+14+....+110)×10
⇒x=10⇒x=10
Vậy x=10x=10
~ Học tốt ~
Ta có : (12+13+14+....+110)×x=19+28+....+82+91(12+13+14+....+110)×x=19+28+....+82+91
Đặt A=19+28+....+82+91A=19+28+....+82+91
=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)
=1+102+103+...+109=102+103+....+109+1010=1+102+103+...+109=102+103+....+109+1010
=(12+13+14+....+110)×10=(12+13+14+....+110)×10 (1)
Từ (1), suy ra:
(12+13+14+...+110)×x=(12+13+14+....+110)×10(12+13+14+...+110)×x=(12+13+14+....+110)×10
⇒x=10⇒x=10
Vậy x=10x=10
~ Học tốt ~