400-861: [372 : 4 x X - 1620 : 9] -199= 198
400-861 : [372 : 4xX -1602 : 9 ] - 199 = 198
bài đó dễ mà
tìm x
x+1/199 + x+2/198 + x+3/197 + x+4/196 + x+5/195 = -5
1/21 +1/28 +1/36 +........+2/ x(x+1)=2/9
Tìm x: \(\dfrac{1+3+5+...+199}{2+4+6+...+198+x}\)=1
a) 1 - 5 + 9 - 13 +...- x = -400
b) -8 + 13 - 18 +...+ x = 250
c) x + (x+1) + (x+2) +...+ 2019 + 2020 = 2020
d) 200 + 199 + 198 +...+ (x+2) + (x+1) + x 200
AI LÀ THẦN ĐỒNG TOÁN HỌC GIẢI GIÚP MIK VỚI Ạ
CẦN SỰ GIÚP ĐỠ VỀ HỌC TÂP
(x+1)/199 + (x+2)/198 + (x+3)/197 + (x+4)/196 + (x+220)/5 = 0 . Ai giúp mik ik , mik cảm ưn
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)
\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)
\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
Đặt: \(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{199}{1}\)là B
Cộng 1 vào mỗi phần số trừ phân số cuối cùng ta sẽ được:
B= \(\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
=> B= \(\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1\)
=> B= \(\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}\)
=> B= \(200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\) => B= \(200\) X A
=> \(\frac{A}{B}\)\(=\frac{1}{200}\)
=> \(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
=>\(x-20\) =\(\frac{1}{2000}:\frac{1}{200}\)
=> \(x-20=\).......................... Bạn tự làm tiếp nhé, chúc bạn học tốt !!!^^\(\)
Tìm x
a) (x+3)/200 + (x+4)/199 + (x+5)/198= -3
b) (x+4)/2000+ (x+6)/999+(x+8)/499+7=0
c) |x-4| - |2x-1|=6
Tìm x
a) (x+3)/200 + (x+4)/199 + (x+5)/198= -3
b) (x+4)/2000+ (x+6)/999+(x+8)/499+7=0
c) |x-4| - |2x-1|=6
x/200+x+1/199+x+2/198+3=0
Tìm x
\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)
\(\frac{x}{200}+1+\frac{x+1}{199}+1+\frac{x+2}{198}+1=0\)
\(\frac{x}{200}+\frac{200}{200}+\frac{x+1}{199}+\frac{199}{199}+\frac{x+2}{198}+\frac{198}{198}=0\)
\(\frac{x+200}{200}+\frac{x+200}{199}+\frac{x+200}{198}=0\)
\(\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\right)=0\)
Vì \(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\) khác 0
Suy ra: x + 200 = 0
x = 0 - 200
x = -200
Vậy x = -200.
Hình như lớp 5 chưa học số âm :v
Bài giải
\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)
\(\left(\frac{x}{200}+1\right)+\left(\frac{x+1}{199}+1\right)+\left(\frac{x+2}{198}+1\right)=0\)
\(\frac{x+200}{200}+\frac{x+200}{199}+\frac{x+198}{198}=0\)
\(\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\right)=0\)
Do \(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\ne0\) nên \(x+200=0\)
\(\Rightarrow\text{ }x=-200\)
(x-20).\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)=\(\frac{1}{2000}\)
Giúp mk với
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)
\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)