Tìm x, biết:\(\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}+\frac{x+5}{2008}+\frac{x+2076}{11}=0\)
Tìm x,biết:
\(\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}+\frac{x+5}{1008}+\frac{x+2076}{11}=0\)
Giúp mình với
Ta có : \(\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}+\frac{x+5}{1008}+\frac{x+2076}{11}=0\)
=> \(\frac{x+2}{2019}+1+\frac{x+3}{2018}+1+\frac{x+4}{2017}+1+\frac{x+5}{1008}+2+\frac{x+2076}{11}-5=0\)
=> \(\frac{x+2}{2019}+\frac{2019}{2019}+\frac{x+3}{2018}+\frac{2018}{2018}+\frac{x+4}{2017}+\frac{2017}{2017}+\frac{x+5}{1008}+\frac{2016}{1008}+\frac{x+2076}{11}-\frac{55}{11}=0\)
=> \(\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}+\frac{x+2021}{1008}+\frac{x+2021}{11}=0\)
=> \(\left(x+2021\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{1008}+\frac{1}{11}\right)=0\)
=> \(x+2021=0\)
=> \(x=-2021\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{-2021\right\}\)
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)
\(\Leftrightarrow\)\(x+1=2019\)
\(\Leftrightarrow\)\(x=2019-1\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
Tìm x:
\(\frac{x+1}{2019}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2010}+\frac{x+11}{2009}+\frac{x+12}{2008}\)
Cho x,y là các số nguyên dương, chứng minh rằng:
\(1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)
Đúng đề nhé bn '' IZUHA ". Bây giờ cậu lấy mỗi phân số cộng với \(1\)để tìm tử chung rồi lầy vế trái trừ vế phải ra bằng \(0\)ta sẽ có hai thừa số , một thừa số có tổng có chứa phần ẩn \(x\)(mình kí hiệu cái này là a), còn thừa số kia cũng có tổng nhưng không chứa ẩn \(x\)( mình cũng kí hiệu là b) . Tiếp theo , cậu sẽ thấy phần b khác \(0\)nên phần a sẽ bằng \(0\). Cuối cùng cậu sẽ có giá trị x cần tìm thỏa mãn nhé !
(Mình kí hiệu a,b là vì phần sau cũng liên quan mà nó hơi dài nên tớ làm vậy cho đỡ phải bấm nhìu nhưng đừng ghi trong vở như tớ nhé ko là chết dở)
Chúc bn học tốt nhé !
Tìm x biết \(\frac{x-1}{2017}+\frac{x-2}{2018}-\frac{x-3}{2019}=\frac{x-4}{2020}\)
Ta có : \(\frac{x-1}{2017}+\frac{x-2}{2018}-\frac{x-3}{2019}=\frac{x-4}{2020}\)
\(\Rightarrow\frac{x-1}{2017}+\frac{x-2}{2018}=\frac{x-4}{2020}+\frac{x-3}{2019}\)
\(\Rightarrow1+\frac{x-1}{2017}+1+\frac{x-2}{2018}=1+\frac{x-4}{2020}+1+\frac{x-3}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}=\frac{2016+x}{2020}+\frac{2016+x}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}-\frac{2016+x}{2019}-\frac{2016+x}{2020}=0\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\text{Mà :
}\)\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)
\(\text{Nên : }\) \(2016+x=0\)
\(\Rightarrow x=-2016\)
Ta có : x−12017 +x−22018 −x−32019 =x−42020
⇒x−12017 +x−22018 =x−42020 +x−32019
⇒1+x−12017 +1+x−22018 =1+x−42020 +1+x−32019
⇒2016+x2017 +2016+x2018 =2016+x2020 +2016+x2019
⇒2016+x2017 +2016+x2018 −2016+x2019 −2016+x2020 =0
⇒(2016+x)(12017 +12018 −12019 −12020 )=0
Mà : 12017 +12018 −12019 −12020 ≠0
Nên : 2016+x=0
⇒x=−2016
tìm x , biết :
\(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
Cho x, y, z thỏa mãn:
\(\frac{x}{2017}+\frac{y}{2018}+\frac{z}{2019}=1\)
\(\frac{2017}{x}+\frac{2018}{y}+\frac{2019}{z}=0\)
CMR:\(\frac{x^2}{2017^2}+\frac{y^2}{2018^2}+\frac{z^2}{2019^2}=1\)
Tìm x biết:
a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Leftrightarrow\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}>0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Leftrightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Leftrightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
a) \(\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
=>\(x+2=0\)
=>\(x=-2\)
nếu có sai thì mong bn thông cảm nha
Giải phương trình
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
Ta có:\(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Mà \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}>0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
\(S=\left\{-2015\right\}\)
gợi ý
2017-x-2=2018-3-x=2019-4-x=2020-5-x
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
\(\Leftrightarrow\left(\frac{x-2}{2017}+1\right)+\left(\frac{x-3}{2018}+1\right)=\left(\frac{x-4}{2019}+1\right)+\left(\frac{x-5}{2020}+1\right)\)
\(\Leftrightarrow\frac{x-2+2017}{2017}+\frac{x-3+2018}{2018}=\frac{x-4+2019}{2019}+\frac{x-5+2020}{2020}\)
\(\Leftrightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2015=0\)
\(\Leftrightarrow x=-2015\)
Tìm x biết:
\(\frac{x+1}{2019}+\)\(\frac{x+2}{2018}\)\(=\frac{x+3}{2017}+\)\(\frac{x+4}{2016}\)
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}-1\right)+\left(\frac{x+2}{2018}-1\right)=\left(\frac{x+3}{2017}-1\right)+\left(\frac{x+4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2020=0:\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)\)
\(\Leftrightarrow x+2020=0\)
Còn lại tự làm :V
Lộn chỗ này , thay chút nha !
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)
Sorry =))
Bài giải
Ta có :
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)
\(\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+3}{2017}+1+\frac{x+4}{2016}+1\)
\(\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)
\(\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)
\(\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{2017}-\frac{x+2020}{2016}=0\)
\(\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
Mà \(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\ne0\text{ }\)
\(\Rightarrow\text{ }x+2020=0\)
\(\Rightarrow\text{ }x=-2020\)