Tìm x, biết: \(\frac{x-18}{2018}=\frac{x-17}{2017} \)
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x-1}{101}+\frac{x-2}{102}\)
Ai giải dùm mình nha, giải thích cho mình luôn thì tốt quá :3
Giải phương trình sau: \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\)
Cộng 1 vào từng phân số ta sẽ đc
\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)
\(\Rightarrow x=-100\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\)
<=> \(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1+\frac{x-3}{103}+1\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)
<=> x + 100 = 0 (vì \(\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)\ne0\))
<=> x = -100
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\\ \Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1+\frac{x-3}{103}+1\\ \Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{101}-\frac{x+100}{102}-\frac{x+100}{103}=0\\ \Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\\ \Leftrightarrow x+100=0\\ \Leftrightarrow x=-100\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-100\right\}\)
tìm x biết:
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
tìm x biết:
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)
\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)
Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)
\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(x=100\)
\(\dfrac{x-1}{99}-\dfrac{x+1}{101}+\dfrac{x-2}{98}-\dfrac{x+2}{102}+\dfrac{x-3}{97}-\dfrac{x+3}{103}+\dfrac{x-4}{96}-\dfrac{x+4}{104}=0\)
<=> \(\dfrac{x-1}{99}-1-\dfrac{x+1}{101}-1+\dfrac{x-2}{98}-1-\dfrac{x-2}{102}-1+\dfrac{x-3}{97}-1-\dfrac{x+3}{103}-1+\dfrac{x-4}{96}-1-\dfrac{x+4}{104}=0\)
tìm x,biết\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=\frac{17}{4}\)mấy bạn giải ra luôn hộ mình
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)(Bạn tự quy đồng chỗ này)
\(\Leftrightarrow\frac{1}{10}x=\frac{37}{8}\)
\(\Leftrightarrow x=\frac{185}{4}\)
\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
Ai giải giúp mình với mình ktick cho :D
\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)
=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396
=>4x+10=-396
4x=-406
x=-406:4=-101,5
Bài 1: Giải phương trình:\(\frac{x+2}{2018}\)+\(\frac{x+3}{2017}\)+\(\frac{x+4}{2016}\)+\(\frac{x+2038}{6}\)= 0
Bài 2: Giải phương trình: \(\frac{x-3}{2018}\)+\(\frac{x-2}{2019}\)=\(\frac{x-2019}{2}\)+\(\frac{x-2018}{3}\)
Bài 3: Giải phương trình: \(\frac{x-90}{10}\)+\(\frac{x-76}{12}\)+\(\frac{x-58}{14}\)+\(\frac{x-36}{16}\)+\(\frac{x-15}{17}\)=15
Mong các bạn giải giúp mình! Mình cần gấp!
MÌNH CẢM ƠN NHIỀU! <3
Gợi ý :
Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)
Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)
Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)
bài 3
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)
=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x=100
\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}+\frac{x+2036}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1+\frac{x+2038}{6}-3=0\)
\(\Leftrightarrow\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}+\frac{x+2020}{6}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\right)=0\)
có : \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\ne0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
\(\frac{x-3}{2018}+\frac{x-2}{2019}=\frac{x-2019}{2}+\frac{x-2018}{3}\)
\(\Leftrightarrow\frac{x-3}{2018}-1+\frac{x-2}{2019}-1=\frac{x-2019}{2}-1+\frac{x-2018}{3}-1\)
\(\Leftrightarrow\frac{x-2021}{2018}+\frac{x-2021}{2019}=\frac{x-2021}{2}+\frac{x-2021}{3}\)
bài 3 thì lần lượt trừ đi 1; 2; 3; 4; 5
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
Cảm ơn bạn rất nhiều mình đã hiểu rồi
Chúc bạn học tốt nhé
Tìm x, biết:
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
Mọi người giúp mình với mình gắp lắm!!!!
like cho những bạn trả lời hợp lý!!!
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)
\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)
\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)
\(\Leftrightarrow\left(x-101\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x-101=0\)
\(\Leftrightarrow x=101\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}-\frac{100}{99}-\frac{99}{98}-\frac{96}{95}=0\)
\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)
\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)
\(\Leftrightarrow\left(x-101\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)
Do \(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\ne0\)
Mà \(x-101=0\Leftrightarrow x=101\)
Vậy x = 101
\(\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
mong các bạn giúp nha mình chuyển bị thi khảo sát cần giải những câu kiểu này bạn nào rảnh thì làm hộ nha , ai có câu kiểu dạng này thì cho mình kèm đáp án thì càng tốt , cảm ơn ! ^^