Giải các phương trình:
a) \(\dfrac{x+2001}{5}+\dfrac{x+1999}{7}+\dfrac{x+1997}{9}+\dfrac{x+1995}{11}=-4;\)
b) \(\dfrac{x-15}{100}+\dfrac{x-10}{105}+\dfrac{x-100}{110}=\dfrac{x-100}{15}+\dfrac{x-105}{10}+\dfrac{x-110}{5}.\)
\(\dfrac{2009-x}{7}+\dfrac{2007-x}{9}+\dfrac{2005-x}{11}+\dfrac{2003-x}{13}=\dfrac{x-17}{-1999}+\dfrac{x-15}{-2001}+\dfrac{x-13}{-2003}+\dfrac{x-11}{-2005}\)
9 - 3 x ( X - 9 ) = 6
4 + 6 x ( X + 1 ) 70
\(\dfrac{X}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{11}{14}-\dfrac{3}{X}=\dfrac{5}{14}\)
5 x ( 3 + 7 x X ) = 40
X x 6 + 12 : 3 = 120
X x 3,7 + X x 6,3 = 120
( 15 x 24 - X ) : 0,25 = 100 : \(\dfrac{1}{4}\)
71 + 65 x 4 = \(\dfrac{X+140}{X}\)+ 260
( X +1 ) + ( X + 4 ) + ( x + 7 ) + ...... + (X + 28 ) = 155
đây là bài tìm X
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
Bài 15:
a)\(\dfrac{-2}{5}\)+\(\dfrac{4}{5}\) . x =\(\dfrac{3}{5}\)
b)\(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\):x = -2
Bài 16
a) x - \(\dfrac{10}{3}\) = \(\dfrac{7}{15}\) . \(\dfrac{3}{5}\)
b) x + \(\dfrac{3}{22}\)= \(\dfrac{27}{121}\) . \(\dfrac{11}{9}\)
c) \(\dfrac{8}{23}\) . \(\dfrac{48}{24}\) - x = \(\dfrac{1}{3}\)
d) 1 - x = \(\dfrac{49}{65}\).\(\dfrac{5}{7}\)
Bài 17: tìm x
a) \(\dfrac{62}{7}\) . x = \(\dfrac{29}{9}\): \(\dfrac{3}{56}\)
b) \(\dfrac{1}{5}\) : x=\(\dfrac{1}{5}\)+\(\dfrac{1}{7}\)
bài 18:
a)\(\dfrac{2}{5}\)+\(\dfrac{3}{4}\): x =\(\dfrac{-1}{2}\)
b)\(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\)x + \(\dfrac{3}{5}\)x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\).x-x = \(\dfrac{-9}{14}\)
bài 19: tính
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+ \(\dfrac{1}{2018.2019}\)
bài 20:tìm x
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{2008}{2009}\)
bài 21: tìm x
\(\dfrac{x+1}{99}\)+\(\dfrac{x+2}{98}\)\(\dfrac{x+3}{97}\)\(\dfrac{x+4}{96}\)=-4
bài 22 : so sánh các phân số sau:
a) \(\dfrac{-1}{5}\)+\(\dfrac{4}{-5}\)và 1
b) \(\dfrac{3}{5}\) và \(\dfrac{2}{3}\)+\(\dfrac{-1}{5}\)
c)\(\dfrac{3}{2}\)+\(\dfrac{-4}{3}\) và \(\dfrac{1}{10}\)+\(\dfrac{-4}{5}\)
d) \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{1}{6}\) và 2
giải phương trình
a) \(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
b) \(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
giải chi tiết giúp e ạ;-;
a: \(\Leftrightarrow x+2016=0\)
hay x=-2016
b: \(\Leftrightarrow x-100=0\)
hay x=100
giải các phương trình sau:
1) \(\dfrac{x-11}{111}+\dfrac{x-12}{112}=\dfrac{x-23}{123}+\dfrac{x-24}{124}\)
2) \(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1980}{15}+\dfrac{x-1990}{5}\)
3) \(\dfrac{109-x}{91}+\dfrac{107-x}{93}+\dfrac{105-x}{95}+\dfrac{103-x}{97}=-4\)
câu 4 giải pt sau
\(\dfrac{x-2003}{16}+\dfrac{x-1997}{11}+\dfrac{x-1992}{9}+\dfrac{x-1991}{7}=10\)
`(x-2003)/16 +(x-1997)/11 +(x-1992)/9 +(x-1991)/7=10`
`<=>((x-2003)/16-1)+((x-1997)/11-2)+((x-1992)/9-3)+((x-1991)/7-4)=0`
`<=>(x-2019)/16+ (x-2019)/11 +(x-2019)/9+(x-2019)/7 =0`
`<=> (x-2019)(1/16+1/11+1/9+1/7)=0`
<=> x-2019=0`
`<=> x=2019`
\(\dfrac{x-2003}{16}+\dfrac{x-1997}{11}+\dfrac{x-1992}{9}+\dfrac{x-1991}{7}=10\)
\(\Leftrightarrow\left(\dfrac{x-2003}{16}-1\right)+\left(\dfrac{x-1997}{11}-2\right)+\left(\dfrac{x-1992}{9}-3\right)+\left(\dfrac{x-1991}{7}-4\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-2003}{16}-\dfrac{16}{16}\right)+\left(\dfrac{x-1997}{11}-\dfrac{22}{11}\right)+\left(\dfrac{x-1992}{9}-\dfrac{27}{9}\right)+\left(\dfrac{x-1991}{7}-\dfrac{28}{7}\right)=0\)
\(\Leftrightarrow\dfrac{x-2003-16}{16}+\dfrac{x-1997-22}{11}+\dfrac{x-1992-27}{9}+\dfrac{x-1991-28}{7}=0\)
\(\Leftrightarrow\dfrac{x-2019}{16}+\dfrac{x-2019}{11}+\dfrac{x-2019}{9}+\dfrac{x-2019}{7}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\dfrac{1}{16}+\dfrac{1}{11}+\dfrac{1}{9}+\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow x-2019=0\) Vì \(\dfrac{1}{16}+\dfrac{1}{11}+\dfrac{1}{9}+\dfrac{1}{7}\ne0\)
\(\Leftrightarrow x=2019\)
Vậy phương trình có 1 nghiệm \(x=2019\)
giải phương trình sau
a) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}+\dfrac{x+4}{97}=-4\)
b) \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
a ) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}+\dfrac{x+4}{97}=-4\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}+\dfrac{x+101}{97}=-1\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{97}\right)=-1\)
Vô lí => Phương trình trên vô nghiệm .
b ) \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-73}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Leftrightarrow x=89\)
Vậy x = 89.
Câu a nhằm tí mà chắc đề sai nếu đúng chỉ bk dùng mày tính để tìm nghiệm .
Giải phương trình sau:\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+10+48}=\dfrac{4}{105}\)
(Giải thích các bước giải)
\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
Giải các bất phương trình sau:
a) \(\dfrac{x-2}{1007}+\dfrac{x-1}{1008}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
b) \(\dfrac{3-x}{100}+\dfrac{4-x}{101}>\dfrac{10-2x}{204}+\dfrac{12-2x}{206}\)
a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)
=>2x-2018<0
=>x<2019
b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)
=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)
=>\(x+97< 0\)
=>x<-97