Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
\(A=-\left|2x-3\right|+1< =1\)
Dấu = xảy ra khi x=3/2
\(C=-\left|5x+2\right|-\left|3y+12\right|+4< =4\)
Dấu = xảy ra khi x=-2/5 và y=-4
\(D=-3\left(x+1\right)^2+5< =5\)
Dấu = xảy ra khi x=-1
\(E=\dfrac{1}{2}\left(x+1\right)^2+3>=3\)
Dấu = xảy ra khi x=-1
\(F=\dfrac{15}{4}+3\left|x-1\right|>=\dfrac{15}{4}\)
Dấu = xảy ra khi x=1
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)
=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)
=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)
=> \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)
=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)
Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)
Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
1.
a) \(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b) \(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
c) \(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}=\frac{23}{2}-\frac{65}{2}=-21\)
d) \(\left(-49,1\right).\frac{13}{27}-58,9.\frac{13}{27}=\frac{13}{27}.\left(-49,1-58,9\right)=\frac{13}{27}.\left(-108\right)=-52\)
e) \(0,375:\left(-4,5\right)=\frac{-1}{12}\)
f) \(3\frac{1}{7}:\left(-1\frac{3}{7}\right)=\frac{22}{7}:\frac{-10}{7}=\frac{-11}{5}\)
g) \(9\frac{1}{3}:4\frac{2}{3}-2=\frac{28}{3}:\frac{14}{3}-2=2-2=0\)
h) \(\left(7\frac{3}{4}:0,3125+4,5.2\frac{2}{45}\right):\left(-8,5\right)=\left(\frac{31}{4}:\frac{5}{16}+\frac{9}{2}.\frac{92}{45}\right):\frac{-17}{2}=\left(\frac{124}{5}+\frac{46}{5}\right):\frac{-17}{2}=34:\frac{-17}{2}=-4\)
Bài 1 : Tính:
a)
\(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b)
\(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{10}{15}+\frac{-2}{15}=\frac{8}{15}\)
c)
\(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}\)\(=\frac{23}{2}-\frac{65}{2}=\frac{-42}{2}=-21\)
....
Tự lm tiếp dạng như v
Bài 2 :
\(A=\frac{-6}{11}.\frac{7}{10}.\frac{11}{-6}.-20=\left(\frac{-6}{11}.\frac{11}{-6}\right).\left(\frac{7}{10}.-20\right)\)\(=1.\left(-14\right)=-14\)
.....
Bài 3 :
\(\frac{3}{7}.x-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{3}{7}-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{1}{35}x=\frac{-17}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}:\frac{1}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}.35=-17\)
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
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\cdot\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{13}\right)\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
d) \(\frac{x+4}{2017}+\frac{x+3}{2018}=\frac{x+2}{2019}+\frac{x+1}{2020}\)
\(\Leftrightarrow\left(\frac{x+4}{2017}+1\right)+\left(\frac{x+3}{2018}+1\right)=\left(\frac{x+2}{2019}+1\right)+\left(\frac{x+1}{2020}\right)\)
\(\Leftrightarrow\frac{x+2021}{2017}+\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
\(\Leftrightarrow\left(x+2021\right)\cdot\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2021=0\)
\(\Leftrightarrow x=-2021\)
a) 2/3.X+4=-12
2/3.X=-12-4
2/3.X=-16
x=-16:2/3
x=-16.3/2
x=-24
vậy x=-24
B) 3/4+1/4:X=-3
1/4:X=-3-3/4
1/4:X=-15/4
X=1/4:-15/4
X=1/4.4/-15
X=-1/15
VẬY X=-1/15
MÌNH MỚI NGHĨ RA HAI Ý THÔI.
THÔNG CẢM CHO MÌNH BẠN NHÁ