|x+12|+|x+13|+|x+14|=4x
Cho mink hỏi x bằng bn
Có cách giả nha!!!!!!! Cảm ơn các bạn nhìu!!!!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
a) 0,2 x 317 x 7 + 0,14 x 3520 + 33,1 x 14
= ( 0,2 x 7 ) x 317 + 0,14 x 3520 + 33,1 x 14
= 1,4 x 317 + 0,14 x 3520 + 33,1 x 14
= 443,8 + 492,8 + 463,4
= 1400
b) 2 + 5 + 8 + ... + 65 - 387: từ 2 đến 65 có 22 số hạng
= { ( 65 + 2 ) x 22 : 2 } - 387
= 737 - 387
= 350
x+32/11 + x+23/12 = x+38/13 + x+27/14
\(\Rightarrow\frac{x+32}{11}-3+\frac{x+23}{12}-2=\frac{x+38}{13}-3+\frac{x+27}{14}-2\)
\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)
\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}-\frac{x-1}{14}=0\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Rightarrow x-1=0\).Do \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
\(\Rightarrow x=1\)
các bạn ơi giúp mink bài 3 với nha mink cần gấp tối nay rồi !
giúp mink với nha mink cảm ơn nhìu lắm
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
|x+12|+|x+13|+|x+14|=4x
|x+12|+|x+13|+|x+14| luôn\(\ge\) 0=>x \(\ge\)0
|x+12|+|x+13|+|x+14| là số dương hay|x+12|+|x+13|+|x+14| =x+12+x+13+x+14=4x
x+12+x+13+x+14=4x
3x+(12+13+14)=4x
4x-3x=12+13+14
x=39
Do \(\left|x+12\right|\ge0,\left|x+13\right|\ge0,\left|x+14\right|\ge0\)
\(\Rightarrow\left|x+12\right|+\left|x+13\right|+\left|x+14\right|>0\)
Do VP>0 suy ra VT>0
Hay \(4x>0\)
\(\Rightarrow x>0\)
Khi đó:\(\left|x+12\right|+\left|x+13\right|+\left|x+14\right|=4x\)
\(\Leftrightarrow3x+39=4x\)(vì x>0)
\(\Rightarrow x=39\)