tìm x biết :
a) 2.( x + 1 ) - 3 ( x - 2 ) + 5. ( x - 1 ) = 25
b ) | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( - 3,2 + \(\frac{2}{5}\))
tìm x biết \(|x-\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Rightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow x-\frac{1}{3}=\hept{\begin{cases}2\\-2\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}\frac{7}{3}\\\frac{-5}{3}\end{cases}}\)
tìm x, biết: \(|x-\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-2,8\right|\)
\(\left|x-\frac{1}{3}\right|=2,8-\frac{4}{5}\)
\(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}}\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{14}{5}+\frac{4}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{19}{5}\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{3}=\frac{19}{5}\Leftrightarrow x=\frac{62}{15}\\x-\frac{1}{3}=-\frac{19}{5}\Leftrightarrow x=-\frac{52}{15}\end{cases}}\)
vây x= \(\frac{62}{15}\)hoặc x=\(-\frac{52}{15}\)
Chúc bn hok tốt
Tìm tập hợp các số nguyên x biết:
\(a,4\frac{5}{9}:2\frac{5}{8}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{1}{2}\right)\)
\(b,\frac{-17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
Giúp mik nha! Ai làm đc mik cho 3 tick!
tìm x:
a)\(|x-\frac{1}{3}|+\frac{4}{5}=|-3,2+\frac{2}{5}|\)
b)\(3|2x+\frac{1}{2}|-\frac{11}{2}=-\frac{11}{3}+\frac{5}{2}\)
\(a,|x-\frac{1}{3}|+\frac{4}{5}=|-3,2+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=|-\frac{7}{10}|\)
\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=\frac{7}{10}\)
\(\Rightarrow|x-\frac{1}{3}|=\frac{7}{10}-\frac{4}{5}\)
\(\Rightarrow|x-\frac{1}{3}|=-\frac{1}{10}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{10}\\x-\frac{1}{3}=-\frac{1}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{30}\\x=\frac{7}{30}\end{cases}}\)
CMR: \(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
tìm x biết:
\(|x-\frac{1}{3}|=|\left(-3,2\right)+\frac{2}{5}|\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)
\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)
\(\Rightarrow|x-\frac{1}{3}|=2.8\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)
tính lại kết quả nhé
\(\left|x-\frac{1}{3}\right|=\left|-3.2+\frac{2}{5}\right|=\left|-\frac{14}{5}\right|\)\(=\frac{14}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{14}{5}\\x-\frac{1}{3}=-\frac{14}{5}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{47}{15}\\x=-\frac{37}{15}\end{cases}}\)
Vậy............
b,
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)
\(\Leftrightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}+1-\frac{x-4}{2008}+1=0\)
\(\Leftrightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}-\frac{x-4-2008}{2008}=0\)
\(\Leftrightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x-2012=0
=>x=2012
Vậy ..............
TK MK NHA
*****CHUC BN HOC GIỎI*****
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
Tìm x \(\in\) Z biết:
a) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\) b) \(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)\)Ta có: 1/2 - (1/3 + 1/4) = 1/2 - 7/12 = -1/12 ;
1/48 - (1/16 - 1/6) = 1/48 + 5/48 = 1/8
Vì \(-\frac{1}{12}< x< \frac{1}{8}\) nên x = 0
b) \(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)\)Ta có :
\(4\frac{5}{9}:2\frac{5}{18}-7=2-7=-5\)
\(\left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)=\left(1+\frac{38}{5}\right):\left(-21\frac{2}{3}\right)=\frac{43}{5}:\frac{-65}{3}=-\frac{129}{325}\)
Vì \(-5< x< -\frac{129}{325}\) nên \(x\in\left\{-4;-3;-2;-1\right\}\)
Tìm x biết :
a. \(|x-\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|\)
b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
a. \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{16}{5}+\frac{2}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{14}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}.}\)
Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{3}\right\}.\)
b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\times\left(x-7\right)^{10}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}.}\)Xét 2 trường hợp:
\(\left(x-7\right)^{x+1}=0\)\(\Leftrightarrow x-7=0\Leftrightarrow x=7.\)\(1-\left(x-7\right)^{10}=0\Leftrightarrow\left(x-7\right)^{10}=1\Leftrightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}.}}\)Vậy \(x\in\left\{6;7;8\right\}.\)
ban nguyen nhat minh giang lai cho mk dong 2 cau b cai
mk cam on
Dòng 2 câu b là đặt nhân tử chung \(\left(x-7\right)^{x+1}\)ra ngoài