1,Tìm GTNN:
a,A=|x-3|+|x-5|+2015
b,B=|x-3|+|x-5|+|x-7|
2 ,Tìm GTLN:
a,Q=\(\frac{2-x}{\left|x\right|}\)
x€Z
b,P=\(\frac{2}{\left|x+2\right|+1}\)
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
Tìm x, biết:
a)\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
b)\(\left|\frac{11}{5}-x\right|+\left|x+\frac{1}{5}\right|+\frac{41}{5}=1,2\)
c)\(2\left|x+\frac{7}{2}\right|+\left|x\right|-\frac{7}{2}=\left|\frac{11}{5}-x\right|\)
1. Tìm x ϵ Q sao cho:
a) (2x-3). (x+1) < 0.
b) \(\left(x-\frac{1}{2}\right).\left(x+3\right)\)> 0.
2. Tính:
S=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{999.1001}\)
3. Tìm x: Biết x không thuộc{-2; -5; -10; -17}
\(\frac{3}{\left(x+2\right).\left(x+5\right)}+\frac{5}{\left(x+5\right).\left(x+10\right)}+\frac{7}{\left(x+10\right).\left(x+17\right)}=\frac{x}{\left(x+2\right).\left(x+17\right)}\)
Bài 1:
a) (2x-3). (x+1) < 0
=>2x-3 và x+1 ngược dấu
Mà 2x-3<x+1 với mọi x
\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)
b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu
Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)
Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
=>....
Bài 2:
\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\cdot\frac{998}{3003}\)
\(=\frac{499}{3003}\)
1. Tìm x ϵ Q sao cho:
a) (2x-3). (x+1) < 0.
b) \(\left(x-\frac{1}{2}\right).\left(x+3\right)>0\)
2.Tính:
S=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{999.1001}\)
3.Tìm x: Biết x không thuộc{-2; -5; -10; -17}
\(\frac{3}{\left(x+2\right).\left(x+5\right)}+\frac{5}{\left(x+5\right).\left(x+10\right)}+\frac{7}{\left(x+10\right).\left(x+17\right)}=\frac{x}{\left(x+2\right).\left(x+17\right)}\)
tự làm nhé. bài cô Kiều cho dễ mừ :)
Tìm x
a/\(\frac{x+7}{2003}+\frac{x+4}{2006}=\frac{x-1}{2011}+\frac{x-5}{2015}\)
b/\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c/\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
Tìm x, biết:
a)\(\left(x+5\right).\left(x+9\right)>0\)
b)\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
TÌM X
a,\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=6\)
b,\(\left(x^2-4\right).\left(2x+\frac{4}{3}\right)=0\)
1)tìm các số nguyên x và y thỏa mãn:\(y^2=x^2+x+1\)
2)cho các số thực x và y thỏa mãn \(\left(x+\sqrt{a+x^2}\right)\left(y+\sqrt{a+y^2}\right)\)=a
tìm giá trị biểu thức \(4\left(x^7+y^7\right)+2\left(x^5+y^5\right)+11\left(x^3+y^3\right)+2016\)
3)cho x;y là các số thực khác 0 thỏa mãn x+y khác 0
cmr \(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)\(=\frac{1}{x^3y^3}\)
4)cho a,b,c là các số dương.cmr\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
TÌM x BIẾT:
a,\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
với x\(\notin\){-2;-5;-10;-17}
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
với x\(\notin\){1;3;8;20}
c, TÌM X BIẾT:
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
GIÚP MÌNH CHÚT NHA MÌNH CẦN NGAY. THANKS!
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)