Tìm x biết :
\(-2.\left|\frac{2}{5}x-\frac{3}{10}\right|=3\frac{1}{5}\)
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Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
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:
\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\cdot\frac{998}{3003}\)
\(=\frac{499}{3003}\)
Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\); \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)
\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\);
=> Phương trình tương đương:
\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
=> x=15
Đáp số: x=15
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
1. \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}.\frac{12}{11}=\frac{72}{55}\)
2. 2x+2 . 3x+1 . 5x = 10800
=> 2x . 22 . 3x . 3 . 5x = 10800
=> ( 2 . 3 . 5 )x . 12 = 10800
=> 30x = 900
=> 30x = 302
=> x = 2
1)\(-\frac{3}{2}\left(\frac{4}{5}-\frac{2}{3}\right)+x=4\left(x-\frac{2}{3}\right)\)
\(\Leftrightarrow-\frac{3}{2}\cdot\frac{2}{15}+x=4x-\frac{8}{3}\)
\(\Leftrightarrow-\frac{1}{5}+x=4x-\frac{8}{3}\)
\(\Leftrightarrow x-4x=-\frac{8}{3}+\frac{1}{5}\)
\(\Leftrightarrow-3x=-\frac{37}{15}\)
\(\Leftrightarrow x=\frac{37}{45}\)
2)\(2\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow3-2x-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow-2x-7x=-\frac{1}{4}-3+\frac{1}{3}\)
\(\Leftrightarrow-9x=-\frac{35}{12}\)
\(\Leftrightarrow x=\frac{35}{108}\)
3)\(\frac{1}{5}\left(-\frac{3}{5}10\right)+5x=x-\frac{2}{3}\)
\(\Leftrightarrow-\frac{6}{5}+5x=x-\frac{2}{3}\)
\(\Leftrightarrow5x-x=-\frac{2}{3}+\frac{6}{5}\)
\(\Leftrightarrow4x=\frac{8}{15}\)
\(\Leftrightarrow x=\frac{2}{15}\)
4)\(-\frac{3}{2}\left(5-\frac{1}{6}\right)+4\left(x-\frac{1}{2}\right)=1\)
\(\Leftrightarrow-\frac{15}{2}+\frac{1}{4}+4x-2=1\)
\(\Leftrightarrow4x=1+\frac{15}{2}-\frac{1}{4}+2\)
\(\Leftrightarrow4x=\frac{41}{4}\)
\(\Leftrightarrow x=\frac{41}{16}\)
\(-2.\left|\frac{2}{5}x-\frac{3}{10}\right|=3\frac{1}{5}\)
\(-2.\left|\frac{2}{5}x-\frac{3}{10}\right|=\frac{16}{5}\)
\(\left|\frac{2}{5}x-\frac{3}{10}\right|=\frac{16}{5}\div\left(-2\right)\)
\(\left|\frac{2}{5}x-\frac{3}{10}\right|=-\frac{8}{5}\)
\(\Rightarrow\)\(\frac{2}{5}x-\frac{3}{10}=\frac{8}{5}\)hoặc \(-\frac{8}{5}\)
Ta xét 2 trường hợp :
Th 1 :
\(\frac{2}{5}x-\frac{3}{10}=\frac{8}{5}\)
\(\frac{2}{3}x=\frac{8}{5}+\frac{3}{10}\)
\(\frac{2}{3}x=\frac{19}{10}\)
\(x=\frac{19}{4}\)
Th 2 :
\(\frac{2}{5}x-\frac{3}{10}=-\frac{8}{5}\)
\(\frac{2}{3}x=-\frac{8}{5}+\frac{3}{10}\)
\(\frac{2}{3}x=\frac{-13}{10}\)
\(x=\frac{-13}{4}\)
Vậy \(x\in\){ \(\frac{19}{4}\); \(\frac{-13}{4}\)}