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HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
\(\begin{array}{l}a)x - \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right) = \dfrac{9}{{20}}\\x = \dfrac{9}{{20}} + \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right)\\x = \dfrac{9}{{20}} + \dfrac{{25}}{{20}} - \dfrac{{28}}{{20}}\\x = \dfrac{{6}}{{20}}\\x = \dfrac{{ 3}}{{10}}\end{array}\)
Vậy \(x = \dfrac{{ 3}}{{10}}\)
\(\begin{array}{*{20}{l}}{b)9 - x = \dfrac{8}{7} - \left( { - \dfrac{7}{8}} \right)}\\\begin{array}{l}9 - x = \dfrac{8}{7} + \dfrac{7}{8}\\9 - x = \dfrac{{64}}{{56}} + \dfrac{{49}}{{56}}\\9 - x = \dfrac{{113}}{{56}}\end{array}\\{x = 9 - \dfrac{{113}}{{56}}}\\{x = \dfrac{{504}}{{56}} - \dfrac{{113}}{{56}}}\\{x = \dfrac{{391}}{{56}}}\end{array}\)
Vậy \(x = \dfrac{{391}}{{56}}\)
VN
1
2 tháng 6 2022
\(\dfrac{3}{7}< \left|x-\dfrac{6}{7}\right|< \dfrac{8}{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{8}{7}< x-\dfrac{6}{7}< -\dfrac{3}{7}\\\dfrac{3}{7}< x-\dfrac{6}{7}< \dfrac{8}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\dfrac{2}{7}< x< \dfrac{3}{7}\\\dfrac{9}{7}< x< 2\end{matrix}\right.\)
HT
25 tháng 7 2017
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
M
5 tháng 1 2016
a) Vì (2x - 5)2000 và (3y + 4)2002 đều có số mũ là chẵn => (2x - 5)2000 \(\ge\) 0; (3y + 4)2002 \(\ge\) 0
Mà tổng trên lại \(\le\) 0
=> (2x - 5)2000 = (3y + 4)2002 = 0
=> 2x - 5 = 3y + 4 = 0
=> x = 2,5; y = \(\frac{-4}{3}\)
b) x = 18 - 0,8 : \(\frac{1,5}{\frac{3}{2}.\frac{4}{10}.\frac{50}{2}}\)+ \(\frac{1}{4}\)+ \(\frac{1+0,5.4}{6-\frac{46}{23}}\)
= 18 - \(\frac{8}{10}:\frac{1,5}{15}+\frac{1}{4}+\frac{3}{4}\)
= \(18-8+1=11\)
30 tháng 3 2020
\(\frac{3}{7}\cdot15\cdot\frac{1}{3}+\frac{3}{7}\cdot5\cdot\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)
\(\Leftrightarrow\frac{15}{7}+\frac{6}{7}\le x\le-6\cdot\frac{-5}{3}\)
\(\Leftrightarrow3\le x\le10\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
20 tháng 9 2020
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
20 tháng 9 2020
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
16 tháng 8 2019
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
16 tháng 8 2019
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
\(\frac{3}{7}< \left|x-\frac{6}{7}\right|< \frac{8}{7}\)
\(\Rightarrow\hept{\begin{cases}\left|x-\frac{6}{7}\right|< \frac{8}{7}\\\left|x-\frac{6}{7}\right|>\frac{3}{7}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-\frac{6}{7}< \frac{8}{7}\\x-\frac{6}{7}< \frac{-8}{7}\end{cases}}\) và \(\orbr{\begin{cases}x-\frac{6}{7}>\frac{3}{7}\\x-\frac{6}{7}>-\frac{3}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x< \frac{14}{7}\\x< \frac{-2}{7}\end{cases}}\) và \(\orbr{\begin{cases}x>\frac{9}{7}\\x>\frac{3}{7}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{9}{7}< x< \frac{14}{7}\\x< \frac{-2}{7}\end{cases}}\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !