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Y
20 tháng 7 2019
Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
21 tháng 7 2019
Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
15 tháng 2 2020
A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
S
2 tháng 7 2019
a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)
\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)
\(=\frac{12}{15}\)
\(=\frac{4}{5}\)
c, \(\frac{3}{8}.3\frac{1}{3}\)
\(=\frac{3}{8}.\frac{10}{3}\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)
\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)
\(=\frac{-3}{5}+\frac{-60}{10}\)
\(=\frac{-3}{5}+\frac{-30}{5}\)
\(=\frac{-33}{5}\)
e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)
\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)
\(=\frac{2}{5}.10\)
\(=4\)
f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.-14\)
\(=-6\)
~Study well~
#KSJ
TA
13 tháng 9 2020
a không có tích để tìm x.
b)\(\frac{1}{12}.x-75\%.x=-1\frac{2}{3}\)
\(x.\left(\frac{1}{12}-\frac{9}{12}\right)=\frac{-1}{3}\)
\(x.\frac{-2}{3}=\frac{-1}{3}\)
\(x=\frac{-1}{3}:\frac{-2}{3}\)
\(x=\frac{-1}{-2}\)
c)\(\left(\frac{-2x}{5}+1\right):-5=\frac{-1}{25}\)
\(\left(\frac{5-2x}{5}\right)=\frac{-1}{25}.\frac{1}{-5}\)
\(\left(\frac{5-2x}{5}\right)=\frac{-1}{-125}\)
\(\frac{2x}{5}=\frac{-1}{-125}-1\)
\(\frac{2x}{5}=\frac{-126}{-125}\)
\(\frac{x.2}{5}=\frac{-126}{-125}\)
\(x=-63\)
Mới cuối cấp I thôi chị ơi.
NC
25 tháng 8 2020
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
25 tháng 8 2020
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
15 tháng 1 2020
bài 1 :
a, A = 3|2x - 1| - 5 = 0
có 3|2x - 1| > 0
=> A > -5
xét A = -5 khi
|2x - 1| = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = 1/2
vậy Min A = -5 khi x = 1/2
b, c, d, làm tương tự
L
17 tháng 1 2020
Bài 1:
\(a)A=3|2x-1|-5\)
Vì \(|2x-1|\ge0\)\(\forall x\)
\(\Rightarrow3|2x-1|\ge0\) \(\forall x\)
\(\Rightarrow3|2x-1|-5\ge-5\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=-5\Leftrightarrow x=\frac{1}{2}\)
\(b)x^2+3|y-2|-1\)
Vì \(\hept{\begin{cases}x^2\ge0\forall x\\3|y-2|\ge0\forall y\end{cases}}\)
\(\Rightarrow x^2+3|y-2|-1\ge-1\) \(\forall x,y\)
Dấu '=' xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x^2=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy \(Min_B=-1\Leftrightarrow x=0,y=2\)
\(c)\left(2x^2+1\right)^4-3\)
Vì \(\left(2x^2+1\right)^4\ge0\)\(\forall x\)
\(\Rightarrow\left(2x^2+1\right)^4-3\ge-3\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x^2+1=0\)
\(\Leftrightarrow2x^2=-1\)
\(\Leftrightarrow x^2=-\frac{1}{2}\left(voli\right)\)
Vậy không tìm được gt x
\(d)D=|x-\frac{1}{2}|+\left(y+2\right)^2+11\)
Vì \(\hept{\begin{cases}|x-\frac{1}{2}|\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\)
\(\Rightarrow|x-\frac{1}{2}|+\left(y+2\right)^2+11\ge11\) \(\forall x,y\)
Dấu '=' xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-2\end{cases}}\)
Vậy \(Min_D=11\Leftrightarrow x=\frac{1}{2},y=-2\)
L
17 tháng 1 2020
Bài 2:
\(a)A=10-5|x-2|\)
Vì \(|x-2|\ge0\)\(\forall x\)
\(\Rightarrow5|x-2|\ge0\)\(\forall x\)
\(\Rightarrow\)\(10-5|x-2|\le10\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_A=10\Leftrightarrow x=2\)
\(b)B=5-|2x-1|^2\)
Vì \(|2x-1|^2\ge0\)\(\forall x\)
\(\Rightarrow5-|2x-1|^2\le5\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Max_B=5\Leftrightarrow x=\frac{1}{2}\)
\(c)C=\frac{1}{|x-2|+3}\)
Vì \(|x-2|\ge0\)\(\forall x\)
\(\Rightarrow|x-2|+3\ge3\) \(\forall x\)
\(\Rightarrow\frac{1}{|x-2|+3}\le\frac{1}{3}\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_C=\frac{1}{3}\Leftrightarrow x=2\)
T
25 tháng 10 2018
\(a.9\cdot3^2\cdot\frac{1}{81}=\frac{3^2.3^2.1}{3^4}=\frac{3^4}{3^4}=1\)
\(b.2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{9}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(\frac{-9}{8}\right)\)
\(=\frac{5}{2}+\frac{-9}{14}=\frac{13}{7}\)
\(c.3,75.\left(7,2\right)+2,8.\left(3,75\right)\)
\(=3,75.\left(7,2+2,8\right)\)
\(=3,75.10=37,5\)
\(d.\left(\frac{-5}{13}\right).\frac{3}{7}+\left(\frac{-8}{13}\right).\frac{3}{7}+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left[\left(\frac{-5}{13}\right)+\left(\frac{-8}{13}\right)\right]+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left(-1\right)+\frac{-4}{7}\)
\(=\frac{-3}{7}+-\frac{4}{7}=-1\)
\(e.\sqrt{81}-\frac{1}{8}.\sqrt{64}+\sqrt{0,04}\)
\(=9-\frac{1}{8}.8+0,2\)
\(=9-1+0,2=8+0,2=8,2\)
(2/3-1/2)x=4/5+7/5
1/6.x=12/5
x=72/5