tìm GTLN:
G= -11+2/-5+/3x-1/
K
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PK
0
TT
1
27 tháng 4 2019
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(=>\frac{3x}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{1}{2}-\frac{1}{14}\right]=\frac{1}{21}\)
\(=>x\cdot\frac{3}{7}=\frac{1}{21}\Leftrightarrow x=\frac{1}{9}\)
NT
3
XO
9 tháng 5 2021
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
=> \(x\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)=\frac{1}{21}\)
=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
=> \(x.\frac{3}{7}=21\)
=> x = 49
Vậy x = 49
PT
4
H
22 tháng 12 2022
`x . 3/7 = 2/3`
`=>x= 2/3 : 3/7`
`=>x= 2/3 . 7/3`
`=>x=14/9`
`-----------`
`x : 8/11 = 11/3`
`=>x= 11/3 . 8/11`
`=>x= 88/33`
`=>x=8/3`
`-----------`
`4/7 . x - 2/3 = 1/5`
`=> 4/7 . x = 1/5 +2/3`
`=>4/7 . x =3/15 + 10/15`
`=>4/7 . x =13/15`
`=>x= 13/15 : 4/7`
`=>x= 13/15 xx 7/4`
`=>x= 91/60`
AH
Akai Haruma
Giáo viên
22 tháng 12 2022
Lời giải:
$x.\frac{3}{7}=\frac{2}{3}$
$x=\frac{2}{3}: \frac{3}{7}=\frac{14}{9}$
-----------
$x: \frac{8}{11}=\frac{11}{3}$
$x=\frac{11}{3}.\frac{8}{11}=\frac{8}{3}$
-----------
$\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}$
$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$
$x=\frac{13}{15}: \frac{4}{7}=\frac{91}{60}$
B
7 tháng 7 2019
\(\left(x+2\right).\left(3x-2\right)-\left(3x-1\right).\left(x-5\right)=11\)
\(\Rightarrow3x^2-2x+6x-4-\left(3x^2-15x-x+5\right)=11\)
\(\Rightarrow3x^2-2x+6x-4-3x^2+15x+x-5=11\)
\(\Rightarrow20x-9=11\)
\(\Rightarrow20x=20\Rightarrow x=1\)
7 tháng 7 2019
(x + 2)(3x - 2) - (3x - 1)(x - 5) = 11
=> 3x2 - 2x + 6x - 4 - 3x2 + 15x + x - 5 = 11
=> 20x - 9 = 11
=> 20x = 11 + 9
=> 20x = 20
=> x = 20 : 20
=> x = 1
NN
3
29 tháng 6 2017
\(\left(\frac{1}{3}-\frac{3}{2}+\frac{5}{6}\right):\left(\frac{11}{3}-3x\right)=1:\left(\frac{1}{4}-\frac{5}{2}+\frac{4}{3}\right)\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{3}{2}+\frac{5}{6}\right):\left(\frac{11}{3}-3x\right)=\frac{-11}{12}\)
\(\Leftrightarrow\frac{-1}{3}:\left(\frac{11}{3}-3x\right)=\frac{-11}{12}\)
\(\frac{11}{3}-3x=\frac{-1}{3}-\frac{-11}{12}\)
\(\frac{11}{3}-3x=\frac{7}{12}\)
\(3x=\frac{11}{3}-\frac{7}{12}\)
\(3x=\frac{27}{12}\)
\(x=\frac{27}{12}:3\)
\(x=\frac{3}{4}\)
DP
29 tháng 6 2017
\(\left(\frac{1}{3}-\frac{3}{2}+\frac{5}{6}\right)\div\left(\frac{11}{3}-3x\right)=1\div\left(\frac{1}{4}-\frac{5}{2}+\frac{4}{3}\right)\)
\(\left(\frac{2}{6}-\frac{9}{6}+\frac{5}{6}\right)\div\left(\frac{11}{3}-3x\right)=1\div\left(\frac{3}{12}-\frac{30}{12}+\frac{16}{12}\right)\)
\(-\frac{1}{3}\div\left(\frac{11}{3}-3x\right)=1\div-\frac{11}{12}\)
\(-\frac{1}{3}\div\frac{11}{3}-3x=-\frac{12}{11}\)
\(\frac{11}{3}-3x=\left(-\frac{1}{3}\right)\div\left(-\frac{12}{11}\right)\)
\(\frac{11}{3}-3x=\frac{11}{36}\)
\(3x=\frac{11}{3}-\frac{11}{36}\)
\(3x=\frac{121}{36}\)
\(\Rightarrow x=\frac{121}{108}\)
4 tháng 4 2015
1) A = 3 - 4x2 - 4x = - (4x2 + 4x +1) + 4 = - (2x+1)2 + 4
Vì - (2x+1)2 \(\le\)0 nên A = - (2x+1)2 + 4 \(\le\) 4 vậy maxA = 4 khi 2x+1 = 0 => x = -1/2
b) ta có x2 + 6x + 11 = x2 + 2.3x + 9 + 2 = (x+3)2 + 2 \(\ge\) 0 + 4 = 4
=> \(B=\frac{1}{x^2+6x+11}\le\frac{1}{4}\) vậy maxB = 1/4 khi x = -3
2) a) 3x2 - 3x + 1 = 3.(x2 - x) + 1 = 3.(x2 - 2.x\(\frac{1}{2}\) + \(\frac{1}{4}\)) + \(\frac{1}{4}\) = 3.(x - \(\frac{1}{2}\) )2 + \(\frac{1}{4}\) \(\ge\)0 + \(\frac{1}{4}\)= \(\frac{1}{4}\)
vậy min(3x2 - 3x + 1) = 1/4 khi x = 1/2
b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b| \(\ge\) |a - b|. dấu = khi a.b < 0
ta có: |3x - 3| + |3x - 5| \(\ge\) |3x - 3 - (3x - 5)| = |2| = 2
vậy min = 2 khi (3x - 3)(3x - 5) < 0 hay 1< x < 5/3
7 tháng 2 2022
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
\(G=-11+\frac{2}{-5}+\left|3x-1\right|\)
+)Ta có:\(\left|3x-1\right|\ge0;\forall x\)
\(\Rightarrow-\frac{2}{5}+\left|3x-1\right|\ge-\frac{2}{5}\)
\(\Rightarrow-11+-\frac{2}{5}+\left|3x-1\right|\le-11-\frac{2}{5}\)
\(\Rightarrow-11+-\frac{2}{5}+\left|3x-1\right|\le-\frac{57}{5}\)
+)GTLN của G bằng \(-\frac{57}{5}\)khi
\(\left|3x-1\right|=0\)
\(\Rightarrow3x-1=0\)
\(3x=1\)
\(x=\frac{1}{3}\)
Vậy \(x=\frac{1}{3}\)
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