3x-8; CHIA HẾT CHO x+1
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MT
0
MV
28 tháng 4 2017
\(\dfrac{3x}{2\cdot5}+\dfrac{3x}{5\cdot8}+\dfrac{3x}{8\cdot11}+\dfrac{3x}{11\cdot14}=\dfrac{1}{21}\\ x\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\right)=\dfrac{1}{21}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ x\cdot\dfrac{3}{7}=\dfrac{1}{21}\\ x=\dfrac{1}{21}:\dfrac{3}{7}\\ x=\dfrac{1}{9}\)
AH
Akai Haruma
Giáo viên
23 tháng 12 2020
Lời giải:
\(3x(3x-8)-9x^2+8=0\)
\(\Leftrightarrow 9x^2-24x-9x^2+8=0\)
\(\Leftrightarrow 8-24x=0\Leftrightarrow x=\frac{1}{3}\)
23 tháng 12 2020
3x ( 3x - 8 ) - 9x2 + 8 = 0
9x2 - 24x - 9x2 + 8 = 0
( 9x2 - 9x2 ) - 24x + 8 = 0
- 24x + 8 = 0
-24x = -8
x = \(\dfrac{1}{3}\)
Vậy x = \(\dfrac{1}{3}\)
LT
1
AH
Akai Haruma
Giáo viên
1 tháng 10 2020
Lời giải:
ĐK: $x\geq \frac{1}{3}$
PT $\Leftrightarrow \sqrt{(3x-1)+6\sqrt{3x-1}+9}+\sqrt{(3x-1)-6\sqrt{3x-1}+9}=3x+4$
$\Leftrightarrow \sqrt{(\sqrt{3x-1}+3)^2}+\sqrt{(\sqrt{3x-1}-3)^2}=3x+4$
$\Leftrightarrow |\sqrt{3x-1}+3|+|\sqrt{3x-1}-3|=3x+4$
Nếu $x\geq \frac{10}{3}$ thì:
$\sqrt{3x-1}+3+\sqrt{3x-1}-3=3x+4$
$\Leftrightarrow 2\sqrt{3x-1}=3x+4$
$\Leftrightarrow 2\sqrt{3x-1}=(3x-1)+5$
$\Leftrightarrow (\sqrt{3x-1}-1)^2=-4< 0$ (vô lý)
Nếu $\frac{1}{3}\leq x< \frac{10}{3}$ thì:
$\sqrt{3x-1}+3+3-\sqrt{3x-1}=3x+4$
$\Leftrightarrow 2=3x\Leftrightarrow x=\frac{2}{3}$ (thỏa mãn)
Vậy.......
ND
1
19 tháng 10 2021
\(\left(3x-8\right)^{10}=\left(3x-8\right)^2\\ \Rightarrow\left(3x-8\right)^{10}-\left(3x-8\right)^2=0\\ \Rightarrow\left(3x-8\right)^2\left[\left(3x-8\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(3x-8\right)^2=0\\\left(3x-8\right)^8=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-8=0\\3x-8=1\\3x-8=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(243\le3^{2x-1}\le3^8\\ \Rightarrow3^5\le3^{2x-1}\le3^8\\ \Rightarrow2x-1\in\left\{5;6;7;8\right\}\\ \Rightarrow x\in\left\{3;\dfrac{7}{2};4;\dfrac{9}{2}\right\}\)
PT
1
5 tháng 8 2018
\(\left(3x+4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)
\(27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)
\(27x^3-27x^3+108x^2+24x^2+144x+72x=64-64=0\)
\(132x^2+216x=0\)
\(x\left(132x+216\right)=0\)
\(\Rightarrow x=\hept{\begin{cases}0\\\frac{216}{132}=\frac{18}{11}\end{cases}}\)
4 tháng 5 2018
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
\(3x.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)
\(3x.\frac{1}{7}=\frac{1}{21}\)
\(\frac{3}{7}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{3}{7}\)
\(x=\frac{7}{81}\)
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
3x-8 chia hết cho x+1
=> 3(x+1)-11 chia hết cho x+1
=> 11 chia hết cho x+1
=> x+1 thuộc Ư(11)={1;-1;11;-11}
=> x thuộc {0;-2;10;-12}
Ta có:
3x - 8 = 3x + 3 - 11 = 3(x + 1) - 11
Để (3x - 8) ⋮ (x - 1) thì 11 ⋮ (x - 1)
⇒ x - 1 ∈ Ư(11) = {-11; -1; 1; 11}
⇒ x ∈ {-10; 0; 2; 12}
(Nếu chỉ tìm x là số tự nhiên thì x ∈ {2; 12})