3x-14=5^5:5^3 giúp mình với moin người !
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22 tháng 12 2021
1) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Rightarrow5\left(x-2\right).3\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
KB
2 tháng 4 2022
1 ) \(lim_{x\rightarrow+\infty}\dfrac{3x^2+5}{x^3-x+2}=lim_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{5}{x^3}}{1-\dfrac{1}{x^2}+\dfrac{2}{x^3}}=0\)
2 ) \(lim_{x\rightarrow-\infty}\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\) \(=lim_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}\left(3-\dfrac{5}{x^2}\right)^3\left(\dfrac{1}{x}-1\right)^5}{3+\dfrac{1}{x^{12}}-\dfrac{1}{x^{14}}}=0\)
3 ) \(lim_{x\rightarrow+\infty}\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}=lim_{x\rightarrow+\infty}\dfrac{\left(7x^2-5\right)}{\left(3x+\sqrt{2x^2+5}\right)\left(x^2-4\right)}\)
\(=lim_{x\rightarrow+\infty}\dfrac{\dfrac{7}{x}-\dfrac{5}{x^3}}{\left(3+\sqrt{2+\dfrac{5}{x^2}}\right)\left(1-\dfrac{4}{x^2}\right)}=0\)
HL
0
KV
23 tháng 2 2023
A) 3x² - x(3x - 5) = 9
3x² - 3x² + 5x = 9
5x = 9
x = 9/5
--------------------
B) 5x² + 9x - 2 = 0
5x² + 10x - x - 2 = 0
(5x² + 10x) - (x + 2) = 0
5x(x + 2) - (x + 2) = 0
(x + 2)(5x - 1) = 0
x + 2 = 0 hoặc 5x - 1 = 0
*) x + 2 = 0
x = -2
*) 5x - 1 = 0
5x = 1
x = 1/5
Vậy x = -2; x = 1/5
---------------------
D) 4(5 - 3x) = 5x - 5
20 - 12x = 5x - 5
-12x - 5x = -5 - 20
-17x = -25
x = 25/17
--------------------
E) 2x² - 11x + 14 = 0
2x² - 4x - 7x + 14 = 0
(2x² - 4x) - (7x - 14) = 0
2x(x - 2) - 7(x - 2) = 0
(x - 2)(2x - 7) = 0
x - 2 = 0 hoặc 2x - 7 = 0
*) x - 2 = 0
x = 2
*) 2x - 7 = 0
2x = 7
x = 7/2
Vậy x = 2; x = 7/2
PT
2
13 tháng 3 2020
\(-7.\left(3x-5\right)+2.\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+7=28\)
\(\Leftrightarrow-7x=21\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3\)
7 tháng 10 2018
\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)(ĐKXĐ: \(\frac{3}{2}\le x\le\frac{5}{2}\))
\(\Leftrightarrow2\sqrt{2x-3}+2\sqrt{5-2x}=6x^2-24x+28\)
\(\Leftrightarrow6x^2-24x+28-2\sqrt{2x-3}-2\sqrt{5-2x}=0\)
\(\Leftrightarrow\left(2x-3-2\sqrt{2x-3}+1\right)+\left(5-2x-2\sqrt{5-2x}+1\right)+6x^2-24x+24=0\)
\(\Leftrightarrow\left(\sqrt{2x-3}-1\right)^2+\left(\sqrt{5-2x}-1\right)^2+6\left(x-2\right)^2=0\)
Do \(\left(\sqrt{2x-3}-1\right)^2\ge0;\left(\sqrt{5-2x}-1\right)^2\ge0;6\left(x-2\right)^2\ge0\forall x\in R\)
Nên \(\hept{\begin{cases}\sqrt{2x-3}-1=0\\\sqrt{5-2x}-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-3=1\\5-2x=1\\x=2\end{cases}}\Leftrightarrow x=2\)(t/m ĐKXĐ)
Vậy pt có nghiệm duy nhất là x=2.
TT
1
27 tháng 4 2019
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(=>\frac{3x}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{1}{2}-\frac{1}{14}\right]=\frac{1}{21}\)
\(=>x\cdot\frac{3}{7}=\frac{1}{21}\Leftrightarrow x=\frac{1}{9}\)
DN
10 tháng 4 2022
a) => 4/3x = 7/9 - 4/9 = 1/3
=> x = 1/3 : 4/3 = 1/4
b) => 5/2 - x = 9/14 : (-4/7) = -9/8
=> x = 5/2 - (-9/8) = 5/2 + 9/8 = 29/8
c) => 3x = 2 và 2/3 - 3/4 = 8/3 - 3/4 = 23/12
=> x = 23/12 : 3 = 23/36
D) => -5/6 - x = 1/4
=> x = -5/6 - 1/4 = -13/12
MH
11 tháng 4 2022
a) \(\dfrac{4}{9}+\dfrac{4}{3}x=\dfrac{7}{9}\)
\(\dfrac{4}{3}x=\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{4}{3}\)
\(x=\dfrac{1}{4}\)
b) \(\left(\dfrac{5}{2}-x\right)\left(-\dfrac{4}{7}\right)=\dfrac{9}{14}\)
\(\dfrac{5}{2}-x=\dfrac{9}{14}:\left(-\dfrac{4}{7}\right)=-\dfrac{9}{8}\)
\(x=\dfrac{5}{2}-\left(-\dfrac{9}{8}\right)\)
\(x=\dfrac{29}{8}\)
c) \(3x+\dfrac{3}{4}=2\dfrac{2}{3}\)
\(3x+\dfrac{3}{4}=\dfrac{8}{3}\)
\(3x=\dfrac{8}{3}-\dfrac{3}{4}=\dfrac{23}{12}\)
\(x=\dfrac{23}{12}:3\)
\(x=\dfrac{23}{36}\)
d) \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(-\dfrac{5}{6}-x=\dfrac{1}{4}\)
\(x=-\dfrac{5}{6}-\dfrac{1}{4}\)
\(x=-\dfrac{13}{12}\)
bằng 13
X=13 NHA