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23 tháng 11 2021
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 10 2023
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 10 2023
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
4 tháng 10 2021
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
18 tháng 8 2022
a: =>5/42-x=11/13-15/28+11/13=421/364
=>x=-1193/1092
b: =>\(\dfrac{7}{2}-2x=7+\dfrac{6}{5}-3-\dfrac{2}{5}-1-\dfrac{4}{5}=3\)
=>2x=1/2
=>x=1/4
c: =>|2x-1/3|=-1/3(vô lý)
d: =>2x-1=-3
=>2x=-2
hay x=-1
e: =>2x=16
hay x=8
15 tháng 5 2022
a: \(\Leftrightarrow\left(2x-2\right)\cdot\dfrac{3}{4}=-6-\dfrac{1}{3}=-\dfrac{19}{3}\)
\(\Leftrightarrow2x-2=-\dfrac{19}{3}:\dfrac{3}{4}=-\dfrac{76}{9}\)
=>2x=-58/9
hay x=-29/9
b: \(\Leftrightarrow\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay x=-2/11
e: \(\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\end{matrix}\right.\)
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 7 2023
\(\dfrac{2}{3}-\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{2}\left(2x+1\right)=5\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{2}-x-\dfrac{1}{2}=5\)
\(\Rightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)
\(\Rightarrow-\dfrac{4}{3}x=\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{13}{3}:-\dfrac{4}{3}\)
\(\Rightarrow x=-\dfrac{13}{4}\)
\(\left(\dfrac{2x-3}{2x}\right)^2=36\\ \left(\dfrac{2x-3}{2x}\right)^2=6^2\\ \dfrac{2x-3}{2x}=6\\ \dfrac{2x}{2x}-\dfrac{3}{2x}=6\\ 1-\dfrac{3}{2x}=6\\ \dfrac{3}{2x}=-5\\ \dfrac{3}{2}.\dfrac{1}{x}=5\\ \dfrac{1}{x}=\dfrac{10}{3}\\ 1:x=\dfrac{10}{3}\\ x=\dfrac{3}{10}=0,3\)
\(\left(\dfrac{2x-3}{2x}\right)^2=36\)
\(\Rightarrow\left(\dfrac{2x-3}{2x}\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2x-3}{2x}=6\\\dfrac{2x-3}{2x}=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{2x}{2x}-\dfrac{3}{2x}=6\\\dfrac{2x}{2x}-\dfrac{3}{2x}=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{3}{2x}=6\\1-\dfrac{3}{2x}=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2x}=1-6\\\dfrac{3}{2x}=1--6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2x}=1-6\\\dfrac{3}{2x}=1+6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2x}=-5\\\dfrac{3}{2x}=7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}\cdot\dfrac{1}{x}=-5\\\dfrac{3}{2}\cdot\dfrac{1}{x}=7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{x}=-5:\dfrac{3}{2}\\\dfrac{1}{x}=7:\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{x}=-\dfrac{10}{3}\\\dfrac{1}{x}=\dfrac{14}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}1:x=-\dfrac{10}{3}\\1:x=\dfrac{14}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1:-\dfrac{10}{3}\\x=1:\dfrac{14}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{10}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{3}{10};\dfrac{3}{14}\right\}\)