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b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
a) Vì \(\left(2.x+3\right)^2=\dfrac{9}{121}\Rightarrow\left\{{}\begin{matrix}2.x+3=\dfrac{3}{11}\\2.x+3=-\dfrac{3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)
b) Vì \(\left(3.x-1\right)^3=-\dfrac{8}{27}\Rightarrow3.x-1=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{9}\)
Bài 1:
\((1-2x)^2=9=3^2=(-3)^2\)
\(\Rightarrow \left[\begin{matrix} 1-2x=3\\ 1-2x=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=2\end{matrix}\right.\)
Bài 2:
\((x+5)^3=-64=(-4)^3\)
\(\Rightarrow x+5=-4\Rightarrow x=-9\)
Bài 3:
\((3x-5)^2=16=4^2=(-4)^2\)
\(\Rightarrow \left[\begin{matrix} 3x-5=4\\ 3x-5=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=\frac{1}{3}\end{matrix}\right.\)
Bài 4:
\((x-1)^3=27=3^3\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
Bài 5:
\(x^2+x=0\Leftrightarrow x(x+1)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.\)
Bài 6:
\(5^{x+2}=625=5^4\)
\(\Rightarrow x+2=4\Rightarrow x=2\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
a) \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{10}{12}\Rightarrow\dfrac{x}{12}=\dfrac{11}{12}\Rightarrow x=11\)
b) \(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\Rightarrow\dfrac{10}{15}-\dfrac{19}{15}x=\dfrac{-3}{5}\Rightarrow\dfrac{-19}{15}x=\dfrac{-13}{15}\Rightarrow x=\dfrac{13}{19}\)
c) \(\dfrac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x=-2187\Rightarrow x=7\)
d) \(2^{x-1}=16\Rightarrow x-1=4\Rightarrow x=5\)
e) \(\left(x-1\right)^2=25\Rightarrow x-1=5\Rightarrow x=6\)
g) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\Rightarrow x=\dfrac{1}{12}\\x+\dfrac{1}{2}=0\Rightarrow x=\dfrac{-1}{2}\end{matrix}\right.\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
a. \(27\left(3x-1\right)^3=-\left|-8\right|\)
\(\Rightarrow27\left(3x-1\right)^3=-8\)
\(\Rightarrow\left(3x-1\right)^3=-8:27\)
\(\Rightarrow\left(3x-1\right)^3=\dfrac{-8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow3x-1=\dfrac{-2}{3}\)
\(\Rightarrow3x=\dfrac{-2}{3}+1=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}:3=\dfrac{1}{9}\)
Vậy...
b. \(\dfrac{2x}{-15}=\dfrac{-120}{x}\)
\(\Leftrightarrow2x.x=\left(-15\right).\left(-120\right)\)
\(\Rightarrow2.x^2=1800\)
\(\Rightarrow x^2=1800:2=900\)
\(\Rightarrow x^2=\left(\pm30\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy...