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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
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
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
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
a) \(\left[\left(\dfrac{3}{5}\right)^2-\left(\dfrac{2}{5}\right)^2\right]\cdot X=\left(\dfrac{1}{5}\right)^3\)
\(\left(\dfrac{3}{5}-\dfrac{2}{5}\right)\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot X=\dfrac{1}{125}\)
\(\dfrac{1}{5}\cdot1\cdot X=\dfrac{1}{125}\)
\(X=\dfrac{1}{125}:\dfrac{1}{5}=\dfrac{1}{25}\)
b) \(1\dfrac{2}{5}\cdot x+\dfrac{3}{7}=\dfrac{-4}{5}\)
\(1\dfrac{2}{5}\cdot x=\dfrac{-4}{5}-\dfrac{3}{7}\)
\(1\dfrac{2}{5}\cdot x=-\dfrac{43}{35}\)
\(x=-\dfrac{43}{35}:1\dfrac{2}{5}=-\dfrac{43}{49}\)
c) \(\left(3x-2\right)^2=9\)
*Nếu \(9=3^2\) thì:
\(3x-2=3\)
\(3x=5\Rightarrow x=\dfrac{5}{3}\)
*Nếu \(9=\left(-3\right)^2\) thì
\(3x-2=-3\)
\(3x=-1\Rightarrow x=-\dfrac{1}{3}\)
d) \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Chúc bạn học giỏi.
a)\(\dfrac{3^2-2^2}{5^2}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow\dfrac{5}{5^2}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow\dfrac{1}{5}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow x=\dfrac{1}{25}\)
b)\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{7}{5}x=-\dfrac{43}{35}\)
\(\Leftrightarrow x=\dfrac{-43}{49}\)
c)\(9x^2-12x+4=9\)
\(\Leftrightarrow9x^2-12x-5=0\)
\(\Leftrightarrow9x^2-15x+3x-5=0\)
\(\Leftrightarrow3x\left(3x-5\right)+3x-5=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
d)\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a) ( x + 5 )3 = -64
x + 5 = - 4
x = - 4 - 5
x = -9
b) (2x - 3)2=9
2x - 3 = 3
2x = 3+3
2x = 6
x = 6 : 2
x = 3
e) \(\dfrac{8}{2x}=4\)
=> 4 . 2x = 8
8x =8
x = 8 : 8
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)
=> x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(\dfrac{1}{4}.x=\dfrac{1}{32}\)
x = \(\dfrac{1}{32}:\dfrac{1}{4}\)
x = \(\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\dfrac{-1}{27}\)
a) (x + 5)3 = -64
=> (x + 5)3 = (-4)3
x + 5 = -4
x = -4 - 5
x = -9
b) (2x - 3)2 = 9
=> (2x - 3)2 = (\(\pm\)3)2
=> 2x - 3 = 3 hoặc 2x - 3 = -3
*2x - 3 = 3
2x = 3 + 3
2x = 9
x = \(\dfrac{9}{2}\)
*2x - 3 = -3
2x = -3 + 3
2x = 0
x = 0 : 2
x = 0
Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)
c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)
=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)
\(\dfrac{x}{2}=8\)
x = 8 : 2
x = 4
d) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)n . (-2)2= (-2)5
(-2)n = (-2)5 : (-2)2
(-2)n = (-2)3
Vậy n = 3
e) \(\dfrac{8}{2x}=4\)
=> 2x . 4 = 8
2x = 8 : 4
2x = 2
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)
\(x=\left(\dfrac{1}{2}\right)^3\)
\(x=\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^3\)
\(x=\dfrac{-1}{27}\).
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10