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a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
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: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
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
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/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)
\(\Leftrightarrow5x+5=4x+6\)
\(\Leftrightarrow5x-4x=6-5\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ...
b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)
Mà với \(\forall x;y;z\) ta có :
\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy ...
c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)
\(\Leftrightarrow x-2=5-3x\)
\(\Rightarrow x+3x=5+2\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
Vậy ......
d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)
\(\Leftrightarrow\left(x+2\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy ...
e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)
\(\Leftrightarrow\left(x-1\right)^2=-100\)
Lại có : \(\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\) k tồn tại x
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
a: =>13/15x=3/4-1/2=1/4
=>x=15/52
b: =>x-3=4
=>x=7
c: =>2x+1=9
=>2x=8
=>x=4
d: =>x+3=-2
=>x=-5
e: =>(x+6)(x-4)=0
=>x=4 hoặc x=-6
f: =>(x-3)(x-7)=0
=>x=3 hoặc x=7