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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
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
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.
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) \(7-\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}=7\)
\(\Rightarrow x=\left(\sqrt{7}\right)^2\)
b) \(5\sqrt{x}+1=40\)
\(\Rightarrow5\sqrt{x}=39\)
\(\Rightarrow\sqrt{x}=7,8\)
\(\Rightarrow x=\left(\sqrt{7,8}\right)^2\)
c) \(\dfrac{5}{12}\sqrt{x}-\dfrac{1}{6}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{12}\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{x}=1,2\)
\(\Rightarrow x=\left(\sqrt{1,2}\right)^2\)
d) \(4x^2-1=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\Rightarrow x=0,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(\sqrt{x+1}-2=0\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Rightarrow x+1=1,414\)
\(\Rightarrow x=0,414\)
f) \(2x^2+0,82=1\)
\(\Rightarrow2x^2=0,18\)
\(\Rightarrow x^2=0,09\)
\(\Rightarrow x=\pm0,3\)
g) Không có kết quả
1,\(\dfrac{a}{b}=\dfrac{x}{y}\) khi ay=bx
2,
a,x=\(\dfrac{-1.12}{4}\)
x=\(\dfrac{-12}{4}=-3\)
b,\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
\(\Rightarrow\)2x-1=5
2x=6
x=6:2=3
c,\(\dfrac{4}{7}\).x=\(\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
x=\(\dfrac{13}{15}.\dfrac{7}{4}=\dfrac{91}{60}\)
3,ta có:\(5^{202}=\left(5^2\right)^{101}\)=\(25^{101}\)
2\(^{505}\)=\(\left(2^5\right)^{101}\)=\(32^{101}\)
vì 25<32 nên \(25^{101}< 32^{101}\) hay \(5^{202}< 2^{505}\)
1) \(\dfrac{a}{b}=\dfrac{x}{y}\) khi \(a.y=b.x\)
2) \(a,\dfrac{x}{12}=\dfrac{-1}{4}\)
\(\Rightarrow4x=-12\)
\(\Rightarrow x=-\dfrac{12}{4}=-3\)
Vậy x = -3
\(b,\left(\dfrac{1}{3}\right)^{2x-1}=\dfrac{1}{243}\)
\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
\(\Rightarrow2x-1=5\)
\(\Rightarrow x=\dfrac{5-1}{2}=2\)
Vậy x = 2
\(c,\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{4}{7}=1\dfrac{31}{60}\)
Vậy \(x=1\dfrac{31}{60}\)
3) So sánh \(5^{202}\) và \(2^{505}\)
\(5^{202}=\left(5^2\right)^{101}=25^{101}\)
\(2^{505}=\left(2^5\right)^{101}=32^{101}\)
\(\Rightarrow25^{101}< 32^{101}\)
\(\Rightarrow5^{202}< 2^{505}\)
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
a: =>x*7/4+3/2=-4/5
=>x*7/4=-4/5-3/2=-8/10-15/10=-23/10
=>x=-23/10:7/4=-23/10*4/7=-92/70=-46/35
b: =>x*9/20=1/7+1/8=15/56
=>x=15/56:9/20=15/56*20/9=25/42
c: |x|=3,5
=>x=3,5 hoặc x=-3,5
d: |x|=-2,7
=>x thuộc rỗng
e: =>|x-1|=3-0,73=2,27
=>x-1=2,27 hoặc x-1=-2,27
=>x=-1,27 hoặc x=3,27
f: \(\Leftrightarrow7\cdot11x+11=0\)
=>77x=-11
=>x=-1/7
l: =>|x+3/4|=-2+5=3
=>x+3/4=3 hoặc x+3/4=-3
=>x=-15/4 hoặc x=9/4