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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) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\Rightarrow x=\dfrac{1}{12}\\x+\dfrac{1}{2}=0\Rightarrow x=\dfrac{-1}{2}\end{matrix}\right.\)
a) x = \(\dfrac{-64}{3}\)
b) x = -3,5
c) x = 80
d) x = -1.162
e) x = 0,9436
g) x \(\in\varnothing\)
a) 16/3 : x = -1/4
=> x = 16/3 : (-1/4)
=> x = 16/3 . (-4)
=> x = -64/3
Vậy x= -64/3
b)2x - 13 = -8
=> 2x = (-8) + 1
=> 2x = -7
=> x = -7/2
d) 0,944 - 2x = 3,268
=> 2x = 0,944 - 3,268
=> 2x = -2,324
=> x = (-2,324) : 2
=> x = -1,162
g) \(\sqrt{5^2-3^2}=-\sqrt{81-x}\)
=> \(\sqrt{25-9}\)= \(-\sqrt{81-x}\)
=> \(\sqrt{16}\)=\(-\sqrt{81-x}\)
=> 4=\(-\sqrt{81-x}\)
tới đây mik bí r hk bt lm nữa
a: Đặt A=0
=>-2/3x=5/9
hay x=-5/6
b: Đặt B(x)=0
=>(x-2/5)(x+2/5)=0
=>x=2/5 hoặc x=-2/5
c: Đặt C(X)=0
\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)
\(\Leftrightarrow x^3=-\dfrac{8}{27}\)
hay x=-2/3
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
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) \(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(\Leftrightarrow A=-5,13:\left(5\dfrac{5}{28}-\dfrac{85}{36}+1\dfrac{16}{63}\right)\)
\(\Leftrightarrow A=-5,13:\left(\dfrac{355}{126}+1\dfrac{16}{63}\right)\)
\(\Leftrightarrow A=-5,13:\dfrac{57}{14}\)
\(\Leftrightarrow A=\dfrac{-63}{50}\)
b) \(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)
\(\Leftrightarrow B=\left(\dfrac{19}{3}+19,5.\dfrac{3}{13}\right).\dfrac{2}{3}\)
\(\Leftrightarrow B=\left(\dfrac{19}{3}+\dfrac{9}{2}\right).\dfrac{2}{3}\)
\(\Leftrightarrow B=\dfrac{65}{6}.\dfrac{2}{3}\)
\(\Leftrightarrow B=\dfrac{65}{9}\)
a: \(\Leftrightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{9}{25}-\dfrac{81}{625}=\dfrac{144}{625}\)
=>x=2
b: =>3x-1=-4
=>3x=-3
hay x=-1
a: \(\Leftrightarrow7^x\cdot49+7^x\cdot\dfrac{2}{7}=345\)
\(\Leftrightarrow7^x=7\)
hay x=1
c: \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
=>x-1=2
hay x=3
d: \(\dfrac{25}{5^x}=\dfrac{1}{125}\)
\(\Leftrightarrow5^x=5^2\cdot5^3=5^5\)
hay x=5
a/ \(\dfrac{\left(-3\right)^x}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^x=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^x=-2187\)
\(\Leftrightarrow\left(-3\right)^x=\left(-3\right)^7\)
\(\Leftrightarrow x=7\)
Vậy ...
b/ \(2^{x-1}=16\)
\(\Leftrightarrow2^{x-1}=2^4\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=5\)
Vậy ....
c/ \(\left(x-1\right)^2=25\)
\(\Leftrightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy ....
d/ \(0,2-\left|4,2-2x\right|=0\)
\(\Leftrightarrow\left|4,2-2x\right|=0,2\)
\(\Leftrightarrow\left[{}\begin{matrix}4,2-2x=0,2\\4,2-2x=-0,2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\)
Vậy ............
e, \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
\(\Leftrightarrow\dfrac{5}{3}.\dfrac{4}{x}=20\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy ..
a) \(\dfrac{\left(-3\right)^x}{81}=-27\)
⇔ \(\dfrac{\left(-3\right)^x}{81}=\dfrac{-2187}{81}\)
⇔ (-3)x = -2187
⇔ (-3)x = (-3)7
⇔ x = 7
b) 2x-1 = 16
⇔ 2x-1 = 24
⇔ x - 1 = 4
⇔ x = 4 + 1
⇔ x = 5
c) (x - 1)2 = 25
⇔ \(\left[{}\begin{matrix}\left(x-1\right)^2=5^2\\\left(x-1\right)^2=\left(-5\right)^2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=5+1\\x=-5+1\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
d) 0,2 - |4,2 - 2x| = 0
⇔ |4,2 - 2x| = 0,2
⇔ \(\left[{}\begin{matrix}4,2-2x=0,2\\4,2-2x=-0,2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}2x=4,2-0,2\\2x=4,2-\left(-0,2\right)\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\)
e) \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
⇔ \(\dfrac{5}{3}.\dfrac{4}{x}=20\)
⇔ \(\dfrac{20}{3x}=20\)
⇔ 3x = 1
⇔ x = \(\dfrac{1}{3}\)
a) \(x=3\)
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