Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}
c , d giải nốt:
c) 2x + 3/6 = 7x - 13/15
=> (2x + 3) . 15 = (7x - 13) x 6
=> 30x +45 = 42x - 78
=> 30x - 42x = -78 - 45
=> -12x = -123
=> x = 41/4
d) 2-x/4 = 3x - 1/ - 3
=> (2-x) . -3 = 4. (3x - 1)
=> -6 - (-3x) = 12x - 4
(-6) + 4 = 12x - - (-3x)
=> -2 = 9x
=> x = -2/9
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow x+3=11\)
hay x=8
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)
\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)
Suy ra: x=5
d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)
\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)
\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)
\(\Leftrightarrow8^x=8^{20}\)
Suy ra: x=20
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
a) \(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\mid:2x-3=\frac{1}{4}\Rightarrow2x=\frac{13}{4}\Rightarrow x=\frac{13}{8}\left(TM\right)\\x< \frac{3}{2}\mid:3-2x=\frac{1}{4}\Rightarrow2x=\frac{11}{4}\Rightarrow x=\frac{11}{8}\left(TM\right)\end{cases}.}\)
b) \(\Leftrightarrow\left|x-1\right|=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}x\ge1\mid:x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\left(TM\right)\\x< 1\mid:1-x=\frac{3}{4}=>x=\frac{1}{4}\left(TM\right)\end{cases}}\)
c) \(\frac{3}{5\left(x-\frac{5}{6}\right)}-\frac{1}{2\left(\frac{3}{2}-1\right)}=-\frac{1}{4}\Leftrightarrow\frac{3}{\frac{5\left(6x-5\right)}{6}}-\frac{1}{2\cdot\frac{1}{2}}=-\frac{1}{4}\Leftrightarrow\frac{18}{5\left(6x-5\right)}=-\frac{1}{4}+1\)
\(\Leftrightarrow\frac{18}{5\left(6x-5\right)}=\frac{3}{4}\Leftrightarrow6x-5=\frac{24}{5}\Leftrightarrow6x=\frac{49}{5}\Leftrightarrow x=\frac{49}{30}\)
d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\Leftrightarrow2\cdot\frac{x+1-2}{2\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow\frac{x-1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow2016x-2016=2015x+2015\Leftrightarrow x=2015+2016=4031\)
Vậy x = 4031.