
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


a) x = \(\frac{-4}{27}:\frac{2}{3}\)
x = \(\frac{-2}{9}\)
b) x = \(\frac{-13}{5}:\frac{11}{15}\)
x = \(\frac{-39}{11}\)
c) \(\frac{1}{2}:x=-4-\frac{1}{3}\)
\(\frac{1}{2}:x=\frac{-13}{3}\)
x = \(\frac{1}{2}:\frac{-13}{3}\)
x = \(\frac{-3}{26}\)
d) \(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-7}{20}\)
\(x=\frac{1}{4}:\frac{-7}{20}\)
x = \(\frac{-5}{7}\)
tk mik nha b


a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
<=> \(x-\frac{1}{2}=\frac{1}{3}\)
<=> x = \(\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
<=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{array}\right.}\)
Vậy...
a)\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{6}\)
b)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\pm\left(\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}\)


| x - 1 | + | x + 3 | = 3 ( * )
xét : x - 1 = 0 => x = 1
x + 3 = 0 => x = -3
x - 1 < 0 => x < 1
x + 3 < 0 => x < -3
x - 1 > 0 => x > 1
x + 3 > 0 => x > -3
Lập bảng xét dấu,ta có :
x -3 1
x+3 - 0 + | +
x-1 - | - 0 +
nếu x < -3 thì * <=> : ( 1 - x ) + ( -3 - x ) = 3
1 - x + ( -3 ) - x = 3
-2x = 5
x = -5/2 ( loại )
nếu -3 \(\le\)x < 1 thì * <=> : ( 1 - x ) + ( x + 3 ) = 3
1 - x + x + 3 = 3
0x = -1 ( ko có GT x thỏa mãn )
nếu x \(\ge\)1 thì * <=> : ( x -1 ) + ( x + 3 ) = 3
x - 1 + x + 3 = 3
2x = 1
x = 1/2 ( ko có GT x thỏa mãn )
Vậy ko có GT x nào thỏa mãn bài trên.
a) 25 < 5n:5 < 625
52 < 5n:5 < 54
2 < n:5 < 4
=> n : 5 = 3
=> n = 15
b) 34 < \(\frac{1}{9}.27^n\)< 310
34 < \(\frac{27^n}{9}\)< 310
34 < 33n-2 < 310
=> 3n - 2 \(\in\) { 5 ; 6 ; 7 ; 8 ; 9 }
Nếu 3n - 2 = 5 thì n = 7/3 ( loại )
Nếu 3n - 2 = 6 thì n = 8/3 ( loại )
Nếu 3n - 2 = 7 thì n = 3 ( thỏa mãn )
Nếu 3n - 2 = 8 thì n = 10/3 ( loại )
Nếu 3n - 2 = 9 thì n = 11/3 ( loại )
Vậy n = 3

\(\text{a)}\)\(2^{x+1}.3^y=2^{2x}.3^x\Leftrightarrow\frac{2^{2x}}{2^{x+1}}=\frac{3^y}{3^x}\)
\(\Leftrightarrow2^{x-1}=3^{y-x}\)
\(\Leftrightarrow x-1=y-x=0\)
\(\Leftrightarrow x=y=1\)

\(\left(\dfrac{1}{3}\right)^{x+3}+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{4}{27}\)
=>\(\left(\dfrac{1}{3}\right)^{x+2}\cdot\left(\dfrac{1}{3}+1\right)=\dfrac{4}{27}\)
=>\(\left(\dfrac{1}{3}\right)^{x+2}\cdot\dfrac{4}{3}=\dfrac{4}{27}\)
=>\(\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{4}{27}:\dfrac{4}{3}=\dfrac{4}{27}\cdot\dfrac{3}{4}=\dfrac{3}{27}=\dfrac{1}{9}\)
=>x+2=2
=>x=0
\(\left(\dfrac{1}{3}\right)^{x+3}+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{4}{27}\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{27}+\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{9}=\dfrac{4}{27}\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^x\cdot\left(\dfrac{1}{27}+\dfrac{1}{9}\right)=\dfrac{4}{27}\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^x\cdot\dfrac{4}{27}=\dfrac{4}{27}\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^x=\dfrac{4}{27}:\dfrac{4}{27}\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^x=1\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)