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3\(x\) - \(\dfrac{12}{5}\) = -0,6
3\(x\) - 2,4 = -0,6
3\(x\) = 0,6 + 2,4
3\(x\) = 3
\(x\) = 1
a ) \(2x:6=5:3\)
\(\Leftrightarrow\frac{2x}{6}=\frac{5}{3}\)
\(\Leftrightarrow2x=5.6:3\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Các câu còn lại tương tự .
\(\left(3x-0,64\right):\frac{3}{5}=0,6\)
\(\Leftrightarrow\left(3x-0,64\right).\frac{5}{3}=0,6\)
\(\Leftrightarrow3x-0,64=0,36\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy ................
\(\left(3x-0,64\right):\frac{3}{5}=0,6\)
\(\left(3x-\frac{16}{25}\right)=0,6.\frac{3}{5}\)
\(3x-\frac{16}{25}=\frac{9}{25}\)
\(3x=1\)
\(x=\frac{1}{3}\)
`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`3^12` và `5^8`
\(3^{12}=\left(3^3\right)^4=9^4\)
\(5^8=\left(5^2\right)^4=25^4\)
Vì `9 < 25` `=> 25^4 > 9^4`
`=> 3^12 > 5^8`
Vậy, `3^12 > 5^8`
`b)`
`(0,6)^9` và `(-0,9)^6`
\(\left(0,6\right)^9=\left(0,6^3\right)^3=\left(0,216\right)^3\)
\(\left(-0,9\right)^6=\left[\left(-0,9\right)^2\right]^3=\left(0,81\right)^3\)
Vì `0,81 > 0,216 => (0,81)^3 > (0,216)^3`
`=> (0,6)^9 < (-0,9)^6`
Vậy, `(0,6)^9<(-0,9)^6`
1.a) Có 312 = 33.4 = 274 ;
58 = 52.4 = 254
Dễ thấy 274 > 254 nên 312 > 58
b) Có \(0,6^9=\dfrac{6^9}{10^9}=\dfrac{6^{3.3}}{10^9}=\dfrac{216^3}{10^9}\)
mà \(\left(-0,9\right)^6=0,9^6=\dfrac{9^6}{10^6}=\dfrac{9^6.10^3}{10^9}=\dfrac{9^{2.3}.10^3}{10^9}=\dfrac{81^3.10^3}{10^9}=\dfrac{810^3}{10^9}\)
Dễ thấy \(\dfrac{216^3}{10^9}< \dfrac{810^3}{10^9}\Rightarrow0,6^9< \left(-0,9\right)^6\)
`@``dn10`
`a,`
`P(x)=-2x^5-3x^4+2x^5+2x-0,6`
`P(x)=(-2x^5+2x^5)-3x^4+2x-0,6`
`P(x)=-3x^4+2x-0,6`
`b,`
Thay `x=1` vào đa thức `B(x)`
`B(1)=-4*1^3+6*1-4=-4*1+6-4=-4+6-4=2-4=-2`
a: =-2x^5+2x^5+3x^4+2x-0,6
=3x^4+2x-0,6
b: B(1)=-4+6-4=-8+6=-2
a, ( 3 - 0,6) - ( 7 + 3\(\dfrac{1}{4}\) - \(\dfrac{8}{5}\)) - ( 9 - 2\(\dfrac{1}{4}\))
= 2,4 - (7 + 3,25 - 1,6) - (9 - 2,25)
= 2,4 - 7 - 3,25 + 1,6 - 9 + 2,25
= (2,4 + 1,6) - (7+ 9) - ( 3,25 - 2,25)
= 4 - 16 - 1
= - 12 - 1
= -13
b, ( - \(\dfrac{5}{8}\) + \(\dfrac{7}{6}\) - \(\dfrac{0}{8}\)) - (\(\dfrac{5}{6}\) - \(\dfrac{7}{8}\) - 1,4) + ( \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\) + \(\dfrac{12}{5}\))
= - \(\dfrac{5}{8}\) + \(\dfrac{7}{6}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{8}\) + \(\dfrac{7}{5}\) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\) + \(\dfrac{12}{5}\)
= (- \(\dfrac{5}{8}\) + \(\dfrac{7}{8}\)) + (\(\dfrac{7}{6}\) - \(\dfrac{5}{6}\)) + ( \(\dfrac{7}{5}\) + \(\dfrac{12}{5}\)) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\)
= \(\dfrac{1}{4}\) + \(\dfrac{1}{3}\) + \(\dfrac{19}{5}\) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + ( \(\dfrac{1}{3}\) + \(\dfrac{5}{3}\)) + \(\dfrac{19}{5}\)
= 1 + 2 + 3,8
= 6,8
a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)
\(\Rightarrow\text{ }4^{100}< 2^{202}\)
b, \(3^0=1< 5^8\)
\(3^0< 5^8\)
c, \(\left(0,6\right)^0=1\)
\(\left(-0,9\right)^6=\left(0,9\right)^6\)
\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)
d,
e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)
\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)
Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)
a. Kiểm tra lại mẫu số vế phải, \(7-5x\) hay \(7-3x\)
b. ĐKXĐ: \(x\ne-\dfrac{5}{3}\)
\(\dfrac{3x+5}{12}=\dfrac{3}{5+3x}\)
\(\Leftrightarrow\dfrac{\left(3x+5\right)^2}{12\left(3x+5\right)}=\dfrac{36}{12\left(3x+5\right)}\)
\(\Rightarrow\left(3x+5\right)^2=36=6^2\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=6\\3x+5=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\) (thỏa mãn)
3x - 12/5 = -0,6
3x = -0,6 + 12/5
3x = -0,6 + 2,4
3x = 1,8
x = 1,8 : 3
x = 0,6
`# \text {04th5.}`
$3x - \dfrac{12}{5} = -0,6$
$\Rightarrow 3x = -0,6 + \dfrac{12}{5}$
$\Rightarrow 3x = \dfrac{9}{5}$
$\Rightarrow x = \dfrac{9}{5} \div 3$
$\Rightarrow x = \dfrac{3}{5}$
Vậy, $x = \dfrac{3}{5}.$