Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{x}{6}=\dfrac{y}{-5}=\dfrac{z}{4}\rightarrow\dfrac{z}{4}=\dfrac{2x}{12}=\dfrac{y}{-5}=\dfrac{2x+y}{12-5}=\dfrac{21}{7}=3\)
-Suy ra: x=6.3=18; y=(-5).3=-15; z=4.3=12
b) \(\dfrac{x}{y}=\dfrac{9}{10}\rightarrow\dfrac{x}{9}=\dfrac{y}{10}\rightarrow\dfrac{x}{27}=\dfrac{y}{30}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\rightarrow\dfrac{z}{4}=\dfrac{y}{3}\rightarrow\dfrac{z}{40}=\dfrac{y}{30}\)
\(\rightarrow\dfrac{x}{27}=\dfrac{y}{30}=\dfrac{z}{40}=\dfrac{x-y-z}{27-30-40}=\dfrac{43}{43}=1\)
\(\rightarrow x=27;y=30;z=40\)
a,
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \)
\(\dfrac{1}{4}:x=\dfrac{8-15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
b,
( 3x + 1)^3 = 64
(3x + 1)^3 = 4^3
(3x + 1) = 4
3x = 4 - 1
3x = 3
x = 3 : 3
x = 1
c,
( 2x - 3)^4 = 81
( 2x - 3) ^4 = 3^4
(2x - 3) = 3
2x = 3 + 3
2x = 6
x = 6: 2
x = 3
Bài 1 :
\(C=\frac{1}{\left|x-2\right|+3}\)
\(C\le\frac{1}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy....
Bài 2 :
a) \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
b) \(2\cdot3^{x-405}=3^{x-1}\)
\(2=3^{x-1}:3^{x-405}\)
\(2=3^{x-1-x+405}\)
\(2=3^{404}\)( vô lí )
=> x thuộc rỗng
c) \(\frac{1}{81}\cdot27^{2x}=\left(-9\right)^4\)
\(\frac{27^{2x}}{81}=9^4\)
\(\frac{\left(3^3\right)^{2x}}{3^4}=\left(3^2\right)^4\)
\(\frac{3^{6x}}{3^4}=3^8\)
\(3^{6x-4}=3^8\)
\(\Rightarrow6x-4=8\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
d) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}4x-1=0\\4x-1=\left\{\pm1\right\}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\left\{\frac{1}{2};0\right\}\end{cases}}\)
`3x+20=0`
`=>3x=0-20`
`=>3x=-20`
`=>x=-20/3`
`---`
`2(-4x+9)=0`
`=>-4x+9=0`
`=>-4x=-9`
`=>x=9/4`
`---`
`2x(x-45)=0`
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-45=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=45\end{matrix}\right.\)
`---`
`-5x(2x+47)=0`
\(\Rightarrow\left[{}\begin{matrix}-5x=0\\2x+47=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{47}{2}\end{matrix}\right.\)
`----`
`x^2 -912=0`
`=>x^2=912`
`=>x∈∅`
1)
`3x+20=0`
`<=>3x=-20`
`<=>x=-20/3`
2)
`2(-4x+9)=0`
<=>-4x+9=0`
`<=>-4x=-9`
`<=>x=9/4`
3)
`2x(x-45)=0`
\(< =>\left[{}\begin{matrix}2x=0\\x-45=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=45\end{matrix}\right.\)
4)
`-x(2x+47)=0`
\(< =>\left[{}\begin{matrix}-x=0\\2x+47=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-\dfrac{47}{2}\end{matrix}\right.\)
5)
`x^2 -912=0`
`<=>x^2=912`
câu 5 xem lại nhé
\(a,2x=\dfrac{4^7}{4^3}=4^{7-3}=4^4\\ =>x=\dfrac{4^4}{2}=\dfrac{256}{2}=128\\ b,3^x=\dfrac{9^4}{81^3}=\dfrac{9^4}{9^{2.3}}\\ =>3^x=\dfrac{9^4}{9^6}=9^{-2}\\ =>3^x=3^{2.\left(-2\right)}=3^{-4}\\ =>x=-4\)