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\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
Mọi người giúp mình bài tìm x
1) (x-2/9)^3 = (8/27)^2
2)2.3^x -405 = 3^x-1
3)4/3 : 0,8 = 0,75x : (-1,5)
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
Mọi người giúp mình bài tìm x
1) (x-2/9)^3 = (8/27)^2
2)2.3^x -405 = 3^x-1
3)4/3 : 0,8 = 0,75x : (-1,5)
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`
`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`
`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`
`=> x^2 . 11 = 2^4 . 11`
`=> x^2 . 11 - 2^4 . 11 = 0`
`=> 11 . (x^2 - 16) = 0`
`=> x^2 - 16 = 0`
`=> x^2 = 16`
`=> x^2 = (+-4)^2`
`=> x = `\(\pm4\)
Vậy, `x \in`\(\left\{4;-4\right\}\)
_____
\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)
`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)
`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)
`=>`\(\dfrac{23}{108}x=48-6^2\)
`=>`\(\dfrac{23}{108}x=48-36\)
`=>`\(\dfrac{23}{108}x=12\)
`=>`\(x=\dfrac{1296}{23}\)
Vậy, `x = `\(\dfrac{1296}{23}\)
\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)
\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)
\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)
\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)
\(\Rightarrow x\in\varnothing\)
\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)
\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
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