(X-1)×(x-1)×(x-1)= 125
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HT
1
PT
1
CM
22 tháng 6 2018
( 1+3+5+7+…+2003+2005) x (125 125 x 127 – 127 127 x 125)
= ( 1+3+5+7+…+2003+2005) x (125 x 1001 x 127 – 127 x 1001x 125)
= ( 1+3+5+7+…+2003+2005) x 0 = 0
16 tháng 10 2019
a)\(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
\(b,\left(2x-1\right)^3=125\)
\(\Rightarrow2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6:2=3\)
MN
16 tháng 10 2019
b) \(\left(2x-1\right)^3=125\)
\(\left(2x-1\right)^3=5^3\)
\(2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
14 tháng 8 2023
x^3-1=(x-1)(x^2+x+1)
8x^3+1=(2x+1)(4x^2-2x+1)
x^3+1=(x+1)(x^2-x+1)
125-x^3=(5-x)(25+5x+x^2)
x^3+8y^3=x^3+(2y)^3
=(x+2y)(x^2-2xy+4y^2)
64y^3-125x^3
=(4y)^3-(5x)^3
=(4y-5x)(16y^2+20xy+25x^2)
\(27x^3-\dfrac{1}{8}=\left(3x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(3x-\dfrac{1}{2}\right)\left(9x^2+\dfrac{3}{2}x+\dfrac{1}{4}\right)\)
\(a^6-b^3=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\cdot\left(a^4+a^2b+b^2\right)\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
\(26\cdot5^{x-1}-5^{x-1}=125\)
\(\Rightarrow5^{^{x-1}}\cdot\left(26-1\right)=125\)
\(\Rightarrow5^{x-1}\cdot25=125\)
\(\Rightarrow5^{x-1}=\dfrac{125}{25}\)
\(\Rightarrow5^{x-1}=5\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=1+1\)
\(\Rightarrow x=2\)
10 tháng 7 2023
\(\Leftrightarrow5^{x-1}\cdot\left(26-1\right)=125\)
=>\(5^{x-1}=5\)
=>x-1=1
=>x=2
16 tháng 2 2021
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
26 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(5\cdot x^3-5=0\)
`=> 5*x^3 = 0+5`
`=> 5*x^3 = 5`
`=> x^3 = 5 \div 5`
`=> x^3 = 1`
`=> x^3 = 1^3`
`=> x=1`
Vậy, `x=1.`
`b)`
\(( x+1)^2 = 16\)
`=> (x+1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy, `x \in {3; -5}`
`c)`
\(( x+1)^3 = 27\)
`=> (x+1)^3 = 3^3`
`=> x+1=3`
`=> x=3-1`
`=> x=2`
Vậy, `x=2.`
`d)`
\(( x-1)^3 = 343\)
`=> (x-1)^3 = 7^3`
`=> x-1=7`
`=> x=7+1`
`=> x=8`
Vậy, `x=8.`
`e)`
\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?
Mình làm cả 2 TH nhé!
`(2x-1^3)=125`
`=> 2x-1=125`
`=> 2x=125+1`
`=> 2x=126`
`=> x=126 \div 2`
`=> x=63`
TH2:
`(2x-1)^3 = 125`
`=> (2x-1)^3 = 5^3`
`=> 2x-1=5`
`=> 2x=5+1`
`=> 2x=6`
`=> x=6 \div 2`
`=> x=3`
Vậy, `x=3.`
TM
26 tháng 6 2023
(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)
(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)
(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)
(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
K viết đề nha!
\(\left(x-1\right)^3\)= 125
=> x - 1 = 5
=> x = 6
Hội con 🐄 chúc bạn học tốt!!!
(x - 1)(x - 1)(x - 1) = 125
(x - 1)3 = 125
(x - 1)3 = 53
x - 1 = 5
x = 5 + 1
x = 6