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\(\left(3x-4\right)^3=7+1^{2019}\)
\(\Leftrightarrow\left(3x-4\right)^3=7+1\)
\(\Leftrightarrow\left(3x-4\right)^3=8\)
\(\Leftrightarrow\left(3x-4\right)^3=2^3\)
\(\Leftrightarrow3x-4=2\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
`(3x-4)^3=7+1^2019`
`(3x-4)^3=7+1`
`(3x-4)^3=8`
`(3x-4)^3=2^3`
`=>3x-4=2`
`3x=6`
`x=2`
\(\left(3x-4\right)^3=7+1^{2019}\)
\(\left(3x-4\right)^3=7+1=8=2^3\)
\(=>3x-4=2\)
\(3x=2+4\)
\(3x=6\)
\(x=6:3\)
\(x=2\)
a) x = 0
b) x = -1
c) x = -9
d) x = 24
e) x = 2 hoặc x = -4
f) x = 5 hoặc x = -3
a, Ta có: 3 x = 3 2 nên x = 2
b, Ta có: 5 x = 5 3 nên x = 3
c, Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1
d, Ta có: 6 x - 1 = 6 2 nên x - 1 = 2, đo đó x = 3
e) Ta có: 3 2 x + 1 = 3 3 nên 2x +1 = 3, do đó x = 1
f) Ta có: x 50 = x nên x 50 - x = 0 , do đó x x 49 - 1 = 0 = 0
Vì thế x = 0 hoặc x = 1
a) \(2x^3-32x=0\)
\(2x\left(x^2-16\right)=0\)
\(2x\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow2x=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(x=0\) hoặc \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
b) \(\left(3x-2\right)^2-\left(x+5\right)^2=0\)
\(\left(3x-2-x-5\right)\left(3x-2+x+5\right)=0\)
\(\left(2x-7\right)\left(4x+3\right)=0\)
\(\orbr{\begin{cases}2x-7=0\\4x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
c) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
d) \(4x^2-25-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(4x^2-25\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x+5\right)\left(2x-5-x-7\right)=0\)
\(\left(2x+5\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
\(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2020^2-1}\right)\)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
13 x 58 x 4 + 32x 36 x2 + 52 x10
= 58 x 52 + 32x 36 x2 + 52 x10
= 52 x ( 58 + 10 ) + 32 x 36 x 2
= 52 x 68 + 2304
= 3536 + 2304 = 5840
Hok tốt
13 . 58 .4 + 32 . 36 . 2 + 52 . 10 = 5840 nhé!
Chúc bạn học tốt!
Có phải đề là như này không : [3x . (-4)] ^3 = 3^2 . 1^2019
\(\left(3x-4\right)^3=3^2.1^{2019}\)
<=> \(\left(3x-4\right)^3=9\)
<=>\(x=\frac{\sqrt[3]{9}+4}{3}\)
chúc bn học tốt