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\(100-6x=8^{42}:8^{40}\)
\(\Leftrightarrow100-6x=64\)
\(\Leftrightarrow6x=36\)
hay x=6
a) x=-72/-8=9
b) x=-54/6=-9
c) x=-40/-4=10
d) x=-66/6=-11
tick nha
a)(-8)*x=-72
x=(-72)/(-8)
x=9
b)6*x=-54
x=-54/6
x=-9
c)(-4)*x=(-40)
x=(-40)/(-4)
x=10
d)6x=-66
x=-66/6
x=-11
\(6x^3-8=40\\ 6x^3=40+8=48\\x^3=\dfrac{48}{6}=8=2^3\\ Vậy:x=2\\ ---\\ 5^{x+1}:5=5^4\\5^{x+1}:5^1=5^4\\ 5^{x+1-1}=5^4\\ 5^x=5^4\\ Vậy:x=4\)
\(a,\) \(6x^3-8=40\)
\(6x^3=40+8\)
\(6x^3=48\)
\(x^3=8\)
\(x^3=8\)
\(x^3=2^3\)
\(x=2\)
Vậy \(x=2\)
\(b,\) \(5^{x+1}:5=5^4\)
\(5^{x+1-1}=5^4\)
\(5^x=5^4\)
\(x=4\)
Vậy \(x=4\)
a) x= -72 : (-8)
x = 9
b) x = -54 : 6 = -9
c) x = -40 : 9-4) = 10
d) x = -66 : 6 = -11
(-8).x= -72
x=(-72):(-8)
x=9
6x=-54
x=(-54):6
x=-9
(-4).x=-40
x=40:(-4)
x= -10
6x=-66
x=(-66):6
x= -11
tick mk nha p!
\(a.5^x=125\)
\(\Leftrightarrow5^x=5^3\)
\(\Leftrightarrow x=3\)
\(b.6x^3-8=40\)
\(\Leftrightarrow6x^3=40+8=48\)
\(\Leftrightarrow x^3=\frac{48}{6}=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
\(c.\left(x+1\right)^3=64\)
\(\Leftrightarrow\left(x+1\right)^3=4^3\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=4-1=3\)
\(d.3^x\div3^2=243\)
\(\Leftrightarrow3^{x-2}=3^5\)
\(\Leftrightarrow x-2=5\)
\(\Leftrightarrow x=5+2=7\)
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
a) (x: 8 - l).2 = 1.14 nên x : 8 - 1 = 7. Do đó x = 64.
b) (2x + 3).30 = 25.6 nên 2x + 3 = 5. Do đó x = 1.
c) 6.(2x - 7) = 9.(x - 3) nên 12x - 42 = 9x - 27.
Do đó 3x = 15. Vậy x = 5.
d) -7.(x + 27) = 6.(x + l) nên -7x - 189 = 6x + 6.
Do đó 13x = -195. Vậy x = -15.
Áp dụng theo quy tắc chuyển vế
a) 5x - 16 = 40 + x
x - 5x = -16 - 40
-4x = -56
x = 14
b) 4x - 10 = 15 - x
-x - 4x = -10 - 15
-5x = -25
x = 5
c) x + 15 = 20 - 4x
-4x - x = 15 - 20
-5 x = -5
x= 1
d) 17x = 7 - 6x
-6x - 17x = -7
-23x = -7
x = 7/23
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
\(6x^3-8=48\\ =>6x^3=40+8\\ =>6x^3=48\\ =>x^3=48:6\\ =>x^3=8\\ =>x^3=2^3\\ =>x=2\)
\(6x^3-8=40\)
\(6x^3=48\)
\(x^3=8\)
\(x^3=2^3\)
\(x=2\)