tìm x , biết
( 3 x - 4 )3 = 7 + 12019
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\(\left(3x-4\right)^3=7+1^{2019}\)
\(\Leftrightarrow\left(3x-4\right)^3=7+1\)
\(\Leftrightarrow\left(3x-4\right)^3=8\)
\(\Leftrightarrow\left(3x-4\right)^3=2^3\)
\(\Leftrightarrow3x-4=2\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
x.8+56+x.4-28=x.3+9+x+4*(1-3)
=>12x+(56-28)=(3x+x)+9+4*(-2)
=>12x+28=4x+9-8=4x+1
=>12x-4x=1-28
8x=-27
x=-27/8
( 3x - 24 ) . 73 = 2 . 74
3x - 16 = 2 . 74 : 73 = 14
3x = 14 + 16 = 30
x = 10
\(\left(3\times x-2^4\right)\times7^3=2\times7^4\)
\(\left(3\times x-2^4\right)\div2=7^4\div7^3\)
\(\left(3\times x-16\right)\div2=7\)
\(3\times x-16=7\times2\)
\(3\times x-16=14\)
\(3\times x=14+16\)
\(3\times x=30\)
\(x=30\div3\)
\(x=10\)
#)Giải :
\(\left(3x-2^4\right).7^3=2.7^4\)
\(\Rightarrow3x-16=2.7^4\div7^3\)
\(\Rightarrow3x-16=2.7\)
\(\Rightarrow3x-16=14\)
\(\Rightarrow3x=30\)
\(\Rightarrow x=10\)
\(\left(3.x-2^4\right).7^3=2.7^4\)
\(\left(3.x-16\right).343=2.2401\)
\(\left(3.x-16\right).343=4802\)
\(3.x-16=4802:343\)
\(3.x-16=14\)
\(3.x=14+16\)
\(3.x=30\)
\(x=30:3\)
\(x=10\)
a: =>x=3/7+3/5=15/35+21/35=36/35
b: =>x/35=4/5-5/7=28/35-25/35=3/35
=>x=3
c: =>x<3/4+8/4=11/4
=>\(x\in\left\{0;1;2;3\right\}\)
d: =>5/3<x<5/6+24/6=29/6
=>\(x\in\left\{2;3;4\right\}\)
e: =>x<10/12-9/12=1/12
=>x=0
f: =>2/3<x<12/6-5/6=7/6
=>x=1
1) <=> 8x-16-21-7x=-4
<=> x-37=-4
<=> x=33
2) <=> 12x-48+6x-12-16x-48=-28
<=> 2x-108=-28
<=> 2x = 80
<=> x = 40
3)<=> 4x-20-35+7x+50-10x=-3
<=> x-5=-3
<=> x=2
4) <=> -3x-15=-45-21
<=> 51=3x
<=> 17=x
5) 4x-28+15=-2x+14-10
<=> 6x-13=4
<=> 6x=17
<=> x = \(\frac{17}{6}\)
Chúc bạn học tốt
a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)
=>2x=44/15
hay x=22/15
`(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`
\(\left(3x-4\right)^3=7+1^{2019}\)
\(\left(3x-4\right)^3=7+1=8=2^3\)
\(=>3x-4=2\)
\(3x=2+4\)
\(3x=6\)
\(x=6:3\)
\(x=2\)