tìm x, biết: (2^x-8)^3+(4^x+13)^3= (4^x+2^x+5)^3
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21 tháng 7 2023
1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
21 tháng 7 2023
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
VN
1
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
7 tháng 1
\(a,\dfrac{2}{3}.x=\dfrac{2}{7}\\ x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{3}{7}\\ ---\\ b,x.\dfrac{3}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\\ ---\\ c,x:\dfrac{8}{13}=\dfrac{13}{7}\\x=\dfrac{13}{7}.\dfrac{8}{13}=\dfrac{8}{7}\\ ----\\ d,\dfrac{3}{2}:x=\dfrac{7}{4}\\ x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{3}{2}.\dfrac{4}{7}=\dfrac{6}{7}\)
NL
1
30 tháng 8 2023
a: 2x(x+1)-135=-200
=>2(x^2+x)=-65
=>2x^2+2x+65=0
=>x^2+x+32,5=0
=>x^2+x+0,25+32,25=0
=>(x+0,5)^2+32,25=0(vô lý)
b: 4x-5(x-1)+15=13
=>4x-5x+5=-2
=>5-x=-2
=>x=5+2=7
c: 2/3x-1/4=3/5-7/8
=>2/3x=3/5-7/8+1/4=24/40-35/40+10/40=-1/40
=>x=-1/40:2/3=-1/40*3/2=-3/80
d: 1/2(2x-3)+105/2=-137/2
=>1/2(2x-3)=-137/2-105/2=-242/2=-121
=>2x-3=-242
=>2x=-239
=>x=-239/2
7 tháng 2 2022
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
15 tháng 4 2023
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm
7 tháng 10 2021
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
(2\(^x\)-8)\(^3\)=(4\(^x\)+2\(^x\)+5)\(^3\)-(4\(^x\)+13)\(^3_{ }\)
(2\(^x\)-8)\(^3\)=[(4\(^x\)+2\(^x\)+5) - (4\(^x\)+13)].[(4\(^x\)... + (4\(^x\)+13)\(^2\)]
(2\(^x\) - 8)\(^3\) = (2\(^x\)-8).[(4\(^x\)+2\(^x\)+5)\(^2\)+(4\(^x\)+... + (4\(_{ }^x\)+13)\(^2\)]
2\(^x\) = 8 \(\Rightarrow\) x = 3
hoặc (2\(^x\)-8)\(^2\) = (4\(^x\)+2\(^x\)+5)\(^2\)+(4\(^x\)+2\(^x\)+5)(4\(^x\)+... + (4\(^x\)+13)\(^2\)
(4\(^x\)+2\(^x\)+5)\(^2\) - (2\(^x\)-8)\(^2\)+(4\(^x\)+2\(_{ }^x\)+5)(4\(^x\)+13) + (4\(^x\)+13)\(^2\) = 0
[(4^x+2^x+5)-(2^x-8)]*[(4^x+2^x+5)+(2^... + (4^x+3)*[(4^x+2^x+5)+(4^x+13)]=0
(4^x+13)*(4^x+2*2^x-3) + (4^x+3)*(2*4^x+2^x+18)=0
(4^x+13)[(4^x+2*2^x-3) + (2*4^x+2^x+18)]=0
4^x+13=0 (VN)
hoặc 3*4^x + 3*2^x +15=0
đặt t = 2\(^x\)( t > 0)
t\(^2\) + t + 5=0 ptvn
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