x : (2/3)4=2/3
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DK
1
15 tháng 4 2020
\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x-9=0
<=> -2x=9
<=> \(x=\frac{-9}{2}\left(tmđk\right)\)
21 tháng 3 2017
\(\frac{x}{6}=\frac{1}{2}+\frac{1}{3}.\frac{3}{4}\)
<=> \(\frac{x}{6}=\frac{1}{2}+\frac{3}{12}\)
<=>\(\frac{x}{6}=\frac{1}{2}+\frac{1}{4}\)
<=>\(\frac{x}{6}=\frac{2}{4}+\frac{1}{4}\)
<=>\(\frac{x}{6}=\frac{3}{4}\)
<=>\(x=\frac{3}{4}.6\)
<=>\(x=\frac{9}{2}\)
kl:
15 tháng 10 2016
C=(1x3+3x5+...+99x101)+(2x4+4x6+...+98x100)
đặt S=1x3+3x5+...+99x101
=>6S=6x(1x3+3x5+...+99x101)
=1x3x(5+1)+3x5x(7-1)+...+97x99x(101-95)+99x101x(103-97)
=1x3x5+1x3x1+3x5x7-1x3x5+....+97x99x101-95x97x99+99x101x103-97x99x101
=1x3x1+99x101x103
=>S=(3+99x101x103):6=171650
=>C=171650+(2x4+4x6+...+98x100)
đặt A=2x4+4x6+...+98x100
=>6A=6x(2x4+4x6+...+98x100)
=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)
=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100
=98x100x102
=>A=98x100x102:6=166600
=>C=166600+171650
=>C=338250
B=2x2+4x4+6x6+...+100x100
=2x(4-2)+4x(6-2)+6x(8-2)+...+100x(102-2)
=2x4-4+4x6-8+6x8-12+...+100x102-200
=(2x4+4x6+6x8+...+100x102)-(4+8+12+...+200)
đặt A=2x4+4x6+...+98x100+100x102
=>6A=6x(2x4+4x6+...+98x100+100x102)
=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)+100x102x(104-98)
=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100+100x102x104-98x100x102
=100x102x104
=>A=100x102x104:6=176800
=>B=176800-(4+8+12+...+200)
đặt S=4+8+12+..+200
Số số hạng của S là:
(200-4):4+1=50 số
S=(200+4)x50:2=5100
=>B=176800-5100
=>B=171700
LV
0
AH
Akai Haruma
Giáo viên
15 tháng 11 2021
Lời giải:
Để $(d)$ song song với $y=-x+3m$ thì:
\(\left\{\begin{matrix} m-4=-1\\ -m+3\neq 3m\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} m=3\\ m\neq \frac{-3}{2}\end{matrix}\right.\Leftrightarrow m=3\)
NM
0
FT
2
22 tháng 7 2017
\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=3\frac{1}{5}\)
\(=\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)
\(\Rightarrow5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)\left(x-2\right)=16\left(x-2\right)\left(x-4\right)\)
\(\Leftrightarrow5x^2-35x+60+5x^2-20x+20=16x^2-96x+128\)
\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)
\(\Leftrightarrow-6x^2+41x-48=0\)
......
DP
22 tháng 7 2017
\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\Leftrightarrow\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)
\(\Leftrightarrow\frac{5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)^2}{5\left(x-2\right)\left(x-4\right)}=\frac{16.\left(x-2\right)\left(x-4\right)}{5\left(x-2\right)\left(x-4\right)}\)
\(\Rightarrow5x^2-20x-15x+60+5x^2-20x+20=16x^2-64x-32x+128\)
\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)
\(\Leftrightarrow6x^2-41x+48=0\)
\(\Leftrightarrow x=\frac{16}{3};x=\frac{3}{2}\)
N
10 tháng 7 2017
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
DP
11 tháng 7 2017
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
NT
4
19 tháng 2 2021
Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)
\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)
\(\Leftrightarrow15x^2-4x-4=0\)
\(\Leftrightarrow15x^2-10x+6x-4=0\)
19 tháng 2 2021
Lỗi :vvvv
\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
Bài làm
\(x:\left(\frac{2}{3}\right)^4=\frac{2}{3}\)
\(x=\frac{2}{3}.\left(\frac{2}{3}\right)^4\)
\(x=\left(\frac{2}{3}\right)^5\)
Vậy \(x=\left(\frac{2}{3}\right)^5\)
x:(2/3)^4=2/3
=>x=(2/3).(2/3)^4
=>x=(2/3)^5
=>x=(2^5)/(3^5)
=>x=32/243