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Từ đề bài=>(x+1)/10+(x+2)/9+(x+3)/8-(-3)=0
<=>......................................(như trên)+3=0
<=>(x+1)/10+1+(x+2)/9+1+(x+3)/8+1=0( vì 1+1+1=3)
<=>(x+1+10)/10+(x+2+9)/9+(x+3+8)/8=0
<=>(x+11)/10+(x+11)/9+(x+11)/8=0
<=>(x+11).(1/10+1/9+1/8)=0
Mà 1/10<1/9<1/8
=>1/10+1/9+1/8 khác 0
<=>x+11=0<=>x=-11
Vậy x=-11
a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
Đề:
Giài:
\(\frac{31}{9}\left|x\right|=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}\left|x\right|=\frac{31}{6}\)
\(\left|x\right|=\frac{31}{6}:\frac{31}{9}\)
\(\left|x\right|=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-3}{2}\end{cases}}\)
Vậy x = \(\frac{3}{2}\)hoặc x = \(\frac{-3}{2}\)
\(\frac{31}{9}\left|x\right|-\frac{5}{2}=\frac{8}{3}\)
\(\frac{31}{9}\left|x\right|=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}\left|x\right|=\frac{8\cdot2}{6}+\frac{5\cdot3}{6}\)
\(\frac{31}{9}\left|x\right|=\frac{16}{6}+\frac{15}{6}\)
\(\frac{31}{9}\left|x\right|=\frac{31}{6}\)
\(\left|x\right|=\frac{31}{6}:\frac{31}{9}\)
\(\left|x\right|=\frac{31}{6}\cdot\frac{9}{31}\)
\(\left|x\right|=\frac{1}{2}\cdot\frac{3}{1}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{3}{2}\right\}\)
Nếu x < 2
=> |x - 2| = -(x - 2) = -x + 2
=> |x - 3| = -(x - 3) = -x + 3
=> |2x - 8| = -(2x - 8) = -2x + 8
Khi đó |x - 2| + |x - 3| + |2x - 8| = 9 (1)
<=> -x + 2 -x + 3 - 2x + 8 = 9
=> -4x + 13 = 9
=> -4x = -4
=> x = 1 (tm)
Nếu 2 \(\le x< 3\)
=> |x - 2| = x - 2
=> |x - 3| = -(x - 3) = -x + 3
=> |2x - 8| = -(2x - 8) = -2x + 8
Khi đó (1) <=> x - 2 - x + 3 - 2x + 8 = 9
=> -2x + 9 = 9
=> -2x = 0
=> x = 0 (loại)
Nếu \(3\le x\le4\)
=> |x - 2| = x - 2
=> |x - 3| = x - 3
=> |2x - 8| = -(2x - 8) = -2x + 8
Khi đó (1) <=> x - 2 + x - 3 - 2x + 8 = 9
=> 0x + 3 = 9
=> 0x = 6 (loại)
Nếu x > 4
=> |x - 2| = x - 2
=> |x - 3| = x - 3
=> |2x - 8| = 2x - 8
Khi đó (1) <=> x - 2 + x - 3 + 2x - 8 = 9
=> 4x - 13 = 9
=> 4x = 22
=> x = 5,5 (tm)
Vậy \(x\in\left\{1;5,5\right\}\)
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
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
\(\frac{x-2}{8}=\frac{x-3}{9}\)
\(\Rightarrow9\left(x-2\right)=8\left(x-3\right)\)
\(\Leftrightarrow9x-18=8x-24\)
\(\Leftrightarrow9x-8x=-24+18\)
\(x=-6\)
\(\frac{x-2}{8}=\frac{x-3}{9}\)( với lớp 7 thì xét tích chéo. Công thức tổng quát : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=cb\))
\(\Leftrightarrow9\left(x-2\right)=8\left(x-3\right)\)
\(\Leftrightarrow9x-18=8x-24\)
\(\Leftrightarrow9x-8x=-24+18\Leftrightarrow x=-6\)