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5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
A, (X+1)+(x+4)+(x+7)+...+(x+25)+(x+28)=155
b,(x+9)+(x-2)+(x+7)
+(x-4)+(x+5)+(x-6)+(x+3)+(x-8)+(x+1)=95
A,(x+1)+(x+4)+(x+7)+...+(x+25)+(x+28)=155
x+1+x+4+x+7+...+x+25+x+28=155
x.[(28-1):3+1]+(1+4+7+...+25+28)=155
10.x+[(28+1).10:2]=155
10.x+145=155
A
<=>x+1+x+4+x+7+...+x+25+x+28=155
<=>10x.(1+4+7+10+13+16+19+22+25+28)=155
<=>10x.145 =155
<=>10x=155:145=\(\frac{31}{29}\)
x=\(\frac{31}{29}:10=\frac{31}{290}\)
Vậy x=\(\frac{31}{290}\)
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
<=> (x + x + ... + x) + (1 + 4 + ... + 28) = 155
<=> 10x + 145 = 155
<=> 10x = 10
<=> x = 1
b, (x + 9) + (x - 2) + (x + 7) + (x - 4) + ... + (x - 8) + (x + 1) = 95
<=> (x + 9 + x + 7 + ... + x + 1) + (x - 2 + x - 4 + .. + x - 8) = 95
<=> (5x + 25) + (4x - 20) = 95
<=> (5x + 4x) + (25 - 20) = 95
<=> 9x + 5 = 95
<=> 9x = 90
<=> x = 10
@láo như cáo
bài 1
a) 155 - 10.(x+1) = 55
=>10 .(x+1) = 100
=>x + 1 = 10
=>x = 9
còn lại tương tự
a: x=-5/11+2/11=-3/11
b: =>x=-3/24+20/24+1/24=18/24=3/4
c: =>5/8-x=1/9+5/4=4/36+45/36=49/36
=>x=5/8-49/36=-53/72
d: =>2/3-x=1/3
=>x=1/3
e: =>1/5:x=12/35
=>x=7/12
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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Giải:
a) \(2^5=4^x\)
\(\Rightarrow2^5=\left(2^2\right)^x\)
\(\Rightarrow2^5=2^{2x}\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
b) \(2.4^2.8^3.16^4=8^x\)
\(\Rightarrow2.\left(2^2\right)^2.\left(2^3\right)^3.\left(2^4\right)^4=\left(2^3\right)^x\)
\(\Rightarrow2.2^4.2^9.2^{16}=2^{3x}\)
\(\Rightarrow2^{30}=2^{3x}\)
\(\Rightarrow3x=30\)
\(\Rightarrow x=30:3\)
\(\Rightarrow x=10\)
c) \(3^3:3^5=9^x\)
\(\Rightarrow3^{-2}=\left(3^2\right)^x\)
\(\Rightarrow3^{-2}=3^{2x}\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-2:2\)
\(\Rightarrow x=-1\)
Chúc bạn học tốt!
a) Ta có: \(2^5=4^x\)
nên \(2^{2x}=2^5\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
b) Ta có: \(2\cdot4^2\cdot8^3\cdot16^4=8^x\)
\(\Leftrightarrow2^{3x}=2\cdot2^5\cdot2^9\cdot2^{16}=2^{31}\)
\(\Leftrightarrow3x=31\)
hay \(x=\dfrac{31}{3}\)
c) Ta có: \(3^3:3^5=9^x\)
\(\Leftrightarrow3^{-2}=3^{2x}\)
\(\Leftrightarrow2x=-2\)
hay x=-1
a) 5.5x = 625
=> 5x+1 = 54
=> x + 1 = 4
=> x = 4 - 1
=> x = 3
Vậy x = 3
b) 8(x + 25) - 155 = 181
=> 8(x + 25) = 181 + 155
=> 8(x + 25) = 336
=> x + 25 = 336 : 8
=> x + 25 = 42
=> x = 42 - 25
=> x = 17
Vậy x = 17
c) (x - 9)4 = (x - 9)2
=> (x - 9)4 - (x - 9)2 = 0
=> (x - 9)2.[(x - 9)2 - 1] = 0
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\\left(x-9\right)^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=1\\x-9=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
a)5*5x=625
\(\Rightarrow5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b)8(x+25)-155 =181
\(\Rightarrow8\left(x+25\right)=336\)
\(\Rightarrow x+25=42\)
\(\Rightarrow x=17\)
c) ( x-9)4 = ( x-9)2
\(\Rightarrow\left(x-9\right)^4-\left(x-9\right)^2=0\)
\(\Rightarrow\left(x-9\right)^2\left[\left(x-9\right)^2-1\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=\pm1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)