tìm x, biết
60%x +2/3x = 1368
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ta có:
x+2x+3x+...+18x=1368
(1+2+3+...+18)x=1368
x=1368:(1+2+3...+18)
x=1368:171
x=8
a 36x 6+ 36x4
=36x[6+4]
=36x10
=360
b 1368 x99+1368
=1368 x[99+1]
=1368x100
=136800
a) 36x6+36x4=36x(6+4)=36x10=360
b) 1368x99+1368=1368x(99+1)=1368x100=136800
a, 70 - 5 ( x - 3 ) = 45
⇔ 5 . x - 3 = 70 - 45
⇔ 5 x - 3 = 35
⇔ x - 3 = 35 : 5 ⇔ x - 3 = 7
⇔ x = 10
b, 10 + 2 x = 4 5 : 4 3
⇔ 10 + 2 x = 4 2 ⇔ 10 + 2 x = 16
⇔ 2 x = 4 ⇔ x = 2
c, 60 - 3 x - 2 = 51
⇔ 3 x - 2 = 60 - 51
⇔ 3 x - 2 = 9
⇔ x - 2 = 3 ⇔ x = 5
d, 4 x - 20 = 2 5 : 2 3
⇔ 4 x - 20 = 2 2 ⇔ 4 x - 20 = 4
⇔ 4 x = 24 ⇔ x = 6
Tách ?
`a, 70 -5.(x-3) =45`
`=> 5.(x-3)= 70-45`
`=> 5.(x-3)=25`
`=>x-3=25:5`
`=>x-3=5`
`=>x= 5+3`
`=>x=8`
______
`b,10 + 2.x = 4^5:4^3`
`=> 10 + 2.x = 4^(5-3)`
`=> 10 + 2.x =4^2=16`
`=> 2.x=16-10`
`=>2.x=6`
`=>x=6:2`
`=>x=3`
_____
`c,60-3.x-2=51`
`=> 60-3.x= 51+2`
`=> 60-3.x=53`
`=>3.x=60-53`
`=> 3.x= 7`
`=>x= 7/3`
____
`d, 4.x-20=2^5:2^3`
`=> 4.x-20=2^(5-3)`
`=> 4.x-20=2^2`
`=> 4.x= 4+20`
`=>4.x=24`
`=>x=24:4`
`=>x=6`
____
`2^x . 4=16`
`=> 2^x=16:4`
`=>2^x= 4`
`=>2^x=2^2`
`=>x=2`
____
`f, 3^x . 3=243`
`=>3^x=243:3`
`=> 3^x=81`
`=> 3^x=3^3`
`=>x=3`
_____
`g, 64. 4^x =16^8`
`=> 4^3 . 4^x=(4^2)^8`
`=> 4^3 . 4^x = 4^(16)`
`=> 4^x= 4^(16-3)`
`=>4^x=4^(13)`
`=>x=13`
_____
`2^x . 16^2 =1024`
`=> 2^x= 1024 : 16^2`
`=>2^x=4`
`=>2^x=2^2`
`=>x=2`
a: =>5(x-3)=25
=>x-3=5
=>x=8
b: =>2x=16-10=6
=>x=3
c: =>58-3x=51
=>3x=7
=>x=7/3
d: =>4x-20=4
=>4x=24
=>x=6
e: =>2^x=4
=>2^x=2^2
=>x=2
f: =>3^x=81
=>3^x=3^4
=>x=4
g: =>4^x*4^3=4^16
=>x+3=16
=>x=13
h: =>2^x=1024/256=4=2^2
=>x=2
a) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)
\(\Leftrightarrow x^3-27+4x-x^3=1\)
\(\Leftrightarrow4x-27=1\Leftrightarrow4x=28\Leftrightarrow x=7\)
b) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)
\(\Leftrightarrow x^3-6x^2+12x+4-x^3-8+6\left(x^2-4\right)=60\)
\(\Leftrightarrow-6x^2+12x-4+6x^2-24=60\)
\(\Leftrightarrow12x-28=60\Leftrightarrow x=\frac{22}{3}\)
a) (x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 + 2x^2 - x^3 + 4x - 2x^2 = 1
4x - 27 = 1
4x = 28
x = 7
b) (x - 2)^3 - (x - 2)(x^2 + 2x + 4) + 6(x - 2)(x + 2) = 60
x^3 - 4x^2 + 4x - 2x^2 + 8x - 8 - x(x^2 + 2x + 4) + 2(x^2 + 2x + 4) + 6x - 24 = 60
x^3 + 12x - 32 - x^2 - 2x^2 - 4x + 2x^2 + 4x + 8 = 60
12x - 24 = 60
12x = 60 + 24
12x = 84
x = 7
Ta có:\(3x=2z-x\Rightarrow4x=2z\Rightarrow2x=z\)
\(x+y+z=60\Rightarrow z=60-x-y\Rightarrow2x=60-x-y\Rightarrow3x=60-y\)
\(\Rightarrow4y=60-y\Rightarrow5y=60\Rightarrow y=12\)
\(\Rightarrow4y=3x=12.4=48\Rightarrow x=\frac{48}{3}=16\)
Mà \(2x=z\Rightarrow z=16.2=32\)
Vậy\(x=16;y=12;x=32\)
Ta có :
3x = 4y = 2z => \(\frac{x}{4}=\frac{y}{3}\)và \(\frac{y}{2}=\frac{z}{4}\)=> \(\frac{x}{8}=\frac{y}{6}\)và \(\frac{y}{6}=\frac{z}{12}\)
=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{12}\). Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\frac{x}{8}=\frac{y}{6}=\frac{z}{12}=\frac{x+y+z}{8+6+12}=\frac{60}{24}\)
Suy ra : \(\frac{x}{8}=\frac{60}{24}\Rightarrow x=\frac{60}{24}.8=20\)
\(\frac{y}{6}=\frac{60}{24}\Rightarrow y=\frac{60}{24}.6=15\)
\(\frac{z}{12}=\frac{60}{24}\Rightarrow x=\frac{60}{24}.12=30\)
Vậy : x = 20 ; y = 15 ; z = 20
(2x + 1)(y - 3) = 10
TH1 : 2x + 1 = 1 => 2x = 0 => x = 0
y - 3 = 10 => y = 13
TH2 : 2x + 1 = 10 => 2x = 9 => x = 4,5
y - 3 = 1 => y = 4
TH3 : 2x + 1 = 2 => 2x = 1 => x = 0,5
y - 3 = 5 => y = 8
TH4 : 2x + 1 = 5 => 2x = 4 => x = 2
y - 3 = 2 => y = 5
TH5 : 2x + 1 = -1 => 2x = -2 => x = -1
y - 3 = -10 => y = -7
TH6 : 2x + 1 = -10 => 2x = -11 => x = -11/2
y - 3 = -1 => y = 2
TH7 : 2x + 1 = -2 => 2x = -3 => x = -3/2
y - 3 = -5 => y = -2
TH8 : 2x + 1 = -5 => 2x = -6 => x = -3
y - 3 = -2 => y = 1
\(\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+2\right)\right].\left[\left(x+5\right)\left(x+6\right)\right]-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+5\right)\right]-60=0\)
\(\Rightarrow\left(x^2+7x+6\right)\left(x^2+7x+10\right)-60=0\left(1\right)\)
Đặt \(x^2+7x+6=a\Rightarrow x^2+7x+10=a+4\)
Thay vào (1), ta có:
\(a\left(a+4\right)-60=0\)
\(\Rightarrow a^2+4a-60=0\)
\(\Rightarrow a^2+10a-6a-60=0\)
\(\Rightarrow a\left(a+10\right)-6\left(a+10\right)=0\)
\(\Rightarrow\left(a-6\right)\left(a+10\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=6\\a=-10\end{cases}}\)
- Nếu \(x^2+7x+6=6\)
\(\Rightarrow x^2+7x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)
- Nếu \(x^2+7x+6=-10\)
\(\Rightarrow x^2+7x+16=0\)
Mà \(x^2+7x+16=x^2+2.x.\frac{7}{2}+\frac{49}{4}+\frac{15}{4}=\left(x+\frac{7}{2}\right)^2+\frac{15}{4}>0\forall x\)
Vậy \(x=0,x=-7\)
Học tốt.
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