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a)120-3(x+4)=23
=>3(x+4)=120-23
=>3(x+4)=97
=>x+4=97:3
=>x+4=\(\frac{97}{3}\)
=>x=\(\frac{97}{3}\)-4
=>x=\(\frac{85}{3}\)
Vậy x=\(\frac{85}{3}\)
b)[(4x+28).3+55]:5=35
=>(4x+28).3+55=35.5
=>(4x+28).3+55=175
=>(4x+28).3=175-55
=>(4x+28).3=120
=>4x+28=120:3
=>4x+28=40
=>4x=40-28
=>4x=12
=>x=12:4
=>x=3
Vậyx= 3
c)\((12x-4^3).8^3=4.8^4\)
=>\(12x-64=4.8^4:8^3\)
=>12x-64=4.8
=>12x-64=32
=>12x=96
=>x=96:12
=>x=8
Vậy x=8
d)720:[41-(2x-5)]=23.5
=>720:[41-2x-5)]=40
=>41-(2x-5)=720:40
=>41-(2x-5)=18
=>2x-5=41-18
=>2x-5=23
=>2x=23+5
=>2x=28
=>x=28:;2
=>x=14
Vậy x=14
a, 120 - 3 [ x+4]=23
3 [x+4]= 120-23 = 97
3x = 97-4
3x= 93
x= 93:3
x=31
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bạn tách các số ra:
(2*8*16)(a^3*a^2*a^3)(x^2*x^3*x^3)(y*y^4*y^3)
=256a^8x^8y^8=(2axy)^8
CHÚ BẠN HỌC TỐT!!!
2x - 138 = 23 . 32
=> 2x - 138 = 72
=> 2x = 72 + 138
=> 2x = 210
=> x = 105
10 + 2x = 43 . 42
10 + 2x =1024
2x = 1024 - 10
2x = 1014
x = 507
(3x - 24) . 73 = 2.74
3x - 24 = 2 . 74 : 73
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 10
x4 = 16
=> x4 = 24
=> x = 2
\(x^2+3x-4=0\)
\(\Leftrightarrow x^2-x+4x-4\)
\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+4\right)\)
\(x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\)
\(a,x^2+3x-4=0\)
\(\Rightarrow x^2-x+4x-4=0\)
\(\Rightarrow x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
\(b,x^2-5x+4=0\)
\(\Rightarrow x^2-4x-x+4=0\)
\(\Rightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)
a ) \(x\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right..\)
Vậy .....
b ) \(\left(3-x\right)\left(x^2+1\right)=0\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy ........
c ) giống câu b
d ) \(\left(2x-1\right)^3=8\Leftrightarrow2x-1=2\Leftrightarrow\Leftrightarrow x=\dfrac{3}{2}\)
e ) \(\left(x+3\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.
4(x - 5) - 23 = 24.3
=> 4(x - 5) - 8 = 16.3
=> 4(x - 5) - 8 = 48
=> 4(x - 5) = 48 + 8
=> 4(x - 5) = 56
=> x - 5 = 56 : 4
=> x - 5 = 14
=> x = 14 + 5
=> x = 19
Vậy x = 19
a) pt <=> \(\frac{x\left(x+1\right)}{2}=500500\)
<=> \(x^2+x=1001000\)
<=> \(x^2-1000x+1001x-1001000=0\)
<=> \(\left(x-1000\right)\left(x+1001\right)=0\)
<=> \(\orbr{\begin{cases}x=1000\\x=-1001\end{cases}}\)
Do \(x>0\)=> \(x=1000\)
b)
<=> \(2x=210\)
<=> \(x=105\)
c)
<=> \(6x-81=3.7\)
<=> \(x=17\)
d)
<=> \(125-5\left(3x-1\right)=5^2\)
<=> \(5\left(3x-1\right)=100\)
<=> \(3x-1=20\)
<=> \(x=7\)
e)
<=> \(4^{x+1}+1=65\)
<=> \(4^{x+1}=64\)
<=> \(x+1=3\)
<=> \(x=2\)
j)
<=> \(2\left(2x-3\right)=14\)
<=> \(2x-3=7\)
<=> \(x=5\)