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Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
Bài 1 :
Gọi 3 số tự nhiên cần tìm là n - 1 ; n ; n + 1 ( với n \(\in\) N* )
Theo bài ra , ta có :
[ ( n -1 )n ] + [ ( n + 1 )n ] + [ (n - 1)( n + 1 ) ] = 242
=> ( \(n^2\) - n ) + ( \(n^2\) + n ) + ( \(n^2\) + n - n - 1 ) = 242
=> \(3n^2\) - 1 = 242
=> \(3n^2\) = 243
=> \(n^2\) = 81
=> n = 9 hoặc n = -9
Mà n là số tự nhiên \(\Rightarrow\) n = 9
Vậy 3 số cần tìm là 8 ; 9 ; 10
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
a) \(\left(3x-1\right)^4=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}+1=\frac{3}{2}\\3x=-\frac{1}{2}+1=\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)
b) \(\left(5x-2\right)^2=\frac{25}{121}\)
\(\Rightarrow\orbr{\begin{cases}5x-2=\frac{5}{11}\\5x-2=-\frac{5}{11}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{5}{11}+2=\frac{27}{11}\\5x=-\frac{5}{11}+2=\frac{17}{11}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{27}{55}\\x=\frac{17}{55}\end{cases}}\)
c) \(\left(5x+6\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow5x+6=-\frac{2}{3}\)
\(\Rightarrow5x=-\frac{2}{3}-6=-\frac{20}{3}\)
\(\Rightarrow x=-\frac{20}{3}:5=-\frac{4}{3}\)
a) (3x — 1) ^4 = 1/16
=>3x-1=-1/2 hoặc 1/2
=>3x=1/2
=>x=1/6
=>3x=3/2
=>x=1/2
b)(5x — 2 )^2 = 25/121
=>5x-2=-5/11 hoặc 5/11
=>5x=17/11
=>x=17/55
=>5x=27/11
=>x=27/55
c)(5x + 6)^3= -8/27
=>5x+6=-2/3
=>5x=-20/3
=>x=-4/3