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a: \(2^{x}\cdot4=128\)
=>\(2^{x}=\frac{128}{4}=32=2^5\)
=>x=5
b: \(x^{15}=x\)
=>\(x^{15}-x=0\)
=>\(x\left(x^{14}-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)
c: \(\left(2x+1\right)^3=125\)
=>\(\left(2x+1\right)^3=5^3\)
=>2x+1=5
=>2x=5-1=4
=>\(x=\frac42=2\)
d: \(\left(x-5\right)^4=\left(x-5\right)^6\)
=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)
=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)
=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)
\(2x+\frac{1}{4}=\frac{3}{2}\)
\(\Rightarrow2x=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)
\(\Rightarrow x=\frac{5}{4}:2=\frac{5}{8}\)
\(\left(x-5\right)-\frac{1}{3}=\frac{2}{5}\)
\(\Rightarrow x-5=\frac{2}{5}+\frac{1}{3}=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}+5=5\frac{11}{15}\)
\(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
\(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
|2x-5|=4
Suy ra: 2x-5=4
hoặc 2x-5 =-4
nếu 2x-5=4 nếu 2x - 5=-4
2x=4+5 2x=(-4)+5
2x=9 2x=-1
x=9:2 x=(-1):2
x=4/5(loại) x=-1/2 (loại)
Số số hạng là :
(2x - 2) : 2 + 1 = x - 1 + 1 = x (số)
Tổng là :
(2x + 2).x : 2 = 210
=> (2x2 + 2x) : 2 = 210
=> x2 + x = 210
=> x(x + 1) = 210
=> x(x + 1) = 20.21
=> x = 20
Vậy x = 20
Ta có : \(\frac{x}{2}=\frac{10}{x+1}\)
=> x(x + 1) = 10.2
=> x(x + 1) = 20
=> sai đề
\(a,234-\left(x-56\right)=789\)
\(\Leftrightarrow x-56=234-789\)
\(\Leftrightarrow x-56=-555\)
\(\Leftrightarrow x=\left(-555\right)+56=-499\)
Vậy x = -499
b) \(\frac{x+3}{-5}=\frac{x-15}{4}\)
\(\Leftrightarrow4\left(x+3\right)=-5\left(x-15\right)\)
\(\Leftrightarrow4x+12=-5x+75\)
\(\Leftrightarrow4x+12-\left(-5x\right)=75\)
\(\Leftrightarrow4x-\left(-5x\right)+12=75\)
\(\Leftrightarrow4x+5x=63\)
\(\Leftrightarrow9x=63\)
\(\Leftrightarrow x=7\)
Vậy x = 7
c) \(8\left(x-1\right)-7=2\left(x+2\right)+5\)
\(\Leftrightarrow8x-8-7=2x+4+5\)
\(\Leftrightarrow8x-8-7-2x+4=5\)
\(\Leftrightarrow8x-2x-8-7+4=5\)
\(\Leftrightarrow8x-2x=5-4+7+8\)
\(\Leftrightarrow4x=16\)
\(\Leftrightarrow x=4\)
Vậy x = 4
d) Đặt \(D=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=\frac{2\left(x-1\right)+5}{x-1}=2+\frac{5}{x-1}\)
=> \(5⋮x-1\)
=> \(x-1\inƯ\left(5\right)\)
=> \(x-1\in\left\{\pm1;\pm5\right\}\)
=> \(x\in\left\{2;0;6;-4\right\}\)
(2\(x+4\)) : 4 = 5
2\(x+4\) = 5 x 4
2\(x\) + 4 = 20
2\(x\) = 20 - 4
2\(x\) = 16
\(x=16:2\)
\(x=8\)
Vậy \(x=8\)
(2x + 4) : 4 = 5
2x + 4 = 5 x 4
2x + 4 = 20
2x = 20 - 4
2x = 16
x = 16 : 2
x = 8
Vậy x = 8


-2x+1/4 = 3/5
-2x = 3/5-1/4
-2x =12/20:5/20
-2x =7/20
x =7/20:-2
x =-7/40