A=7 B=1 OR =0 C=2 D=4.5.6

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: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)

=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

\(\left|x+5\right|=5\)

<=> \(\hept{\begin{cases}x+5=5\\x+5=-5\end{cases}}\)

<=> \(\hept{\begin{cases}x=0\\x=-10\end{cases}}\)

\(\left|x+1\right|+7=10\)

<=> \(\left|x+1\right|=3\)

<=> \(\hept{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)

<=> \(\hept{\begin{cases}x=2\\x=-4\end{cases}}\)

\(\left|x-3\right|-6=5\)

<=> \(\left|x-3\right|=11\)

<=> \(\hept{\begin{cases}x-3=11\\x-3=-11\end{cases}}\)

<=> \(\hept{\begin{cases}x=14\\x=-8\end{cases}}\)

\(\left|x+2\right|-6\left(x-4\right)=20-6x\)

<=> \(\left|x+2\right|-6x+24=20-6x\)

<=> \(\left|x+2\right|=-4\)

<=> \(\hept{\begin{cases}x+2=-4\\x+2=4\end{cases}}\)

<=> \(\hept{\begin{cases}x=-2\\x=2\end{cases}}\)

a) \(|x+5|=5\)

\(\Rightarrow\orbr{\begin{cases}x+5=5\\x+5=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)

Vậy x = 0 hoặc x = -10

b) \(|x+1|+7=10\)

\(\Rightarrow|x+1|=10-7\)

\(\Rightarrow|x+1|=3\)

\(\Rightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)

Vậy x = 2 hoặc x = -4

c) \(|x-3|-6=5\)

\(\Rightarrow|x-3|=5+6\)

\(\Rightarrow|x-3|=11\)

\(\Rightarrow\orbr{\begin{cases}x-3=11\\x-3=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=14\\x=-8\end{cases}}\)

Vậy x = 14 hoặc x = -8

d) \(|x+2|-6\left(x-4\right)=20-6x\)

\(\Rightarrow|x+2|-6x+24=20-6x\)

\(\Rightarrow|x+2|=20-6x-24+6x\)

\(\Rightarrow|x+2|=\left(20-24\right)+\left(-6x+6x\right)\)

\(\Rightarrow|x+2|=-4\)

Vì \(|x|\ge0\)mà \(|x+2|=-4\)

\(\Rightarrow\)Không có giá trị x thỏa mãn 

_Chúc bạn học tốt_

a) \(2^{3x+2}=4^{x+5}\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\Leftrightarrow2^{3x+2}=2^{2x+10}\)

\(\Rightarrow3x+2=2x+10\Leftrightarrow3x+2-2x-10\)

\(\Leftrightarrow x-8=0\Leftrightarrow x=8\) vậy \(x=8\)

câu : b ; c mk cảm thấy đề sai

a) Ta có : \(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)

\(\Rightarrow\frac{x}{3}-\frac{1}{5}=\frac{4}{y}\)

\(\Rightarrow\frac{x.5}{15}-\frac{3}{15}=\frac{4}{y}\)

\(\Rightarrow\frac{x.5-3}{15}=\frac{4}{y}\)

\(\Rightarrow\left(x.5-3\right).y=15.4\)

\(\Rightarrow x.5.y-3.5=60\)

\(\Rightarrow xy5-15=60\)

 \(\Rightarrow xy5=60+15\)

\(\Rightarrow xy5=75\) 

\(\Rightarrow xy=75\div5\)

\(\Rightarrow xy=15\)

\(\Rightarrow xy=1.15=3.5=\left(-15\right)\left(-1\right)=\left(-3\right)\left(-5\right)=\left(-5\right)\left(-3\right)=\left(-1\right)\left(-15\right)=5.3=15.1\)

Do đó x = 1 thì y = 15

x = 3 thì y =5

x = -15 thì y = -1

x = -3 thì y = -5

x = -5 thì y = -3

x = -1 thì y = -15

x = 5 thì y = 3

x = 15 thì y = 1