32:(3x-2)=2^3
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a) \(x-2=-6\)
\(x=-6+2\)
\(x=-4\)
b) \(15-\left(x-7\right)=-21\)
\(x-7=36\)
\(x=43\)
c) \(4.\left(3x-4\right)-2=18\)
\(4\left(3x-4\right)=20\)
\(3x-4=5\)
\(3x=9\)
\(x=3\)
d) \(\left(3x-6\right)+3=32\)
\(3x-6=29\)
\(3x=29+6\)
\(3x=35\)
\(x=\frac{35}{3}\)
e) \(\left(3x-6\right).3=32\)
\(3x-6=\frac{32}{3}\)
\(3x=\frac{32}{3}+6\)
\(3x=\frac{50}{3}\)
\(x=\frac{50}{9}\)
f) \(\left(3x-6\right):3=32\)
\(3x-6=96\)
\(3x=102\)
\(x=34\)
g) \(\left(3x-6\right)-3=32\)
\(3x-6=35\)
\(3x=41\)
\(x=\frac{41}{3}\)
h) \(\left(3x-2^4\right).7^3=2.7^4\)
\(\left(3x-2^4\right)=2.7=14\)
\(\left(3x-16\right)=14\)
\(3x=14+16=30\)
\(x=10\)
i) \(\left|x\right|=\left|-7\right|\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
k) \(\left|x+1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
l) \(\left|x-2\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
m) \(x+\left|-2\right|=0\)
\(x+2=0\)
\(x=-2\)
o) \(72-3\left|x+1\right|=9\)
\(3\left|x-1\right|=63\)
\(\left|x-1\right|=21\)
\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)
p) Ta có: \(\left|x-1\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)
mà \(x+1< 0\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-2\)
q) \(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
hok tốt!!


a)
\(3^x+3^{x+1}+3^{x+2}=117\\ \Leftrightarrow3^x+3.3^x+9.3^x=117\\ 13.3^x=117\\ \Leftrightarrow3^x=9\\ \Leftrightarrow3^x=3^2\\ \Leftrightarrow x=2\)
b)
\(3+4\left(x-10\right)=3^2+6\\ \Leftrightarrow3+4\left(x-10\right)=15\\ \Leftrightarrow4\left(x-10\right)=12\\ \Leftrightarrow x-10=3\\ \Leftrightarrow x=13\)
a) \(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x.\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(x=2\)
b) \(3+4\left(x-10\right)=3^2+6\)
\(3+4x-40=9+6\)
\(4x=15+40-3\)
\(4x=52\)
\(x=13\)

Hàm số y = 2 + 3 2 − 3 + 2 − 3 2 + 3 x − 5 có:
a = 2 + 3 2 − 3 + 2 − 3 2 + 3 = 2 + 3 2 + 2 − 3 2 2 − 3 2 + 3 = 4 + 4 3 + 3 + 4 − 4 3 + 3 4 − 3 = 14 > 0
nên là hàm số đồng biến trên
Đáp án cần chọn là: B

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35

\(\left(3x-2\right)^3=2.32\)
\(=>\left(3x-2\right)^3=64\)
\(=>\left(2x-3\right)^3=4^3\)
\(=>2x-3=4\)
\(=>2x=7\)
\(=>x=\frac{7}{2}\)
Ủng hộ nha
\(\left(3x-2\right)^3=2.32\)
\(\left(3x-2\right)^3=64\)
Cách 1: \(3x-2=\sqrt[3]{64}\)(cách này lớp 7)
\(3x-2=4\)
\(3x=4+2\)
\(3x=6\)
\(\Rightarrow x=6:3=2\)
Cách 2:
\(\Rightarrow4^3=64\)
\(\Rightarrow3x-2=4\)
(giải y như cách trên)
\(32:\left(3x-2\right)=2^3\\ \Leftrightarrow32:\left(3x-2\right)=8\\ \Leftrightarrow3x-2=\frac{32}{8}\\ \Leftrightarrow3x-2=4\\ \Leftrightarrow3x=4+2\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=\frac{6}{3}\\ \Leftrightarrow x=2\)
Vậy \(x=2\)
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