\(a,2^{x+1}=32\\ 2^{x+1}=2^5\\ x+1=5\\ x=4\\ b,2^{2x}+2^{2x+1}=48\\ 2^{2x}+2\cdot2^{2x}=48\\ 3\cdot2^{2x}=48\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=2\)

\(c,3^x+5\cdot3^{x+1}=144\\ 3^x+15\cdot3^x=144\\ 16\cdot3^x=144\\ 3^x=9\\ 3^x=3^2\\ x=2\\ d,3^{x+5}=9^{x+1}\\ 3^{x+5}=3^{2x+2}\\ x+5=2x+2\\ x=3\)

 2x² - x  - 2x² - 3x + 4x + 6 

x = 6

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow9x^3-18x^2+9x-29=0\)

\(\Leftrightarrow x=2,208024627\)

a) Điều kiện: x > 0

\(\sqrt{\left(1-2x\right)^2}=x\) => (1 - 2x)² = x² => 1 - 2x = x hoặc 1 - 2x = - x

+) 1 - 2x = x => 1 = 3x => x = 1/3 (Thỏa mãn)

+) 1 - 2x = - x => 1 = x (Thỏa mãn)

Vậy....

b) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

=> \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\left(x+1\right)=0\)

=> \(\frac{43}{60}\left(x+1\right)=0\)=> x + 1 = 0 => x = - 1

Vậy....

Bài đầu tiên hình như bạn ghi sai đề

Bài thứ hai là x = -1

ĐKXĐ: \(x\ge0\)

\(x+2\sqrt{x}+1=0\)

\(\Rightarrow\left(\sqrt{x}+1\right)^2=0\)

\(\Rightarrow\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=-1\) (vô nghiệm)

                                                            Vậy  \(x\in\phi\)