ĐK: \(x\ge3\)

ta có:

\(\log_5^{\left(x+5\right)^{\frac{1}{2}}}+\log_5^{\sqrt{x-3}}=\log_5^{\sqrt{2x+1}}\Rightarrow\log_5^{\sqrt{\left(x+5\right)\left(x-3\right)}}=\log_5^{\sqrt{2x+1}}\) 

suy ra \(\sqrt{\left(x+5\right)\left(x-3\right)}=\sqrt{2x+1}\Rightarrow\left(x+5\right)\left(x-3\right)=2x+1\Leftrightarrow x^2+2x-15=2x+1\Leftrightarrow x^2=16\Rightarrow x=\pm4\)

mà \(x\ge3\)

suy ra x=4 là nghiệm của pt

a: ĐKXĐ: \(4x-3>0\)

=>x>3/4

\(log_5\left(4x-3\right)=2\)

=>\(log_5\left(4x-3\right)=log_525\)

=>4x-3=25

=>4x=28

=>x=7(nhận)

b: ĐKXĐ: \(x\ne0\)

\(log_2x^2=2\)

=>\(log_2x^2=log_24\)

=>\(x^2=4\)

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

c: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{3}{2}\right\}\)

\(\log_52x+1=\log_5-2x+3\)

=>2x+1=-2x+3

=>4x=2

=>\(x=\dfrac{1}{2}\left(nhận\right)\)

d: ĐKXD: \(x\notin\left\{3\right\}\)

\(ln\left(x^2-6x+7\right)=ln\left(x-3\right)\)

=>\(x^2-6x+7=x-3\)

=>\(x^2-7x+10=0\)

=>(x-2)(x-5)=0

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\notin\left\{\dfrac{1}{5};2\right\}\)

\(log\left(5x-1\right)=log\left(4-2x\right)\)

=>5x-1=4-2x

=>7x=5

=>\(x=\dfrac{5}{7}\left(nhận\right)\)

Điều kiện :  

                 \(\log_{\frac{1}{5}}\left(\log_5\frac{x^2+1}{x+3}\right)\ge0\)

           \(\Leftrightarrow0< \log_{\frac{1}{5}}\left(\log_5\frac{x^2+1}{x+3}\right)\le1\)

           \(\Leftrightarrow\log_51< \log_5\frac{x^2+1}{x+3}\le\log_55\)

\(\Leftrightarrow1< \frac{x^2+1}{x+3}\le5\)\(\Leftrightarrow\begin{cases}\frac{x^2-x-2}{x+3}>0\\\frac{x^2-5x-14}{x+3}\le0\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}-3< x< -1\\x>2\end{array}\right.\) và \(\left[\begin{array}{nghiempt}x< -3\\-2\le x\le7\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}-2\le x< -1\\2< x\le7\end{array}\right.\)

Vậy tập xác định là D = [-2;-1) U (2;7]

Bạn coi lại đề câu a, chỗ \(\log_5-x\) đó

b.

\(\Leftrightarrow9^x-3^x-2.3^x-2=0\)

\(\Leftrightarrow3^x\left(3^x-1\right)-2\left(3^x-1\right)=0\)

\(\Leftrightarrow\left(3^x-2\right)\left(3^x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3^x=2\\3^x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\log_32\\x=0\end{matrix}\right.\)

\(E=16\left[\log_{3^{-2}}3^{\frac{3}{2}}\right]^2+23\log_{2^{\frac{9}{2}}}2^{\frac{5}{2}}-12\log_55^{-3}=16\left(-\frac{3}{4}\right)^2+9\frac{5}{9}-12\left(-3\right)=50\)

Điều kiện x>0. Nhận thấy x=2 là nghiệm. 

Nếu x>2 thì

\(\frac{x}{2}>\frac{x+2}{4}>1\); \(\frac{x+1}{3}>\frac{x+3}{5}>1\)

Suy ra 

\(\log_2\frac{x}{2}>\log_2\frac{x+2}{4}>\log_4\frac{x+2}{4}\)hay :\(\log_2x>\log_2\left(x+2\right)\)

\(\log_3\frac{x+1}{3}>\log_3\frac{x+3}{5}>\log_5\frac{x+3}{5}\) hay \(\log_3\left(x+1\right)>\log_5\left(x+3\right)\)

Suy ra vế trái < vế phải, phương trình vô nghiệm.

Đáp số x=2

Điều kiện \(x+y>0\)

Từ hệ phương trình \(\Leftrightarrow\begin{cases}5^{\frac{x-y}{3}}.3^{y-x}=\frac{5}{27}\\x+y=5^{\frac{x-y}{3}}\end{cases}\) \(\Leftrightarrow\begin{cases}\left(\frac{\sqrt[3]{5}}{3}\right)^{x-y}=\left(\frac{\sqrt[3]{5}}{3}\right)^3\\x+y=5^{\frac{x-y}{3}}\end{cases}\)  \(\Leftrightarrow\begin{cases}x-y=3\\x+y=5\end{cases}\) \(\Leftrightarrow\begin{cases}x=4\\y=1\end{cases}\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(4;1\right)\)

1/ ĐKXĐ: \(x>0\)

\(log_{5x}5-log_{5x}x+log_5^2x=1\)

\(\Leftrightarrow\dfrac{1}{log_55x}-\dfrac{1}{log_x5x}+log_5^2x=1\)

\(\Leftrightarrow\dfrac{1}{1+log_5x}-\dfrac{1}{1+log_x5}+log_5^2x-1=0\)

\(\Leftrightarrow\dfrac{1}{1+log_5x}-\dfrac{log_5x}{1+log_5x}+\left(log_5x-1\right)\left(log_5x+1\right)=0\)

\(\Leftrightarrow\dfrac{1-log_5x}{1+log_5x}-\left(1-log_5x\right)\left(1+log_5x\right)=0\)

\(\Leftrightarrow\left(1-log_5x\right)\left(\dfrac{1}{1+log_5x}-\left(1+log_5x\right)\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1-log_5x=0\\\dfrac{1}{1+log_5x}=1+log_5x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}1-log_5x=0\\1+log_5x=1\\1+log_5x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=\dfrac{1}{25}\end{matrix}\right.\)

2/ ĐKXĐ: \(x>0\)

\(log_5\left(5^x-1\right).log_{25}\left(5^{x+1}-5\right)=1\)

\(\Leftrightarrow log_5\left(5^x-1\right).log_{5^2}5\left(5^x-1\right)=1\)

\(\Leftrightarrow log_5\left(5^x-1\right)\left(1+log_5\left(5^x-1\right)\right)=2\)

\(\Leftrightarrow log_5^2\left(5^x-1\right)+log_5\left(5^x-1\right)-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}log_5\left(5^x-1\right)=1\\log_5\left(5^x-1\right)=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5^x-1=5\\5^x-1=\dfrac{1}{25}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5^x=6\\5^x=\dfrac{26}{25}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=log_56\\x=log_5\dfrac{26}{25}\end{matrix}\right.\)

3/ ĐKXĐ: \(x>0\)

\(2log_3^2x-log_3x.log_3\left(\sqrt{2x+1}-1\right)=0\)

\(\Leftrightarrow log_3x\left(2log_3x-log_3\left(\sqrt{2x+1}-1\right)\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}log_3x=0\Rightarrow x=1\\2log_3x-log_3\left(\sqrt{2x+1}-1\right)=0\left(1\right)\end{matrix}\right.\)

Xét (1): \(log_3x^2=log_3\left(\sqrt{2x+1}-1\right)\Leftrightarrow x^2=\sqrt{2x+1}-1\)

\(\Leftrightarrow x^2+1=\sqrt{2x+1}\Leftrightarrow x^4+2x^2+1=2x+1\)

\(\Leftrightarrow x^4+2x^2-2x=0\Leftrightarrow x\left(x^3+2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x^3+2x-2=0\end{matrix}\right.\) ????

Pt bậc 3 kia có nghiệm rất xấu, chỉ giải được bằng công thức Cardano mà bậc phổ thông không học, nên bạn có chép đề sai không vậy?