ĐK: x>0
ta có
\(\log_x=\log_{\left(x+1\right)^{\frac{1}{2}}}\Rightarrow x=\left(x+1\right)^{\frac{1}{2}}\Rightarrow x^2-x-1=0\Rightarrow x=\frac{1+\sqrt{5}}{2};x=\frac{1-\sqrt{5}}{2}\)(loại)
vậy \(x=\frac{1+\sqrt{5}}{2}\)
tìm x
\(\frac{3}{2}\log_{\frac{1}{2}}\left(x+2\right)^2-3=\log_{\frac{1}{4}}\left(4-x\right)^3+\log_{\frac{1}{4}}\left(6+x\right)^3\)
giải pt sau
\(\frac{1}{2}\log_5^{\left(x+5\right)}+\log_5^{\sqrt{x-3}}=\frac{1}{2}\log_5^{\left(2x+1\right)}\)
tìm x
\(\log_2\left(3x-1\right)+\frac{1}{\log_{\left(x+3\right)}2}=2+\log_2\left(x+1\right)\)
tìm x
\(2\log_9^{\left(x^2-5x+6\right)}=\log_{\sqrt{3}}^{\left(\frac{x-1}{2}\right)}+\log_3^{\left(x-3\right)^2}\)
tìm giá trị lớn nhất , nhỏ nhất trên \(\left[\frac{1}{4};4\right]\)của \(y=\frac{1}{3}log_{\frac{1}{2}}^3x+log^2_{\frac{1}{2}}x-\left(3log_{\frac{1}{2}}x\right)+1\)
Giải hệ pt:
1.\(\sqrt[4]{x}\left(\left\{\left\{\frac{1}{4}+\frac{2\sqrt{x}+\sqrt{y}}{x+y}\right\}\right\}\right)=2\)
2.\(\sqrt[4]{y}\left(\frac{1}{4}-\frac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=1\)
SOS
Bất phương trình logarit
$$1) \sqrt{log_{1/2}^{2} \frac{2x}{4-x} - 4} \leq \sqrt{5}$$
$$2)log_{2}(x-1)^{2} > 2log_{2} (x^{3} +x +1)$$
$$3)\frac{1}{log_{2}(4x)^{2} +3 } + \frac{1}{log_{4} 16x^{3}-2} <-1$$
$$4)log_{2} (4^{x}+4) < log_{\frac{1}{2}} (2^{x+1} -2)$$
1 / giải phương trình sau:
\(\frac{1}{\left(x+2000\right).\left(x+2001\right)}+\frac{1}{\left(x+2001\right).\left(x+2002\right)}...\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)
Giải pt :
\(\left(8x-4x^2-1\right)\left(x^2+2x+1\right)=4\left(x^2+x+1\right)\)
Không nhân hết ra nhé!
