a: ĐKXĐ: 1-x>0

=>x<1

=>TXĐ là D=(-∞;1)

c: ĐKXĐ: \(x^2-x-12>0\)

=>\(x^2-4x+3x-12>0\)

=>(x-4)(x+3)>0

=>x>4 hoặc x<-3

Vậy: TXĐ là D=(4;+∞)\(\cup\) (-∞;-3)

b: ĐKXĐ: \(\frac{2x+1}{x+5}>0\) và \(\log_{0,5}\frac{2x+1}{x+5}-2\ge0\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(\log_{0,5}\frac{2x+1}{x+5}\ge2\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(\frac{2x+1}{x+5}\le\frac14\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(\frac{2x+1}{x+5}-\frac14\le0\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(\frac{4\left(2x+1\right)-\left(x+5\right)}{4\left(x+5\right)}\le0\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(\frac{8x+4-x-5}{4\left(x+5\right)}\le0\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(\frac{7x-1}{x+5}\le0\)

=>\(\left[\begin{array}{l}x>-\frac12\\ x<-5\end{array}\right.\) và \(-5

=>\(-\frac12

=>TXĐ là D=(-1/2;1/7]

1.

\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)

\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)

\(=\dfrac{27}{4}log_2a\)

2.

\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)

\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)

cho e hỏi tại sao \(3\log_{2^2}\sqrt{a}\) lại bằng \(3.\dfrac{1}{2}.\dfrac{1}{2}\log_2a\) và \(2\log_{a^{\dfrac{1}{2}}}a=2.2.\log_2a\)

\(\dfrac{1}{5}\)

-1

giải thích ra đc k ạ

hello

ban solo voi minh khong

\(\left[log_24x\right]^2-log_{\sqrt{2}}2x=5\)

=>\(\left[log_2\left(2\cdot2x\right)\right]^2-log_{2^{\dfrac{1}{2}}}2x=5\)

=>\(\left[1+log_22x\right]^2-1:\dfrac{1}{2}\cdot log_22x=5\)

=>\(\left(log_22x\right)^2+2\cdot log_22x+1-2\cdot log_22x=5\)

=>\(\left(log_22x\right)^2=4\)

=>\(\left[{}\begin{matrix}log_22x=2\\log_22x=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow log_22x=2\)

=>\(2x=2^2=4\)

=>x=2

Bài 1:

a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=8\sqrt{7}\)

Bài 3: 

a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)

b: \(\sqrt{37^2-35^2}=12\)

c: \(\sqrt{65^2-63^2}=16\)

d: \(\sqrt{117^2-108^2}=45\)

a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)

b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\\ =\sqrt{\sqrt{5^2}+2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\sqrt{5^2}-2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left|\sqrt{5}+\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\\ =2\sqrt{3}\)

\(b,\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\\ =\sqrt{\sqrt{2^2}+2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}+\sqrt{\sqrt{2^2}-2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\\ =\left|\sqrt{2}+\sqrt{3}\right|+\left|\sqrt{2}-\sqrt{3}\right|\\ =\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{3}=2\sqrt{3}\)

a) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

\(=\sqrt{5-2\cdot\sqrt{5\cdot3}+3}-\sqrt{5+2\cdot\sqrt{5\cdot3}+1}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)

\(=-2\sqrt{3}\)

b. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{3-2\cdot\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)