\(A=2\sqrt{1}+2\sqrt{3}+...+2\sqrt{21}\)

\(A=2.\left(\sqrt{1}+\sqrt{3}+...+\sqrt{21}\right)\)

\(B=2\sqrt{2}+2\sqrt{4}+....2\sqrt{22}\)

\(B=2.\left(\sqrt{2}+\sqrt{4}+...+\sqrt{22}\right)\)

Có \(\sqrt{1}+\sqrt{3}+...+\sqrt{21}\) Có 11 số hạng.

\(\sqrt{2}+\sqrt{4}+...+\sqrt{22}\) Có 11 số hạng.

Mà \(\hept{\begin{cases}\sqrt{1}< \sqrt{2}\\....\\\sqrt{21}< \sqrt{22}\end{cases}}\)

=> \(2.\left(\sqrt{1}+\sqrt{3}+...+\sqrt{21}\right)< 2.\left(\sqrt{2}+\sqrt{4}+...+\sqrt{22}\right)\)

\(\Rightarrow A< B\)

fffffffffffffff

thua bạn luôn rồi đấy

ko hỉu sao ! 

trên đời có nhìu người zai dữ ~

Haiz...

bị trừ điểm giờ ! 

b) Ta có: \(B=3\sqrt{50}-7\sqrt{8}+12\sqrt{18}\)

\(=15\sqrt{2}-14\sqrt{2}+36\sqrt{2}\)

\(=37\sqrt{2}\)

c) Ta có: \(C=2\sqrt{80}-2\sqrt{245}+2\sqrt{180}\)

\(=8\sqrt{5}-14\sqrt{5}+12\sqrt{5}\)

\(=6\sqrt{5}\)

d) Ta có: \(D=2\sqrt{12}-\sqrt{48}+3\sqrt{27}-\sqrt{108}\)

\(=4\sqrt{3}-4\sqrt{3}+9\sqrt{3}-6\sqrt{3}\)

\(=3\sqrt{3}\)

thx bn

a) \(\sqrt{1-4x+4x^2}=5\) 

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)

\(\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2+6x+9}=3x-1\)

\(\Leftrightarrow\sqrt{\left(x+3\right)^2=3x-1}\)

\(\Leftrightarrow\left|x+3\right|=3x-1\)

\(\Leftrightarrow x+3=3x-1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)

\(TH_1:x\le\dfrac{1}{2}\)

\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)

\(TH_2:x\ge\dfrac{1}{2}\)

\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)

\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)

\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)

a) \(\sqrt{-\dfrac{2t}{3}}.\sqrt{-\dfrac{3t}{8}}=\sqrt{-\dfrac{2t}{3}.\left(-\dfrac{3t}{8}\right)}=\sqrt{\dfrac{1}{4}t^2}=\left|\dfrac{1}{2}t\right|=-\dfrac{1}{2}t\left(t< 0\right)\)

b) \(\sqrt{\dfrac{2a^3}{b}}:\sqrt{\dfrac{ab^3}{8}}=\sqrt{\dfrac{2a^3}{b}:\dfrac{ab^3}{8}}=\sqrt{\dfrac{2a^3}{b}.\dfrac{8}{ab^3}}=\sqrt{16\dfrac{a^2}{b^4}}\)

\(=\left|4\dfrac{a}{b^2}\right|=-\dfrac{4a}{b^2}\left(a< 0\right)\)

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

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

Câu 5:

a: Thay m=5 vào hệ, ta được:

\(\begin{cases}x+5y=5+1=6\\ 5x+y=2\cdot5=10\end{cases}\Rightarrow\begin{cases}5x+25y=30\\ 5x+y=10\end{cases}\)

=>\(\begin{cases}5x+25y-5x-y=30-10\\ x+5y=6\end{cases}\Rightarrow\begin{cases}24y=20\\ x=6-5y\end{cases}\)

=>\(\begin{cases}y=\frac56\\ x=6-5\cdot\frac56=6-\frac{25}{6}=\frac{11}{6}\end{cases}\)

b: Để hệ có nghiệm duy nhất thì \(\frac{1}{m}<>\frac{m}{1}\)

=>\(m^2<>1\)

=>m∉{1;-1}

\(\begin{cases}x+my=m+1\\ mx+y=2m\end{cases}\Rightarrow\begin{cases}mx+m^2y=m^2+m\\ mx+y=2m\end{cases}\)

=>\(\begin{cases}mx+m^2y-mx-y=m^2+m-2m\\ x+my=m+1\end{cases}\Rightarrow\begin{cases}y\left(m^2-1\right)=m^2-m\\ x=m+1-my\end{cases}\)

=>\(\begin{cases}y=\frac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\frac{m}{m+1}\\ x=m+1-m\cdot\frac{m}{m+1}=\frac{m^2+2m+1-m^2}{m+1}=\frac{2m+1}{m+1}\end{cases}\)

x+5y=4

=>\(\frac{2m+1}{m+1}+\frac{5m}{m+1}=4\)

=>7m+1=4(m+1)=4m+4

=>3m=3

=>m=1(loại)