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1/ a/ \(\sqrt{0,9.0,16.0,4}=\sqrt{\frac{9.16.4}{10000}}=\sqrt{\frac{\left(3.4.2\right)^2}{10^4}}=\frac{24}{1010}=\frac{6}{25}\)
b/ \(\sqrt{0,0016}=\sqrt{\frac{16}{100}}=\frac{4}{10}=\frac{2}{5}\)
c/ \(\frac{\sqrt{72}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{36}}{\sqrt{2}}=\sqrt{36}=6\)
d/ \(\frac{\sqrt{2}}{\sqrt{288}}=\frac{\sqrt{2}}{\sqrt{2}.\sqrt{144}}=\frac{1}{\sqrt{144}}=\frac{1}{12}\)
2.
a/ \(\frac{2}{a}.\sqrt{\frac{16a^2}{9}}=\frac{2}{a}.\frac{4\left|a\right|}{3}=-\frac{8a}{3a}=-\frac{8}{3}\) (Vì a<0)
b/ \(\frac{3}{a-1}.\sqrt{\frac{4a^2-8a+4}{25}}=\frac{3}{a-1}.\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3.2\left|a-1\right|}{5.\left(a-1\right)}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{\sqrt{243a}}{\sqrt{3a}}=\frac{9\sqrt{3a}}{\sqrt{3a}}=9\)
d/ \(\frac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}=\frac{3.3\sqrt{2}.\left|a\right|.\left|b\right|^2}{\sqrt{2}.\left|a\right|.\left|b\right|}=9\left|b\right|\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c = 3
Chứng minh rằng với mọi k > 0 ta luôn có....
.
Cho a,b,c là các số thực dương thỏa mãn a+b+c = 3
Chứng minh rằng với mọi k > 0 ta luôn có
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
a) \(A=\sqrt{9a}-\sqrt{16a}-\sqrt{49a}=3\sqrt{a}-4\sqrt{a}-7\sqrt{a}=-8\sqrt{a}\)
b) \(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=2+\sqrt{3}+\sqrt{2}+1-\sqrt{3}-\sqrt{2}=3\)
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)
\(=\left(4\sqrt{3}-\sqrt{2}\right)2\sqrt{3}\)
\(=24-2\sqrt{6}\)
Câu 3:
a: =>|2x-1|=4
=>2x-1=4 hoặc 2x-1=-4
=>x=-3/2 hoặc x=5/2
b: \(\Leftrightarrow2\sqrt{x+1}+3\sqrt{x+1}-2\sqrt{x+1}=5\)
=>3căn x+1=5
=>x+1=25/9
=>x=16/9
Đặt: \(x_1=\sqrt{a^2}\)
\(x_2=\sqrt{a^2+\sqrt{a^2}}\)
\(x_3=\sqrt{a^2+\sqrt{a^2+\sqrt{a^2}}}\)
...
\(x_n=\sqrt{a^2+\sqrt{a^2+...+\sqrt{a^2}}}\) ( n dấu căn )
Ta có: \(a\ne0\Rightarrow0< x_1< x_2< x_3< ...< x_{n-1}< x_n\)
Từ: \(x_n=\sqrt{a^2+\sqrt{a^2+...+\sqrt{a^2}}}\Rightarrow x_n^2=a^2+\sqrt{a^2+...+\sqrt{a^2}}\) (n-1 dấu căn ) \(=a^2+x_{n-1}\)
\(\Rightarrow x_n^2-a^2=x_{n-1}< x_n\Rightarrow x_n^2-a^2< x_n\Rightarrow x_n^2-x_n-a^2< 0\)
\(\Rightarrow\left(x_n-\frac{1}{2}\right)^2-\frac{1}{4}-a^2< 0\Rightarrow\left(x_n-\frac{1}{2}\right)^2< \frac{1+4a^2}{4}\Rightarrow x_n< \frac{1}{2}+\frac{\sqrt{1+4a^2}}{2}\) (1)
Ta cần chứng minh: \(\frac{1}{2}+\frac{\sqrt{1+4a^2}}{2}< \frac{1}{2}+\frac{1}{8}\left(\sqrt{1+16a^2}+\sqrt{9+16a^2}\right)\) (2)
Thật vậy, ta có: \(\left(2\right)\Leftrightarrow\frac{\sqrt{1+4a^2}}{2}< \frac{1}{8}\left(\sqrt{1+16a^2}+\sqrt{9+16a^2}\right)\)
\(\Leftrightarrow4\sqrt{1+4a^2}< \sqrt{1+16a^2}+\sqrt{9+16a^2}\)
\(\Leftrightarrow16\left(1+4a^2\right)< 10+32a^2+2\sqrt{\left(1+16a^2\right)\left(9+16a^2\right)}\)
\(\Leftrightarrow32a^2+6< 2\sqrt{\left(1+16a^2\right)\left(9+16a^2\right)}\)
\(\Leftrightarrow16a^2+3< \sqrt{\left(1+16a^2\right)\left(9+16a^2\right)}\)
\(\Leftrightarrow256a^4+96a^2+9< 9+160a^2+256a^4\)
\(\Leftrightarrow-64a^2< 0\) ( luôn đúng với mọi a khác 0)
=> Bất đẳng thức (2) đúng
Từ \(\left(1\right),\left(2\right)\Rightarrow x_n< \frac{1}{2}+\frac{1}{8}\left(\sqrt{1+16a^2}+\sqrt{9+16a^2}\right)\)
\(\Leftrightarrow\sqrt{a^2+\sqrt{a^2+...+\sqrt{a}}}< \frac{1}{2}+\frac{1}{8}\left(\sqrt{1+16a^2}+\sqrt{9+16a^2}\right)\)
Ngọc bổ sung một cách khác nhé :))
Ta xét vế trái, vì dễ thấy \(\sqrt{a^2+\sqrt{a^2+...+\sqrt{a^2}}}\) (n dấu căn) \(< \sqrt{a^2+\sqrt{a^2+\sqrt{a^2+...}}}\)(vô hạn dấu căn)
Ta đặt \(\sqrt{a^2+\sqrt{a^2+\sqrt{a^2+...}}}=t,t\ge0\)
\(\Rightarrow t^2=t+a^2\Rightarrow t^2-t-a^2=0\)
Ta đưa phương trình trên về phương trình bậc hai ẩn t , khi đó \(\Delta=1+4a^2>0\Rightarrow t=\frac{1+\sqrt{1+4a^2}}{2}\) (vì \(t\ge0\))
Do vậy ta chỉ cần chứng minh \(\frac{1+\sqrt{1+4a^2}}{2}< \frac{1}{2}+\frac{1}{8}\left(\sqrt{1+16a^2}+\sqrt{9+16a^2}\right)\)
\(\Leftrightarrow4\sqrt{1+4a^2}< \sqrt{1+16a^2}+\sqrt{9+16a^2}\)
\(\Leftrightarrow16\left(1+4a^2\right)< 32a^2+10+2\sqrt{1+16a^2}.\sqrt{9+16a^2}\)
\(\Leftrightarrow16a^2+3< \sqrt{1+16a^2}.\sqrt{9+16a^2}\)
\(\Leftrightarrow\left(16a^2+3\right)^2< \left(16a^2+1\right)\left(16a^2+9\right)\)
\(\Leftrightarrow16^2a^4+96a^2+9< 16^2a^4+160a^2+9\)
\(\Leftrightarrow0< 64a^2\) (luôn đúng với \(a\ne0\))
Vậy ta có đpcm.