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Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.
A= ( \(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\))
= (1+1,4+1,7)+(4,4+6,3+7,7)
= 4,1+18,4
=22,5
Ta sẽ chứng minh 1 bđt sau:
\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
\(\Rightarrow a+2\sqrt{ab}+b\ge a+b\)
\(\Rightarrow a+2\sqrt{ab}+b-a-b\ge0\)
\(\Rightarrow2\sqrt{ab}\ge0\) *đúng*
Dấu "=" xảy ra khi: \(ab=0\)
Trở lại bài toán,vì không có thừa số nào bằng 0,nên ta dễ dàng có: \(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
Hay \(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}=\left(\sqrt{1}+\sqrt{20}\right)+\left(\sqrt{40}+\sqrt{2}\right)+\left(\sqrt{60}+\sqrt{3}\right)>\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}=A\)
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)
\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)
\(=\frac{4}{5}+11-2\)
\(=\frac{4}{5}+9\)
\(=\frac{49}{9}\)
b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+4-5+64\)
= 55
c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)
\(=\frac{\sqrt{48}}{91-7}\)
\(=\frac{4\sqrt{3}}{84}\)
\(=\frac{\sqrt{3}}{41}\)
d) Xem lại đề nhé em!
e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)
\(=5-3.\frac{2}{3}\)
= 5 - 2
= 3
h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)
\(=-9.\frac{1}{3}-7+125:5\)
\(=-3-7+25\)
= 15
pn lấy đề ở đâu vậy ?
Ở lớp học thêm c ạ