Bài 3: Liên hệ giữa phép nhân và phép khai phương

Đề đánh bị lỗi.

Áp dụng bất đẳng thức Bunhiacopski:

\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{\left[\sqrt{c}^2+\sqrt{\left(a-c\right)}^2\right]\left[\sqrt{c}^2+\sqrt{\left(b-c\right)}^2\right]}\)

\(=\sqrt{\left(c+a-c\right)\left(c+b-c\right)}=\sqrt{ab}\)

\(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2\sqrt{10}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)

\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\) ( ĐK : \(-4\le x\le1\) )

\(\Leftrightarrow x+4-2\sqrt{\left(x+4\right)\left(1-x\right)}+1-x=1-2x\)

\(\Leftrightarrow-2\sqrt{\left(x+4\right)\left(1-x\right)}=-2x-4\)

\(\Leftrightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=x+2\) ( ĐK : \(x\ge-2\) )

\(\Leftrightarrow\left(x+4\right)\left(1-x\right)=\left(x+2\right)^2\)

\(\Leftrightarrow-x^2-3x+4=x^2+4x+4\)

\(\Leftrightarrow-2x^2-7x=0\)

\(\Leftrightarrow x\left(-2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-\dfrac{7}{2}\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

Chúc bạn học tốt !

\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\\ \Leftrightarrow-\sqrt{1-x}=\sqrt{1-2x}-\sqrt{x+4}\\ \Leftrightarrow1-x=1-2x-2\sqrt{\left(1-2x\right)\left(x+4\right)}+x+4\Leftrightarrow-x=-2x-2\sqrt{x+4-2x^2-8x}+x+4\\ \Leftrightarrow-x=-x-2\sqrt{-7x+4-2x^2}+4\\ \Leftrightarrow0=-2\sqrt{-7x+4-2x^2}+4\\ \Leftrightarrow2\sqrt{-7x+4-2x^2}=4\\ \Leftrightarrow\sqrt{-7x+4-2x^2}=2\\ \Leftrightarrow-7x+4-2x^2=4\\ \Leftrightarrow-7x-2x^2=0\\ \Leftrightarrow-x\left(7+2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x=0\\7+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-\dfrac{7}{2}\left(ktm\right)\end{matrix}\right.\)

ĐK: x > -2

\(\dfrac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\Leftrightarrow3x+2=2\left(x+2\right)\Leftrightarrow3x+2=2x+4\Leftrightarrow3x-2x=4-2\Leftrightarrow x=2\left(tm\right)\)

Vậy x = 2

\(\dfrac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\) . ĐKXĐ : \(x>-2\)

\(\Leftrightarrow\) \(\dfrac{3x+2}{\sqrt{x+2}}=\dfrac{2\sqrt{x+2}.\sqrt{x+2}}{\sqrt{x+2}}\)

\(\Leftrightarrow3x+2=2x+4\)

\(\Leftrightarrow x=2\) ( Thỏa mãn )

Vậy \(S=\left\{2\right\}\)

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

⇔ \(3x+2=2\left(x+2\right)\)

⇔ \(3x+2=2x+4\)

⇔ \(x=2\left(TM\right)\)

KL.........

a) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)

\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)

\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)

\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)

\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)

\(=\sqrt{\left(4-3\right)\cdot4}\)

\(=\sqrt{1\cdot4}\)

\(=\sqrt{4}\)

\(=2\)

b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)

\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)

\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)

\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)

\(=0+14\)

\(=14\)

c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D

\(1a.2\sqrt{25xy}+\sqrt{225x^3y^3}-3y\sqrt{16x^3y}=10\sqrt{xy}+15xy\sqrt{xy}-12xy\sqrt{xy}=10\sqrt{xy}+3xy\sqrt{xy}=\sqrt{xy}\left(10+3xy\right)\left(x,y\ge0\right)\)

\(b.-\sqrt{36b}-\dfrac{1}{3}\sqrt{54b}+\dfrac{1}{5}\sqrt{150b}=-6\sqrt{b}-\sqrt{6b}+\sqrt{6b}=-6\sqrt{b}\left(b\ge0\right)\)

\(2.\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)

\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}-\sqrt{3x}-3\sqrt{1-2x}=0\)

\(\Leftrightarrow\sqrt{1-2x}=4\sqrt{3x}\left(x\ge\dfrac{1}{2}\right)\)

\(\Leftrightarrow1-2x=48x\)

\(\Leftrightarrow x=\dfrac{1}{50}\left(KTM\right)\)

KL....

2)

\(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}\)

\(=\sqrt{12,1.36.10}\)

= \(\sqrt{121.36}\)

\(=\sqrt{4356}\)

\(=66\)

3)

\(\sqrt{5a}.\sqrt{45a}-3a\)

\(=\sqrt{5.45a^2}-3a\)

\(=\sqrt{225a^2}-3a\)

\(=\sqrt{\left(15a\right)^2}-3a\)

\(=-15a-3a\) ( vì \(a\le0\))

\(=-18a\)

5)

\(\sqrt{0,36a^2}\)

\(=\sqrt{\left(0,6a\right)^2}\)

\(=-0,6a\) ( vì \(a< 0\) )

Để tối mình rảnh lên coi có làm tiếp được nữa hông thì mình làm ha.

Chúc bạn học tốt!

1)

\(\sqrt{3a^3}.\sqrt{12}\)

\(=\sqrt{3}.\sqrt{a^3}.\sqrt{12}\)

\(=\sqrt{3.12}.\sqrt{a^3}\)

\(=6\sqrt{a^3}\)

4)

\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

\(=9.6a.a^2-\sqrt{0,2}.\sqrt{18}.\sqrt{10}.\sqrt{a^2}\)

\(=54a^3-\sqrt{2}.\sqrt{18}.\sqrt{a^2}\)

\(=34a^3-\sqrt{2.18}.\sqrt{a^2}\)

\(=54a^3-6\sqrt{a^2}\)

\(=54a^3-6a^2\) ( vì a<0)

6)

\(\sqrt{a^4.\left(3-a^{ }\right)^2}\)

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

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

\(=\left|a^2\right|\left|3-a\right|\) ( vì a>3 => a>3 nên 3-a<0)

Mà

\(=a^2\left(a-3\right)\)

\(=a^3-3a^2\)

Còn lại bạn làm tương tự nha, trể quá rùi :)))))

a: \(=\dfrac{\left(1-\sqrt{2}\right)^2}{1-\sqrt{2}}=1-\sqrt{2}\)

b: \(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{x-y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

d: \(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x-y}=\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

Mình làm được câu đầu với câu cuối thui bạn :(

\(2\sqrt{10}.5\sqrt{8}.\sqrt{2}\)

\(=\sqrt{10.4}.\sqrt{2.100}.\sqrt{2}\)

\(=\sqrt{40}.\sqrt{200}.\sqrt{2}\)

\(=\sqrt{40.200.2}\)

\(=40\sqrt{10}\)

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

\(=\left(3\sqrt{2}\right)^2+2.3\sqrt{3}.\sqrt{3}+\left(\sqrt{3}\right)^2\)

\(=18+18+3\)

\(=39\)

Chúc bạn học tốt! :)

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1-\sqrt{3}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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