Bài 4: Liên hệ giữa phép chia và phép khai phương

Lời giải:

a. Bạn xem lại đề. Phân thức này rút gọn cũng không đẹp lắm

b. \(2|x+y|\sqrt{\frac{1}{x^2+2xy+y^2}}=2|x+y|\sqrt{\frac{1}{(x+y)^2}}=2|x+y|.|\frac{1}{x+y}|=2|(x+y).\frac{1}{x+y}|=2\)c. 

\(=\frac{(x-5)^4}{(x-4)^2}-\frac{(x-5)(x+5)(x-4)}{(x-4)^2}=\frac{(x-5)^4-(x-5)(x+5)(x-4)}{(x-4)^2}=\frac{(x-5)(x^3-16x^2-105)}{(x-4)^2}\)

\(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\left(1-\dfrac{3}{\sqrt{x}}\right)\)

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

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

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) và ghi đầy đủ yêu cầu đề để được hỗ trợ tốt hơn nhé.

Bài 4:

a) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)

\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

\(=-\sqrt{3}\)

b) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

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

\(=2\sqrt{2}\)

c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)

\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)

\(=6\sqrt{a}\)

Bài 1:

a) \(\sqrt{7x^2}=\left|x\right|\sqrt{7}=x\sqrt{7}\) (Do \(x>0\Rightarrow\left|x\right|=x\)) 

b) \(\sqrt{8y^2}=\sqrt{2^2\cdot2\cdot y^2}=\left|2y\right|\sqrt{2}=-2y\sqrt{2}\) (Do \(y< 0\Rightarrow\left|2y\right|=-2y\))

c) \(\sqrt{25x^3}=\sqrt{5^2\cdot x^2\cdot x}=\left|5x\right|\sqrt{x}=5x\sqrt{x}\) (Do \(x>0\Rightarrow\left|5x\right|=5x\)

d) \(\sqrt{48y^4}=\sqrt{4^2\cdot3\cdot\left(y^2\right)^2}=\left|4y^2\right|\sqrt{3}=4y^2\sqrt{3}\) (Do \(y< 0\Rightarrow\left|4y^2\right|=4y^2\left(y^2\ge0\right)\))

e) \(\sqrt{75a^3}=\sqrt{5^2\cdot3\cdot a^2\cdot a}=\left|5a\right|\sqrt{3a}=5a\sqrt{3a}\) (Do \(a>0\Rightarrow\left|5a\right|=5a\))

f) \(\sqrt{98a^5\left(b^2-6b+9\right)}=\sqrt{7^2\cdot2\cdot\left(a^2\right)^2\cdot a\left(b-3\right)^2}=\left|7a^2\left(b-3\right)\right|\sqrt{2a}=7a^2\left|b-3\right|\sqrt{2a}\)

3:

a: 3*căn 7=căn 63

2*căn 15=căn 60

mà 63>60

nên 3*căn 7>2*căn 15

b: -4căn 5=-căn 80

-5căn 3=-căn 75

mà 80>75

nên -4căn 5<-5 căn 3

\(P=\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\) với \(\dfrac{1}{4}< x< \dfrac{1}{2}\)

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

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

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

Do \(\dfrac{1}{4}< x< \dfrac{1}{2}\Leftrightarrow0< \sqrt{4x-1}< 1\)

\(\Rightarrow P=\dfrac{1}{\sqrt{2}}\left(\sqrt{4x-1}+1+1-\sqrt{4x-1}\right)=\sqrt{2}\)

Vậy \(P=\sqrt{2}\).

\(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}-\sqrt{x}-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\left(\text{đ}pcm\right)\)

`(\sqrtx-1)/(\sqrtx)+(2\sqrtx+1)/(x+\sqrtx)`

`=(\sqrtx-1)/(\sqrtx)+(2\sqrtx+1)/(\sqrtx(\sqrtx+1))`

`=((\sqrtx-1)(\sqrtx+1))/(\sqrtx(\sqrtx+1))+(2\sqrtx+1)/(\sqrtx(\sqrtx+1))`

`=(x-1)/(\sqrtx(\sqrtx+1))+(2\sqrtx+1)/(\sqrtx(\sqrtx+1))`

`=(x-1+2\sqrt+1)/(\sqrtx(\sqrtx+1))`

`=(x+2\sqrtx)/(\sqrtx(\sqrtx+1))`

`=(\sqrtx(\sqrtx+2))/(\sqrtx(\sqrtx+1))`

`=(\sqrtx+2)/(\sqrtx+1)` (đpcm)

1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)

2: \(=6+3=9\)

3: \(=\sqrt{3}+1-\sqrt{3}=1\)

4: \(=-\sqrt{2}-\sqrt{2}=-2\sqrt{2}\)

5: \(=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

6: \(=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)

7: \(=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)

8: \(=3\sqrt{2}-2\sqrt{2}=\sqrt{2}\)

9: \(=-3-3\sqrt{3}-\sqrt{3}=-3-4\sqrt{3}\)

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

\(=\dfrac{2\sqrt{2}-1-3\sqrt{2}-3+2+\sqrt{2}}{5+4\sqrt{2}}=\dfrac{-2}{5+4\sqrt{2}}\)

2: \(=6\sqrt{3}+3\sqrt{3}-\sqrt{3}=8\sqrt{3}\)

4:

\(=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{16-7}=8-2\cdot3=2\)

6: \(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)

8: \(=3\sqrt{2}-3\sqrt{2}+2\sqrt{2}+6\sqrt{2}=8\sqrt{2}\)

10: \(=3\sqrt{3}+5\sqrt{3}-20\sqrt{3}+2\sqrt{3}=-10\sqrt{3}\)

12: \(=9\sqrt{2}-10\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

\(\left(\sqrt{x}+7-3\sqrt{2}\right)\sqrt{3}=\sqrt{6}\\ \Rightarrow\left(\sqrt{x}+7-3\sqrt{2}\right)=\sqrt{3.6}=\sqrt{18}=3\sqrt{2}\\ \Rightarrow\sqrt{x}=3\sqrt{2}+3\sqrt{2}-7=6\sqrt{2}-7\\ \Rightarrow x=\left(6\sqrt{2}-7\right)^2=72-84\sqrt{2}+49=121-84\sqrt{2}\)

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

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

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

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

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

\(=\dfrac{3\sqrt{3}+9}{\sqrt{3}\left(\sqrt{3}+1\right)}\)

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

\(=3\)

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

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

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

\(=\dfrac{3\sqrt{3}+9}{3+\sqrt{3}}\)

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

\(=3\)