K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

24 tháng 8 2017

a, \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

\(=\sqrt{x}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{y}\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

b, \(x^2+x\sqrt{x}+\sqrt{x}+1\)

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

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

c, \(\sqrt{6x}-x2\sqrt{3}\)

\(=\sqrt{6x}-\sqrt{12x^2}\)

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

22 tháng 4 2017

Để học tốt Toán 9 | Giải bài tập Toán 9

Để học tốt Toán 9 | Giải bài tập Toán 9

20 tháng 7 2017

\(a)\) \(xy-y\sqrt{x}+\sqrt{x}-1\)

= \(y\sqrt{x}.(\sqrt{x}-1)+\sqrt{x}-1\)

=\((\sqrt{x}-1).(y\sqrt{x}+1)\).

\(b)\)\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

=\(\sqrt{a}.\sqrt{x}-\sqrt{b}.\sqrt{y}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}\)

=\(\sqrt{a}.\sqrt{x}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}-\sqrt{b}.\sqrt{y}\)

=\(\sqrt{x}.(\sqrt{a}+\sqrt{b})-\sqrt{y}.(\sqrt{a}+\sqrt{b})\)

=\((\sqrt{x}-\sqrt{y}).(\sqrt{a}+\sqrt{b})\).

\(c)\)\(\sqrt{a+b}+\sqrt{a^2-b^2}\)

=\(\sqrt{a+b}+\sqrt{(a+b).(a-b)}\)

=\(\sqrt{a+b}+\sqrt{a+b}.\sqrt{a-b}\)

=\(\sqrt{a+b}.\left(1+\sqrt{a-b}\right)\).

\(d)\) \(12-\sqrt{x}-x\)

=\(12-4\sqrt{x}+3\sqrt{x}-x\)

=\(4.\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)

=\(\left(3-\sqrt{x}\right).\left(4+\sqrt{3}\right)\).

20 tháng 8 2015

a>  = \(y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(y\sqrt{x}-1\right)\)

20 tháng 8 2015

a) \(xy-y\sqrt{x}+\sqrt{x}-1\)

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

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

b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

\(=\left(\sqrt{ax}-\sqrt{ay}\right)+\left(-\sqrt{by}+\sqrt{bx}\right)\)

\(=\sqrt{a}.\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}.\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

13 tháng 6 2019

a/ \(=\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

b/ \(=\sqrt{\left(a-b\right)\left(a+b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\)

\(=\sqrt{a+b}\left(\sqrt{\left(a-b\right)\left(a^2-ab+b^2\right)}\right)\)

c/ \(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{81}{4}=\left(\sqrt{x}-\frac{3}{2}-\frac{9}{2}\right)\left(\sqrt{x}-\frac{3}{2}+\frac{9}{2}\right)=\left(\sqrt{x}-6\right)\left(\sqrt{x}+3\right)\)

13 tháng 6 2019

\(a.\sqrt{ax}+\sqrt{by}-\sqrt{ay}-\sqrt{bx}\\ =\left(\sqrt{ax}-\sqrt{ay}\right)-\left(\sqrt{bx}-\sqrt{by}\right)\\ =\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\\ =\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

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

\(c.x-3\sqrt{x}-18=x-6\sqrt{x}+3\sqrt{x}-18\\ =\sqrt{x}\left(\sqrt{x}-6\right)+3\left(\sqrt{x}-6\right)\\ =\left(\sqrt{x}+3\right)\left(\sqrt{x}-6\right)\)

\(d.x\sqrt{x}+4x-12\sqrt{x}-27=\left(\sqrt{x^3}-27\right)+\left(4x-12\sqrt{x}\right)\\ =\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\\ =\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\\ =\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)

(có gì sai mong mọi người góp ý)

11 tháng 10 2019

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

b,\(x\sqrt{x}+\sqrt{x}-x-1\)

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

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

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

c,\(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)=\(\sqrt{a}\left(\sqrt{b}+2\right)\)+\(3\left(\sqrt{b}+2\right)\)

=\(\left(\sqrt{b}+2\right)\left(3+\sqrt{a}\right)\)

d,\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

=\(\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\)

=\(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

14 tháng 10 2019

thank you

Bài 2: 

a: \(\sqrt{ax}+\sqrt{by}-\sqrt{bx}-\sqrt{ay}\)

\(=\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

b: \(\sqrt{a-b}-\sqrt{a^2-b^2}\)

\(=\sqrt{a-b}-\sqrt{a-b}\cdot\sqrt{a+b}\)

\(=\sqrt{a-b}\left(1-\sqrt{a+b}\right)\)

Bài 2: 

a: \(\sqrt{ax}-\sqrt{bx}+\sqrt{by}-\sqrt{ay}\)

\(=\sqrt{x}\left(\sqrt{a}-\sqrt{b}\right)-\sqrt{y}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

b: \(\sqrt{a-b}-\sqrt{a^2-b^2}\)

\(=\sqrt{a-b}-\sqrt{a-b}\cdot\sqrt{a+b}\)

\(=\sqrt{a-b}\left(1-\sqrt{a+b}\right)\)

5 tháng 7 2015

\(\text{a) }\sqrt{a^3+b^3}+\sqrt{a^2-b^2}=\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}+\sqrt{\left(a+b\right)\left(a-b\right)}\)

\(=\sqrt{a+b}\left(\sqrt{a^2-ab+b^2}+\sqrt{a-b}\right)\)

\(\text{b) }\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{xy}\text{ không phân tích được.}\)

\(\text{c) }=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\left(\sqrt{x}-\sqrt{y}\right).\sqrt{xy}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+2\sqrt{xy}\right)\)\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(\text{d) }a+5\sqrt{a}+4=\sqrt{a}.\sqrt{a}+\sqrt{a}+4\sqrt{a}+4=\sqrt{a}\left(\sqrt{a}+1\right)+4\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+4\right)\)

16 tháng 7 2019

\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)

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

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

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

\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

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

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

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

\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)

\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)

\(\text{f)}x-2\sqrt{x-1}-a^2\)

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

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

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

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