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20 tháng 6 2016

A)=a+\(2\sqrt{a}+2\sqrt{a}\)+4

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

b)= \(\left(a-\sqrt{7}\right)\left(a+\sqrt{7}\right)\)

c) \(\sqrt{a}\left(\sqrt{b}-4\right)+3\cdot\left(\sqrt{b}-4\right)=\left(\sqrt{a}+3\right)\left(\sqrt{b}-4\right)\)

14 tháng 8 2019

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

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

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

14 tháng 8 2019

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

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

16 tháng 7 2019

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

14 tháng 6 2019

\(x\sqrt{x}+4x-12\sqrt{x}-27\)

\(=\left(x\sqrt{x}-27\right)+\left(4x-12\sqrt{x}\right)\)

\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\)

\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\)

\(=\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)

14 tháng 6 2019

a, \(\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\)

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

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

19 tháng 7 2018

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

b ) \(x-4\sqrt{x}+3=\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2-1=\left(\sqrt{x}-2\right)^2-1\)

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

19 tháng 7 2018

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

\(x-4\sqrt{x}+3=\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2\right]-1^2=\left(\sqrt{x}-2\right)^2-1^2\)

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

\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)\)

2 tháng 8 2018

a)  \(5+\sqrt{10}-\sqrt{5}=\sqrt{5}.\left(\sqrt{5}+\sqrt{2}-1\right)\)

b)  ĐK:  \(a\ge0\)

\(a-4\sqrt{a}-5=a+\sqrt{a}-5\sqrt{a}-5=\left(\sqrt{a}+1\right)\left(\sqrt{a}-5\right)\)

c)  ĐK:  \(a\ge0\)

\(a+12\sqrt{a}+32=a+8\sqrt{a}+4\sqrt{a}+32=\left(\sqrt{a}+8\right)\left(\sqrt{a}+4\right)\)

d)  ĐK: \(a\ge0\)

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

3 tháng 7 2015

Toán mà cũng có chữ Giải trí nữa

3 tháng 7 2015

đinh tuấn việt đầu bảng mà giải đi

4 tháng 10 2020

a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)

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

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

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

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

4 tháng 10 2020

b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

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

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

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