Bạn làm đc bài này chưa chỉ mình với

a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)

b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)

\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)

\(=19ab\sqrt{ab}\)

c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)

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

\(=\sqrt{ab}\)

d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)

\(=-4\sqrt{5a}+9\sqrt{a}\)

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

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

c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)

a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)

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

c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)

\(\left(a\right)\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\\ =\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}\\ =\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)

\(\left(b\right) \frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2\right)}{\sqrt{2}+\sqrt{3}+2}\\=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\\ =1+\sqrt{2}\)

\(\left(c\right)\sqrt{9\left(3-a\right)^2}vớia>3\\ =\sqrt{9}.\sqrt{\left(3-a\right)^2}\\ =3.\left|3-a\right|\\ =-3\left(3-a\right)vì.a>3\\ =3a-9\)

\(\left(d\right)\sqrt{a^2.\left(a-2\right)^2}vớia< 0\\ =\sqrt{\left[a\left(a-2\right)\right]^2}\\ =\left|a\left(a-2\right)\right|=-a.\left[-\left(a-2\right)\right]=a\left(a-2\right)=a^2-2a\)

Chúc bạn học tốt !

a) \(\sqrt{16x-8}+\sqrt{36x-18}-\sqrt{64x-32}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{8\left(2x-1\right)}+\sqrt{18\left(2x-1\right)}-\sqrt{32\left(2x-1\right)}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{8}.\sqrt{2x-1}+\sqrt{18}.\sqrt{2x-1}-\sqrt{32}.\sqrt{2x-1}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}.\left(\sqrt{8}+\sqrt{18}-\sqrt{32}\right)=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}.\sqrt{2}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}=\sqrt{5}\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

Vậy ...

b) \(\sqrt{x^2-6x+9}=x+3\)

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

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

\(\Leftrightarrow\left|x-3\right|=x+3\)

\(\Leftrightarrow x-3=x+3\) hoặc \(x-3=-x-3\)

\(\Leftrightarrow x=0\)

Vậy ...

bài 2 :

A = \(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)

\(=\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a+\sqrt{b}}\right)}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)

\(=\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{ab}-a-b}\right)\)

\(=\left(\dfrac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-a+\sqrt{ab}-b}\right)\)

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

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

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

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

cuối cùng cũng xong, mong bn phù hộ độ trì cho mk

a \(\sqrt{16x-8}+\sqrt{36x-18}-\sqrt{64x-3x}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{8\left(2x-1\right)}+\sqrt{18\left(2x-1\right)}-\sqrt{32\left(2x-1\right)}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{8}+\sqrt{18}-\sqrt{32}\right)=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}\left(2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\right)=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2\left(2x-1\right)}=\sqrt{10}\)

\(\Rightarrow2\left(2x-1\right)=10\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2-6x+9}=x+3\)

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

\(\Rightarrow x-3=x+3\) ( vô nghiệm )

\(\sqrt{10-2\sqrt{21}}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

=\(\sqrt{7}-\sqrt{3}\)

=> a=7 và b=3

=> a-b=7-3=4

\(\sqrt{10-2\sqrt{21}}=\sqrt{7}-\sqrt{3}\)

\(\Rightarrow\sqrt{7}-\sqrt{3}=\sqrt{a}-\sqrt{b}\)

Suy ra \(\sqrt{7}=\sqrt{a}\rightarrow a=7\)

\(\sqrt{3}=\sqrt{3}\rightarrow b=3\)

Vậy \(a-b=7-3=4\)

Bunyakovsky:

\(\sqrt{a+b}+\sqrt{a-b}\le\sqrt{2.2a}=2\sqrt{a}\)