a,-6a

b,3a³

^c,20ab

Từ kết quả bài toán suy ngược ra thôi

Muốn giải thích thì cứ phá 2 vế ra rồi so sánh là tìm ra cách tách biểu thức

Câu 4 mình ko biết giải quyết kiểu lớp 9 (mặc dù chắc chắn là biểu thức sẽ được biến đổi như vầy)

Đó là kiểu trình bày của lớp 11 hoặc 12 để bạn tham khảo thôi

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

b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=\sqrt{\left[3\left(b-2\right)\right]^2}=3b-6\)

c) bạn xem lại đề

d)
\(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\sqrt{\left(15a\right)^2}-3a=15a-3a=12a\)

e) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{\sqrt{16}}{\sqrt{x^2}}=\dfrac{4}{x}\)

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

b) \(\sqrt{7-2\sqrt{6}}=\sqrt{6-2\cdot\sqrt{6}+1}=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

c) \(\sqrt{4x^6}=\sqrt{\left(2x^3\right)^2}=2\left|x^3\right|=-2x^3\)

d) \(\sqrt{\frac{12x^2y^4}{25}}=\frac{xy^2\sqrt{12}}{5}\)

Các bạn giải giúp mình vss mình đang cần gấpp

a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

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

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

\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)

Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)

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

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

\(=-2\sqrt{b}\)

c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

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

\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)

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

\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)

\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

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

e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)

\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)

Câu 1 :

a, Ta có : \(A=x-2\sqrt{3}+3\)

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

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

b, Ta có : \(B=x+2\sqrt{x}-3\)

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

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

c, Ta có : \(C=x\sqrt{x}-1\)

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

d, Ta có : \(D=2x-3\sqrt{xy}-5y\)

\(=2x+2\sqrt{xy}-5\sqrt{xy}-5y\)

\(=2\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)-5\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(2\sqrt{x}-5\sqrt{y}\right)\)

Bài 1: 

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

b: \(-\sqrt{\left(-2\right)^4}=-\left|\left(-2\right)^2\right|=-4\)

c: \(=-\sqrt{5^2}=-\left|5\right|=-5\)

d: \(=\sqrt{\left(-3\right)^6}=\sqrt{3^6}=\left|3^3\right|=27\)

e: \(-\sqrt{\left(-1\right)^8}=-\left|\left(-1\right)^4\right|=-1\)

a) Với \(a\ge0\) :

\(M=\sqrt{16a^2}-5a=\sqrt{\left(4a\right)^2}-5a=\left|4a\right|-5a=4a-5a=-a\)

b) Với \(b\le0\) :

\(N=\sqrt{25b^2}+3b=\sqrt{\left(5b\right)^2}+3b=\left|5b\right|+3b=-5b+3b=-2b\)

c) Với \(x\ge5\) :

\(P=\sqrt{x^2-10x+25}=\sqrt{\left(x-5\right)^2}=\left|x-5\right|=x-5\)

d) Với \(x>\frac{1}{3}\) :

\(Q=3x+2-\sqrt{9x^2+6x+1}=3x+2-\sqrt{\left(3x+1\right)^2}=3x+2-\left|3x+1\right|=3x+2-\left(3x+1\right)=1\)