a. \(7x-x^2-6=0\)

\(\Rightarrow-x^2+7x-6=0\)

\(\Rightarrow-x^2+x+6x-6=0\)

\(\Rightarrow-x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-x+6\right)=0\)

+) Nếu \(x-1=0\Rightarrow x=1\)

+) Nếu \(-x+6=0\Rightarrow x=6\)

Vậy x=1 hoặc x=6

b. \(8x^3-36x^2+57x-27=0\)

\(\Rightarrow\left(2x\right)^2-3.2^2.x^2.3+3.2x.3^2-3^3=0\)

\(\Rightarrow\left(2x-3\right)^3=0\)

\(\Rightarrow2x-3=0\Rightarrow x=\frac{3}{2}\)

Vậy...........

a)\(\dfrac{x^2-4xy+4y^2}{xy-2y^2}\)

=\(\dfrac{x^2-4xy+\left(2y\right)^2}{y\left(x-2y\right)}\)

=\(\dfrac{\left(x-2y\right)^2}{y\left(x-2y\right)}\)

=\(\dfrac{x-2y}{y}\)

b)\(\dfrac{x^3-36x}{x^2+6x}\)

=\(\dfrac{x\left(x^2-6^2\right)}{x\left(x+6\right)}\)

=\(\dfrac{x\left(x+6\right)\left(x-6\right)}{x\left(x+6\right)}\)

= \(x-6\)

#Fiona 

Chúc bạn học tốt !

\(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-36x+81}\)

\(=\sqrt{\left(2x\right)^2-2.2x.1+1^2}+\sqrt{\left(2x\right)^2-2.2x.9+9^2}\)

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

\(=\left|2x-1\right|+\left|2x-9\right|\)

\(=2x-1+9-2x=8\)

Thanks bn♥

Bài 2: 

a: \(A=a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2-2ab-2bc+2ac\)

\(=2a^2+2b^2+2c^2-4bc\)

\(=2+2\cdot9+2\cdot1-4\cdot3\cdot\left(-1\right)=22+12=34\)

b: \(B=\left(a+b-a+b\right)\left(a+b+a-b\right)=4ab=4\cdot2\cdot5=40\)

a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)

b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)

c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)

\(y\ne0\)

Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)

e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)

Vì y < 0 nên \(\left|y\right|=-y\)

Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)

f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)