BÀi 1:

a: \(\left(a+b\right)^2-\left(a-b\right)^2\)

=(a+b-a+b)(a+b+a-b)

\(=2b\cdot2a=4ab\)

b: \(\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)

\(=a^2+4a+4-\left(a^2-4\right)\)

\(=a^2+4a+4-a^2+4=4a+8\)

c: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

=>\(4x^2+12x+9-4\left(x^2-1\right)=49\)

=>\(4x^2+12x+9-4x^2+4=49\)

=>12x+13=49

=>12x=36

=>x=3

d: \(Q=\left(x+3\right)^2+\left(x+3\right)\left(x-3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-4x+2x-8\right)\)

\(=2x^2+6x-2\left(x^2-2x-8\right)=2x^2+6x-2x^2+4x+16=10x+16\)

Khi x=1/2 thì Q=10*1/2+16=5+16=21

Bài 2:

a: \(A=\left(4x^2+y^2\right)\left(2x+y\right)\left(2x-y\right)\)

\(=\left(4x^2+y^2\right)\left(4x^2-y^2\right)=16x^4-y^4\)

b: \(\left(7x+1\right)^2-\left(x+7\right)^2\)

\(=49x^2+14x+1-\left(x^2+14x+49\right)\)

\(=49x^2+14x+1-x^2-14x-49=48x^2-48=48\left(x^2-1\right)\)

c: \(16x^2-\left(4x-5\right)^2=15\)

=>\(16x^2-\left(16x^2-40x+25\right)=15\)

=>\(16x^2-16x^2+40x-25=15\)

=>40x=40

=>x=1

d: \(A=-x^2+2x+3\)

\(=-x^2+2x-1+4\)

\(=-\left(x-1\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

\(A=2x^2+2xy+x^2-2xy+y^2-4y^2=3x^2-3y^2\)

\(A=2x^2+2xy+x^2-2xy+y^2-4y^2=3x^2-3y^2\)

a: \(A=\frac{x^2+2x}{x^2-4x+4}:\left(\frac{x+2}{x}-\frac{1}{2-x}+\frac{6-x^2}{x^2-2x}\right)\)

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

\(=\frac{x\left(x+2\right)}{\left(x-2\right)^2}\cdot\frac{x\left(x-2\right)}{x^2-4+x+6-x^2}=\frac{x^2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2}{x-2}\)

b: |2x+1|=3

=>2x+1=3 hoặc 2x+1=-3

=>2x=2 hoặc 2x=-4

=>x=1(nhận) hoặc x=-2(loại)

Thay x=1 vào A, ta được:

\(A=\frac{1^2}{1-2}=\frac{1}{-1}=-1\)

c: A<0

=>\(\frac{x^2}{x-2}<0\)

=>x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: x<2 và x∉{0;-2}

d: Để A nguyên thì \(x^2\vdots x-2\)

=>\(x^2-4+4\vdots x-2\)

=>4⋮x-2

=>x-2∈{1;-1;2;-2;4;-4}

=>x∈{3;1;4;0;6;-2}

Kết hợp ĐKXĐ, ta được: x∈{3;1;4;6}

1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+3}+\dfrac{2-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+3\sqrt{x}+2\sqrt{x}-2+2-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

(x-1)³+(x+1)³+6.(x+1).(x-1)

= (x³-3x²+3x-1)+(x³+3x²+3x+1)+(6x+6).(x-1)

= x³-3x²+3x-1+x³+3x²+3x+1+6x²-6x+6x-6

= x³+x³+3x+3x+6x²-6

= 2x³+6x+6x²-6

tới đây thôi muốn làm thành nhân tử hay gì đó thì tùy!

đk \(\left\{{}\begin{matrix}x\ne1\\x>0\end{matrix}\right.\)

A= \(\dfrac{-x\left(1+\sqrt{x}\right)}{\sqrt{x}\left(1-x\right)}\)+\(\dfrac{3\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-x\right)\sqrt{x}}\)+\(\dfrac{\left(6\sqrt{x}-4\right)\sqrt{x}}{\left(1-x\right)\sqrt{x}}\)

=\(\dfrac{-x-x\sqrt{x}+3\sqrt{x}-3x+6x-4\sqrt{x}}{\left(1-x\right)\sqrt{x}}\)

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

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3\sqrt{x}}{x+\sqrt{x}}+\dfrac{6\sqrt{x}-4}{1-x}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

a: \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6}{\sqrt{x}-1}-\dfrac{2\sqrt{3}}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-6-2\sqrt{3}}{\sqrt{x}-1}\)

b: \(=\dfrac{3-\sqrt{x}-1+\sqrt{x}+5\sqrt{x}}{\sqrt{x}-2}=\dfrac{5\sqrt{x}+2}{\sqrt{x}-2}\)

c: \(=\dfrac{2-6\sqrt{x}-1+\sqrt{x}-3+\sqrt{x}}{\sqrt{x}-4}\)

\(=\dfrac{-4\sqrt{x}-4}{x-4}\)

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

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

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