a: \(\left(a-b-c\right)-\left(-c+b+a\right)-\left(a-b\right)\)

\(=a-b-c+c-b-a-a+b\)

\(=-a-b\)

b: \(a\left(b+c\right)-a\left(b+d\right)-\left(1+ac-ad\right)\)

\(=ab+ac-ab-ad-1-ac+ad\)

=-1

Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

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

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

a) (x+9)(x-9)-x²=x²-81-x²=-81

b) (10x-1)(10x+1)-(10x-1)²=100x²-1-100x²+20x-1=20x-2

d) (x-1)(x-2)-(x-2)(x+2)=x²-3x+2-x²+4=-3x+6

1:Sửa đề: \(A=\left(\frac{2x+1}{x\cdot\sqrt{x}-1}-\frac{\sqrt{x}}{x-\sqrt{x}}\right)\left(x-\frac{x-4}{\sqrt{x}-2}\right)\)

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

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

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

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

2: Thay x=2 và y=1 vào y=ax+b, ta được:

\(a\cdot2+b=1\)

=>b=1-2a

Thay x=1 và y=2 vào y=ax+b, ta được:

\(a\cdot1+b=2\)

=>a+b=2

=>a+1-2a=2

=>1-a=2

=>a=1-2=-1

b=1-2a=1-2*(-1)=1+2=3

\(1,A=\dfrac{2x+1-x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\left(x-\sqrt{x}-2\right)\\ A=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\left(x+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+1}\\ 2,\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-3\end{matrix}\right.\Leftrightarrow y=-x-3\)

a) A= 3.(x²-2xy+y²)- 2. (x²+2xy+y²) - x²-y²

A= 3.x²-2xy+y²-2. x²+2xy+y²-x²-y²

m

m

1:

a: ĐKXĐ: 1-x>=0

=>x<=1

b: ĐKXĐ: 2/x>=0

=>x>0

c: ĐKXĐ: 4/x+1>=0

=>x+1>0

=>x>-1

d: ĐKXĐ: x^2+2>=0

=>x thuộc R

Câu 2:

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

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

2a) pt <=> (x + 6)^2 = 0

<=> x = -6

b) pt <=> (4x - 1)^2 = 0

<=> x = 1/4

c) pt<=> (x + 1)^3 = 0

<=> x = -1

Bài 1:

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)