a) (x² + 2xy + y²) : (x + y) = (x + y)² : (x + y) = x + y.

b) (125x³ + 1) : (5x + 1) = [(5x)³ + 1] : (5x + 1)

= (5x)² – 5x + 1 = 25x² – 5x + 1.

c) (x² – 2xy + y²) : (y – x) = (x – y)² : [-(x – y)] = - (x – y) = y – x

Hoặc (x² – 2xy + y²) : (y – x) = (y² – 2xy + x²) : (y – x)

= (y – x)² : (y – x) = y - x.

Bài giải:

a) (x² + 2xy + y²) : (x + y) = (x + y)² : (x + y) = x + y.

b) (125x³ + 1) : (5x + 1) = [(5x)³ + 1] : (5x + 1)

= (5x)² – 5x + 1 = 25x² – 5x + 1.

c) (x² – 2xy + y²) : (y – x) = (x – y)² : [-(x – y)] = - (x – y) = y – x

Hoặc (x² – 2xy + y²) : (y – x) = (y² – 2xy + x²) : (y – x)

= (y – x)² : (y – x) = y - x.

a) (x² +2xy + y² ) : (x +y)

= (x +y)² : (x +y)

= x + y

b) (125x³ + 1) : (5x + 1)

= (5x + 1)(25x² - 5x + 1) : (5x + 1)

= 25x² - 5x + 1

c) \(\left(x^2-2xy+y^2\right)\left(y-x\right)\)

= \(\left(x-y\right)^2:\left(y-x\right)\)

= \(x-y\)

a) Biến đổi VT . Mẫu chung là ( a + 2b )( a - 2b )

\(VT=\frac{a+2b-6b-2\left(a-2b\right)}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 1 )

Biến đổi VP 

\(-\frac{1}{2a}\left(\frac{a^2+4b^2}{a^2-4b^2}+1\right)=-\frac{1}{2a}\cdot\frac{a^2+4b^2+a^2-4b^2}{a^2-4b^2}\)

\(=-\frac{1}{2a}\cdot\frac{2a^2}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP ( đpcm )

b) \(a^3+b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)^3\)

<=> \(b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)^3=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)-a^3\)( * )

Biến đổi VT của ( * ) ta có :

\(VT=\left[b+\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right]\left[b^2-\frac{b^2\left(2a^3+b^3\right)}{a^3-b^3}+\frac{b^2\left(2a^3+b^3\right)^2}{\left(a^3-b^3\right)^2}\right]\)

\(=\frac{3a^3b}{a^3-b^3}\cdot\frac{3a^6b^2+3a^3b^5+3b^8}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 1 )

\(VP=\left[\frac{a\left(a^3+2b^3\right)}{a^3-b^3}-a\right]\left[\frac{a^2\left(a^3+2b^3\right)^2}{\left(a^3-b^3\right)^2}+\frac{a^2\left(a^3+2b^3\right)}{a^3-b^3}+a^2\right]\)

\(=\frac{3ab^3}{a^3-b^3}\cdot\frac{3a^8+3a^5b^3+3a^2b^6}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP => ( * ) đúng 

=> Hằng đẳng thức đúng 

a/ \(\left(m+n\right)\left(m^3-mn+n^2\right)=m^3+n^3\)

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

\(\left(1+x+x^2\right)\left(1-x\right)\left(1+x\right)\left(1-x+x^2\right)=\left(\left(1+x+x^2\right)\left(1-x\right)\right)\left(\left(1-x+x^2\right)\left(1+x\right)\right)=\left(1-x^3\right)\left(1+x^3\right)=1-x^6\)

a) m³+n³

b)  (a -b-c+a-b+c)(a-b-c-a+b-c)

= -4c(a-b)

c) (1-x³)(1+x³)

=1-x⁶

a) ta có : \(\left(2+i\sqrt{3}\right)^2=2^2+2.2.i\sqrt{3}+\left(i\sqrt{3}\right)^2\)

\(=4+4\sqrt{3}i-3=1+4\sqrt{3}i\)

b) ta có : \(\left(1+2i\right)^3=1^3+3.1^2.2i+3.1.\left(2i\right)^2+\left(2i\right)^3\)

\(=1+6i-6-8i=-5-2i\)

c) \(\left(3-i\sqrt{2}\right)^3=3^3-3.3^2.i\sqrt{2}+3.3.\left(i\sqrt{2}\right)^2+\left(i\sqrt{2}\right)^3\)

\(=27-27\sqrt{2}i-18-2\sqrt{2}i=9-29\sqrt{2}i\)

d) \(\left(2-i\right)^3=2^3-2.2^2.i+2.2.i^2-i^3\)

\(=8-8i-4+i=4-7i\)

\(C=2sin3x.cos3x.\frac{cos3x}{sin3x}-\left(cos^23x-sin^23x\right)\)

\(=2cos^23x-cos^23x+sin^23x=cos^23x+sin^23x=1\)

\(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)\)

\(=\sqrt{2}\left(sinx.sin\frac{\pi}{4}-cosx.cos\frac{\pi}{4}\right)=-\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

Câu này bạn ghi nhầm đề (lưu ý rằng \(sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\))

Câu 2b bạn cũng xem lại đề, chắc chắn ko đúng

\(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow sin\frac{A}{2}=cos\left(\frac{B}{2}+\frac{C}{2}\right)=cos\frac{B}{2}cos\frac{C}{2}-sin\frac{B}{2}sin\frac{C}{2}\)

Câu 3 bạn cũng ghi sai đề luôn

Trong 1 ngày đẹp trời thì câu 4 cũng sai luôn cho đỡ lạc lõng đồng đội:

\(sin\left(a+b-b\right)=sin\left(a+b\right)cosb-cos\left(a+b\right)sinb=2sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(a+b\right)\left[cosb-2\right]=cos\left(a+b\right).sinb\)

\(\Leftrightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{sinb}{cosb-2}\Leftrightarrow tan\left(a+b\right)=\frac{sinb}{cosb-2}\)

4 câu bạn ghi đúng đề bài duy nhất câu 1, kinh thiệt :(

a: \(=a^2-b^4\)

b: \(=\left(a^2+2a\right)^2-9\)

c: \(=a^2-\left(2a+3\right)^2\)

d: \(=a^4-\left(2a-3\right)^2\)

e: \(=\left(-a^2-2a+3\right)^2\)

g: \(=4a^2-a^4\)