a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)

\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)

\(=\dfrac{3a-1}{2}\)

\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)

b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)

\(=\dfrac{1}{b-3}\)

\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)

\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)

\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)

\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)

\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)

b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24

a) Ta có:

\(VT=\left(a-b\right)^2\)

\(=a^2-2\cdot a\cdot b+b^2\)

\(=a^2-2ab+b^2\)

\(=a^2-4ab+2ab+b^2\)

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

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

⇒ Đpcm

b) Ta có:

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

\(=x^2+2\cdot x\cdot y+y^2+x^2-2\cdot x\cdot y+y^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

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

\(=2x^2+0+2y^2\)

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

\(=2\left(x^2+y^2\right)=VP\)

⇒ Đpcm

a: (a-b)^2

=a^2-2ab+b^2

=a^2+2ab+b^2-4ab

=(a+b)^2-4ab

b: (x+y)^2+(x-y)^2

=x^2+2xy+y^2+x^2-2xy+y^2

=2x^2+2y^2

=2(x^2+y^2)

a) a > b

⇒ 2a > 2b (nhân hai vế với 2 > 0)

⇒ 2a - 3 > 2b - 3 (cộng hai vế với -3)

b) a < b

⇒ -3a > -3b (nhân hai vế với -3 < 0)

⇒ -3a + 2 > -3b + 2 (1) (cộng hai vế với 2)

5 > 2

⇒ -3a + 5 > -3a + 2 (2) (cộng hai vế với -3a)

Từ (1) và (2) ⇒ -3a + 5 > -3b + 2

a, \(\dfrac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow\)a^2+2ab+b^2>=4ab

\(\Leftrightarrow\)a^2-2ab+b^2>=0

\(\Leftrightarrow\)(a-b)^2>=0 (luôn đúng)

b,\(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\) 

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) luôn đúng

\(\Leftrightarrow a^4-a^3b+b^4-ab^3>=0\)

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)>=0\)

\(\Leftrightarrow\left(a-b\right)^2\cdot\left(a^2+ab+b^2\right)>=0\)(luôn đúng)

a) \(\sqrt[3]{a^3b}=\sqrt[3]{a^3}\sqrt[3]{b}=a\sqrt[3]{b}\)

b) \(\sqrt[3]{\dfrac{a}{b^2}}=\sqrt[3]{\dfrac{ab}{b^3}}=\dfrac{\sqrt[3]{ab}}{\sqrt[3]{b^3}}=\dfrac{1}{b}\sqrt[3]{ab}\)