\(8x^3+12x^2y+6xy^2+y^3-z^3\)

\(=\left(2x+y\right)^3-z^3\)

\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

\(920+3^2.\left(2^2.5^3-6.5^2\right)\\ \\ \\ =920+9.\left(4.125-6.25\right)\\ \\ \\ =920+9.\left(500-150\right)\\ \\ \\ =920+9.350=920+3150=4070\)

   920 + 3² . (2² . 5³ - 6 . 5²)

= 920 + 9 . (4 . 125 - 6 . 25)

= 920 + 9 . (500 - 150)

= 920 + 9 . 350

= 920 + 3150

= 4070

    64:2^x=2

⇒ 2⁶:2^x=2

⇒ 2^6-x=2¹

⇒ 6-x=1

⇒ x=6-1

⇒ x=5

5

Ta có:\(8^{12}=\left(2^3\right)^{12}=2^{3.12}=2^{36}\\ \\ \\ 32^6=\left(2^5\right)^6=2^{5.6}=2^{30}\) Mà \(2^{36}>2^{30}\)

⇒ Chọn A

8¹² = (2³)¹² = 2³⁶

32⁶ = (2⁵)⁶ = 2³⁰

Vì 2³⁶ > 3³⁰ ⇒ 8¹² > 32⁶ ⇒ Chọn A

2³⁰=2^3.10=(2³)¹⁰=8¹⁰

số phần C góp là 1-1/3-1/2 = 1/6

số tiền mua truyện là 20 000 :1/6 = 120 000

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)