` a,` 

`1/3 + 3/4 xx 5/9 `

`=1/3 + (3/4 xx 5/9) `

`= 1/3 + 5/12 `

`= 4/12 + 5/12`

`= 9/12 `

`= 3/4 .`

`b,`

` 7/12 - 5/4 xx 2/15 `

`= 7/12 - (5/4 xx 2/15)`

`= 7/12 - 1/6`

`= 7/12 - 2/12 `

`= 5/12 .`

\(a,\dfrac{6}{11}+6+\dfrac{5}{7}=\dfrac{42+462+55}{77}=\dfrac{559}{77}\)

\(b,\dfrac{9}{8}\times\dfrac{3}{12}:\dfrac{5}{9}=\dfrac{9}{8}\times\dfrac{3}{12}\times\dfrac{9}{5}=\dfrac{243}{480}=\dfrac{81}{160}\)

\(c,\dfrac{8}{7}:4+2=\dfrac{8}{7}\times\dfrac{1}{4}+2=\dfrac{8}{28}+2=\dfrac{2}{7}+2=\dfrac{16}{7}\)

\(d,\dfrac{3}{5}+4:\dfrac{6}{4}=\dfrac{3}{5}+4\times\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{8}{3}=\dfrac{49}{15}\)

\(a,\dfrac{6}{11}+6+\dfrac{5}{7}=\dfrac{6}{11}+\dfrac{66}{11}+\dfrac{5}{7}=\dfrac{72}{11}+\dfrac{5}{7}=\dfrac{504}{77}+\dfrac{55}{77}=\dfrac{559}{77}\)

\(b,\dfrac{9}{8}\times\dfrac{3}{12}:\dfrac{5}{9}=\dfrac{9}{32}:\dfrac{5}{9}=\dfrac{9}{32}\times\dfrac{9}{5}=\dfrac{81}{160}\)

\(c,\dfrac{8}{7}:4+2=\dfrac{8}{7}\times\dfrac{1}{4}+2=\dfrac{2}{7}+2=\dfrac{2}{7}+\dfrac{14}{7}=\dfrac{16}{7}\)

\(d,\dfrac{3}{5}+4:\dfrac{6}{4}=\dfrac{3}{5}+4\times\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{8}{3}=\dfrac{9}{15}+\dfrac{40}{15}=\dfrac{49}{15}\)

ai lm tr mình cho tích

Ta có : \(B\text{=}4x^2-12x+9\)

\(B\text{=}\left(2x-3\right)^2\)

Với \(x\text{=}\dfrac{1}{2}\)

\(\Rightarrow B\text{=}\left(2.\dfrac{1}{2}-3\right)^2\)

\(B\text{=}\left(-2\right)^2\text{=}4\)

Ta có : \(A\text{=}5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5\left(x^2-9\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5x^2-45+4x^2+12x+9+x^2-12x+36\)

\(A\text{=}10x^2\)

Với \(x\text{=}-\dfrac{1}{5}\)

\(\Rightarrow A\text{=}10.\left(-\dfrac{1}{5}\right)^2\text{=}\dfrac{2}{5}\)

B = 4x² - 12x + 9

= (2x - 3)²

Tại x = 1/2 ta có:

B = (2.1/2 - 3)²

= (-2)²

= 4

-------------------

A = 5(x + 3)(x - 3) + (2x + 3)² + (x - 6)²

= 5x² - 45 + 4x² + 12x + 9 + x² - 12x + 36

= 10x²

Tại x = 1/5 ta có:

A = 10.(1/5)²

= 2/5

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

\(B=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

\(B=\left(2x-3\right)^2\)

Thay \(x=\dfrac{1}{2}\) vào B ta có: 

\(B=\left(2\cdot\dfrac{1}{2}-3\right)^2=4\)

_______________________

\(A=5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A=5\left(x^2-9\right)+4x^2+12x+9+x^2-12x+36\)

\(A=5x^2-45+5x^2+45\)

\(A=10x^2\)

Thay \(x=\dfrac{1}{5}\)vào A ta có:

\(A=10\cdot\left(-\dfrac{1}{5}\right)^2=\dfrac{2}{5}\)

ủa sao ko một bóng ai giúp mik vậy , mik cầu xin mn đó , làm ơn giúp mik ik . Thật sự là mik đang cần rất gấp mà là cực kì gấp mà sao chẳng ai giúp mik , mong mn sẽ giúp mik . Cảm ơn mn nhìu 

ủa sao ko ai giúp mik hết vậy . Mik đang cần gấp bài này mà 

Bài 1:

ta có: 333<3333; 444<4444

=> 333⁴⁴⁴<3333⁴⁴⁴⁴

Bài 2:

\(A=\frac{21^5}{81}=\frac{\left(3.7\right)^5}{3^4}=\frac{3^5.7^5}{3^4}=3.7^5=50421\)

\(B=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{\left(3.0,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{3^5.\left(0,5\right)^5}=\frac{1}{3^2}=\frac{1}{9}\)

\(C=2^2.\frac{1}{128}.45.2^{-6}=\frac{2^2.45}{128.64}=\frac{2^2.45}{2^7.2^6}=\frac{45}{2^{11}}=\frac{45}{2048}\)

\(D=\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+2^2.3^3+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}\)\(=3^3.\left(-1\right)=-27\)