\(9-9x^2+2x-\dfrac{2}{9}\\ =-\left(9x^2-2x+\dfrac{1}{9}-\dfrac{80}{9}\right)\\ =-\left(3x+\dfrac{1}{3}\right)^2+\dfrac{80}{9}\le\dfrac{80}{9}\)

Dấu "=" xảy ra khi \(-\left(3x+\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow3x+\dfrac{1}{3}=0\\ \Leftrightarrow3x=-\dfrac{1}{3}\\ \Leftrightarrow x=-\dfrac{1}{9}\)

Vậy \(Max=\dfrac{80}{9}\Leftrightarrow x=-\dfrac{1}{9}\)

9 - 9x² + 2x - \(\dfrac{2}{9}\)
=\(\dfrac{80}{9}\)-[(3x)²-2x+(\(\dfrac{1}{3}\))²]
=\(\dfrac{80}{9}\)-(3x-\(\dfrac{1}{3}\))²
Vì (3x-\(\dfrac{1}{3}\))²≥0 ⇒-(3x-\(\dfrac{1}{3}\))²≤0⇒\(\dfrac{80}{9}\)-(3x-\(\dfrac{1}{3}\))²≤\(\dfrac{80}{9}\)
Trường hợp dấu bằng xảy ra khi: (3x-\(\dfrac{1}{3}\))²=0⇒3x-\(\dfrac{1}{3}\)=0⇒3x=\(\dfrac{1}{3}\)⇒x=\(\dfrac{1}{9}\)
Vậy GTLN của biểu thức là \(\dfrac{80}{9}\) khi x=\(\dfrac{1}{9}\)

 

\(9!-8!-7!.8^2\)

\(=362880-40320-5040.64\)

\(=362880-40320-322560\)

\(=322560-322560\)

\(=0\)

9!-8!-7!.8^2=0

362880-40320-5040.64=0

`25 - 15 : 5= 25 - 3 =22`

`101 × (16 – 7)= 101 xx 9=909`

`40 + 8 : 2= 40 + 4=44`

`48 : (8 : 2)= 48 : 4=12`

a: \(A=0x^2y^4z+\dfrac{7}{2}x^2y^4z-\dfrac{2}{5}x^2y^4z=\dfrac{31}{10}x^2y^4z=\dfrac{31}{10}\cdot2^2\cdot\dfrac{1}{16}\cdot\left(-1\right)=-\dfrac{31}{40}\)

a: \(=\dfrac{7}{5}x^4z^3y=\dfrac{7}{5}\cdot2^4\cdot\left(-1\right)^3\cdot\dfrac{1}{2}=-\dfrac{56}{5}\)

b: \(=-xy^3\)

[(195+35/7)/8+195]*2-400

=[(195+5)/8+195]*2-400

=[200/8+195]*2-400

=[25+195]*2-400

=220*2-400

=440-400

=40

8/35+2/7:2/4.

=8/35+2/7*4/2.

=8/35+4/7.

=8/35+20/35.

=28/35.

=4/5.

=8/35+2/7x2

=8/35x4/7

=32/245

8/35+2/7:2/4

=8/35+4/7

=28/35=4/5

chúc bạn học tốt

Lời giải:

a.

$\frac{5}{15}-\frac{1}{6}\times \frac{2}{5}=\frac{5}{15}-\frac{1}{15}=\frac{4}{15}$

b.

$\frac{8}{24}+\frac{3}{4}:\frac{1}{8}=\frac{1}{3}+6=\frac{19}{3}$

c.

$\frac{1}{7}: \frac{2}{8}-\frac{1}{7}=\frac{1}{7}\times 4-\frac{1}{7}$

$=\frac{1}{7}\times (4-1)=\frac{1}{7}\times 3=\frac{3}{7}$