a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

a) \(\left(x-10\right)^2-x\left(x+8\right)=-12x+100=-11,76+100=88,24\)

b) \(x^3-9x^2+27x-27=\left(x-3\right)^3=\left(5-3\right)^3=8\)

c) \(6x\left(2x-7\right)-\left(3x-5\right)\left(4x+7\right)=-43x+35=121\)

\(a)\) \(\left(x-10\right)^{^2}-x.\left(x+8\right)\) \(với\) \(x=0,98\)

\(=-12x+100\)

\(=-11,76+100\)

\(=88,24\)

\(b)\) \(x^3-9x^2+27.x-27\) \(với\) \(x=5\)

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

\(=\left(5-3\right)^3\)

\(=8\)

\(c)\)\(6x.\left(2x-7\right)-\left(3x-5\right).\left(4x+7\right)\) \(tại\) \(x=-2\)

\(=-43+35\)

\(=121\)

Chúc bạn hôc tốt nha ❤

a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)

\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

`= x^2 - 2xy - y^3 + 2xy`

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

\(=x^2-2xy-y^3+2xy\)

\(=x^2-y^3\)

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

a: =2+6*(-1)^2019+2026

=2028-6

=2022

b: \(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}...\cdot\dfrac{625}{624}\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2}{\left(4-1\right)\left(4+1\right)}...\cdot\dfrac{625}{\left(25-1\right)\left(25+1\right)}\)

\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{1\cdot2\cdot3\cdot...\cdot48}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\)

\(=\dfrac{49}{1}\cdot\dfrac{2}{50}=\dfrac{98}{50}=\dfrac{49}{25}\)

`A=sqrt{(2-sqrt5)^2}+sqrt{(2sqrt2-sqrt5)^2}`

`A=|2-sqrt5|+|2sqrt2-sqrt5|`

`A=\sqrt5-2+2sqrt2-sqrt5`

`A=2sqrt2-2`

`b)B=sqrt{(sqrt7-2sqrt2)^2}+sqrt{(3-2sqrt2)^2}`

`B=|sqrt7-2sqrt2|+|3-2sqrt2|`

`A=2sqrt2-sqrt7+3-2sqrt2`

`A=3-sqrt7`

a,=> A=\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\sqrt{2}\right)^2}=2-\sqrt{5}+\sqrt{5}-2\sqrt{2}=2-2\sqrt{2}\)

b tương tự

a) Ta có: \(A=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)

\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)

\(=\sqrt{2}\)

b) Ta có: \(B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)

\(=2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\)

\(=3-\sqrt{7}\)

ta có :

`x^2 = 4`

`=> x = 2 ;-2`

TH1 :

thay `x=2 ; y = 5` ta có :

`2(3.5 -1) = 2.14 = 28`

TH2 :

thay `x= -2 , y = 5` ta có:

`(-2)(3.5-1) = (-2).14 = -28`

`b)`

ta có : `y^2 =1 `

`=> y = 1 ; -1;`

TH1:

thay `x=5 ; y=1` vào ta có:

`(5-3)(1-4)`

`=2.(-3)`

`=-6`

TH2:

thay `x = 5 ; y = -1` vào ta có :

`(5-3)(-1-4) `

`= 2 . (-5)`

`= -10`

a. \(x^2=4\\ \Leftrightarrow x=\sqrt{4}=2\)

Thay \(x=2;y=5\) vào ta được:

\(2\left(3\cdot5-1\right)\)

\(30-2=28\)

b. \(y^2=1\\ \Leftrightarrow y=\sqrt{1}=1\)

Thay \(x=5;y=1\) vào ta được:

\(\left(5-3\right)\left(1-4\right)\)

\(1\cdot\left(-3\right)=-3\)