Tìm x,y,z nhá

a) -75-(x+20)+95=0

<=> 20-(x+20)=0

<=> 20 - x - 20 = 0

<=> -x = 0

b) (135 + x ) -135 = 0

<=> 135 + x = 135

<=> x = 0

c)(-562) - ( x - 562 ) = 0

<=> -562 - x + 562 = 0

<=> x =0

d ) (61 + 9x -23) = 91

<=> 38+9x = 91

<=> 9x = 53

<=> x =\(\dfrac{53}{9}\)

h) 63-(51+3x) = 237 

<=> 63 - 51-3x = 237

<=> 12 - 3x = 237

<=> -3x = 225

<=> x = -75

làm tới đó hoi nha tự lực gánh sinh đi :))

`-75-(x+20)+95=0`

`-75-(x+20)=0-95`

`-75-(x+20)=-95`

`x+20=-75-(-95)`

`x+20=-75+95`

`x+20=20`

`x=20-20`

`x=0`

___________________

`(135+x)-135=0`

`135+x=0+135`

`135+x=135`

`x=135-135`

`x=0`

_________________

`-562-(x-562)=0`

`x-562=-562-0`

`x-562 = -562`

`x=-562+562`

`x=0`

___________________

`-(2x+24)+211=-123`

`-2x-24+211=-123`

`-2x+187=-123`

`-2x=-123-187`

`-2x=-310`

`-x=-310:(-2)`

`x=310:2`

`x=155`

giup

\(x^6-32x=0\)

\(\Leftrightarrow x\left(x^5-32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^5-32=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^5=2^5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Theo đề bài => 1 trong hai số bị trừ ( x^6) hoặc 32x =0

Nếu x^6=0 => x=0 => x^6 - 32x = 0 (1)

Nếu 32x=0 => x=0 => x^6 - 32x = 0 (2)

Vậy từ (1) và( 2) => x = 0

`1-1/2+2-2/3+3-3/4-1/4-3-1/3-2-1/2-1`

`=(1+2+3-3-2-1)+(-1/2-2/3-3/4-1/4-1/3-1/2)`

`=(1+2+3-3-2-1)+[(-1/2 - 1/2)-(2/3-1/3)-(3/4 - 1/4)]`

`=0 + (-1 - 1 - 1)`

`=0 + (-3)`

`=0-3`

`=-3`

 A = 1 -1/2 + 2 - 2/3 + 3 -3/4 -1/4 -3 -1/3 -2 -1/2 -1

A = ( 1 - 1/2 -1/2)  + (2-2) + ( 3-3) - (2/3 -1/3) - (3/4+1/4) - 1

A = 0 + 0 + 0 -1 - 1-1

A = -3

A = - (x + 18/1203)² - 183/121

-(x+ 18/1203)² ≤ 0 

⇔ A(max) = -183/121 ⇔ x = -18/1203 = - 6/401

B = \(\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\)

để B(max) thì (x+1/3)² + 5 nhỏ nhất 

đặt C = (x+1/3)² + 5  thì  B(max) ⇔ C(min)

vì (x+1/3)² ≥ 0 ⇔ C(min) = 5 ⇔ x =-1/3

B(max) = 4/5 ⇔ x = -1/3

a) \(A=-\left(\dfrac{x+18}{1203}\right)^2-\dfrac{183}{121}\le-\dfrac{183}{121}\)

Dấu bằng xảy ra 

\(\Leftrightarrow x+18=0\Leftrightarrow x=-18\)

b) \(B=\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\)

Ta có : \(\left(x+\dfrac{1}{3}\right)^2+5\ge5\)

\(\Leftrightarrow\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\le\dfrac{4}{5}\)

Dấu bằng xảy ra 

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

B = (x-1/4)⁴ + (y-3)² + 9

(x-1/4)⁴ ≥ 0 ; (y-3)² ≥ 0 ⇒ A(min) = 9 ⇔ x =1/4; y =3