Decomposition factor (1) x & # 178; (X-Y) + (Y-X) (2) (x-3) (x + 2) + 4 / 25 Evaluate (1) x / x + 1 △ (x - X / 1) where x = root 2 + 1 (2) solve the system of inequalities ① 2x-3 ≥ 1 ② 2 / X-1 + 2 ≥ - X (PS: 2 / X-1 + 2 ≥ - x is x-1) ——+2 ≥ - x is like this, two

Decomposition factor (1) x & # 178; (X-Y) + (Y-X) (2) (x-3) (x + 2) + 4 / 25 Evaluate (1) x / x + 1 △ (x - X / 1) where x = root 2 + 1 (2) solve the system of inequalities ① 2x-3 ≥ 1 ② 2 / X-1 + 2 ≥ - X (PS: 2 / X-1 + 2 ≥ - x is x-1) ——+2 ≥ - x is like this, two

x"( x - y ) + ( y - x )
= x"( x - y ) - ( x - y )
= ( x - y )( x" - 1 )
= ( x - y )( x - 1 )( x + 1 )
( x - 3 )( x + 2 ) + 25/4
= x" + 2x - 3x - 6 + 6 + 1/4
= x" - x + 1/4
= x" - ( 2 / 2 )x + ( 1/2 )"
= ( x - 1/2 )"
[ ( x + 1 )/x ] / ( x - 1/x )
= [ ( 1/x )( x + 1 ) ] / [ ( 1/x )( x" - 1 ) ]
= ( x + 1 ) / [ ( x + 1 )( x - 1 ) ]
= 1 / ( x - 1 )
= 1 / [ ( √2 + 1 ) - 1 ]
= 1 / ( √2 )
= √2 / 2
2x - 3 >= 1
2x >= 4
x >= 2
( x - 1 )/2 + 2 >= -x
x - 1 + 4 >= -2x
3x + 3 >= 0
3x >= -3
x >= -1
be careful:
If the numerator and denominator are written on the same line,
It should be the numerator divided by the denominator, such as [numerator / denominator]

When √ 2-x is meaningful, simplify the results of √ X & # 178; - 4x + 4 - √ X & # 178; - 6x + 9
How to continue the reduction when it is reduced to | X-2 | - | x-3 |

From √ 2-x, we can get x ＜ 2
Then X-2 ＜ 0, x-3 ＜ 0
So √ X & # 178; - 4x + 4 - √ X & # 178; - 6x + 9
=|x﹣2|﹣|x﹣3|
=2-x-3+x=-1

The value of (X & # 178; Y & # 179;), # 8319; has been determined by X &; = 5, Y &; = 3

(x²y³)ⁿ
=(x²ⁿ)(y³ⁿ)
=(xⁿ)²×(yⁿ)³
=5²×3³
=675