Đáp án:

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Bài `6`

`a,`

`x^2 + (2y-3/5)^{10}=0`

Với mọi `x,y` có : `x^2 ≥ 0, (2y-3/5)^{10} ≥0`

`⇔ x^2 + (2y-3/5)^{10} ≥0`

Dấu "`=`" xảy ra khi :

`⇔x^2=0, (2y-3/5)^{10}=0`

`⇔x=0,2y-3/5=0`

`⇔x=0,2y=3/5`

`⇔x=0,y=3/10`

Vậy `(x;y) = (0;3/10)`

`b,`

`( (-3)/2x+2)^{20} + (y^2 - 4/9)^2 ≤ 0`

Với mọi `x,y` có : `( (-3)/2x + 2)^{20} ≥0, (y^2 - 4/9)^2 ≥0`

`⇔ ( (-3)/2x+2)^{20} + (y^2 - 4/9)^2 ≥0 ∀x,y`

Dấu "`=`" xảy ra khi :

`⇔ ( (-3)/2x+2)^{20}=0, (y^2 -4/9)^2 =0`

`⇔ (-3)/2x+2=0,y^2-4/9=0`

`⇔ (-3)/2x=-2,y^2=4/9`

`⇔ x=4/3, y=±2/3`

Vậy `(x;y) = (4/3; ±2/3)`

Bài `7.`

`(3x-5)^{2010} + (y^2-1)^{2012} + (x-z)^{2014}=0`

Với mọi `x,y,z` có : `(3x-5)^{2010} ≥0, (y^2-1)^{2012}≥0, (x-z)^{2014}≥0`

`⇔ (3x-5)^{2010} + (y^2-1)^{2012} + (x-z)^{2014} ≥0 ∀x,y,z`

Dấu "`=`" xảy ra khi :

`⇔ (3x-5)^{2010}=0, (y^2-1)^{2012}=0, (x-z)^{2014}=0`

`⇔ 3x-5=0,y^2-1=0, x-z=0`

`⇔ 3x=5,y^2=1,x=z`

`⇔ x=5/3, y=±1,z=5/3`

Vậy `(x;y;z) = (5/3; ±1; 5/3)`

Bài `8`

`a,`

Xét vế trái ta có :

`32^8 × 625^{10}`

`= (2^5)^8 × (5^4)^{10}`

`= 2^{5×8) × 5^{4×10}`

`= 2^{40} × 5^{40}`

`= (2×5)^{40}`

`= 10^{40}` (Bằng vế phải)

`-> 32^8 × 625^{10} = 10^{40}` (đpcm)

`b,`

`A = (4^2 × 25^2 + 32 × 125)/(2^3 × 5^2)`

`⇔ A = ( (2^2)^2 × (5^2)^2 + 2^5 × 5^3)/(2^3×5^2)`

`⇔ A = (2^4 × 5^4 + 2^5 × 5^3)/(2^3 × 5^2)`

`⇔ A = (2^3 × 5^2 × (2 × 5^2 + 2^2 × 5) )/(2^3 × 5^2)`

`⇔ A = 2 × 5^2 + 2^2 × 5`

`⇔ A = 2 × 25 + 4 ×5`

`⇔A=50+20`

`⇔A=70`

Vậy `A=70`

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