`1)<=> -4x-3 + 5x+ 2 =0`

`<=> 5x-4x = -2+3`

`<=> x =1`

`2)<=> -5x +2-3x+6 =4`

`<=> -5x-3x = 4-6-2`

`<=> -8x=-4`

`<=> x=1/2`

`3) <=> -7x^2 +2 +7x^2 +14x =8`

`<=> 14x +2 =8`

`<=> 14x = 6`

`<=> x=3/7`

`1)sqrt[(2x-1)^2]=3`

`<=> |2x-1| =3`

`<=>[(2x-1=3),(2x-1=-3):}`

`<=>[(x=2),(x=-1):}`

`2)5/3sqrt(15x) -sqrt(15x)-2 =1/3sqrt(15x)(đk:x>=0)`

`<=> (5/3-1-1/3)sqrt(15x) = 2`

`<=> 1/3sqrt(15x)=2`

`<=> sqrt{15x} =6`

`<=> 15x =36`

`=>x = 12/5(t//m)`

3)`sqrt(x-6) (x^2-1)=0(đk:x>=6)`

`<=> [(sqrt(x-6)=0),(x^2-1)=0:}`

`<=> [(x-6=0),(x^2=1):}`

`<=> [(x=6(t//m)),(x=+-1(loại)):}`

4)`2sqrt(x-3)=1/3sqrt(4x-12)(đk:x>=3)`

`<=>2sqrt(x-3) = 2/3sqrt(x-3)`

`<=> 4/3sqrt(x-3)=0`

`<=> x-3=0`

`<=> x=3(t//m)`

`5)sqrt(4(x+1))=sqrt(8)(đk:x>=-1)`

`<=> 4(x+1)=8`

`<=> x+1=2`

`<=> x=1(t//m)`

\(P=\dfrac{2\sqrt{x}+5}{\sqrt{x}+1}=\dfrac{2\sqrt{x}+2+3}{\sqrt{x}+1}=2+\dfrac{3}{\sqrt{x}+1}\)`(đk:x >=0)`

Để `P in ZZ`

`=> sqrt(x) +1 in Ư(3)={+-1;+-3}`

Mà `sqrt{x} +1 >0 AA x>=0`

`=> sqrt(x) +1 in {1;3}`

`@ sqrt{x} +1 =1 => x =0`

`@ sqrt(x) +1 =3 => x=4`

`j) 2/x^2 >=0`

`=> x^2 >0`

`=> x ne 0`

`k)1/(-1+x)>=0`

`=> -1+x >0`

`=> x>1`

`l) 4/(x+3) >=0`

`=> x+3 >0`

`=> x>-3`

`m) 4x^2 >=0`

`=> x^2 >=0 `

`=> x in NN`

`n) -3x^2 >=0`

`=> x^2 <=0`

`=> x^2 =0`

`=> x=0`

`0) x^2 -2x +1 >=0`

`=> (x-1)^2 >=0 ( luôn -đúng)`

`=> x in NN`

P)`-x^2 -2x -1 >=0`

`<=> -(x^2 +2x+1)>=0`

`<=> -(x+1)^2 >=0`

`=> (x+1)^2 <=0`

Mà `(x+1)^2 >=0 AAx`

`=> x+1 =0`

`=> x=-1`

câu h là thuộc N thôi nha

a) `-2x +3 >= 0`

`<=> -2x >= -3`

`<=> x <= 3/2`

`b) -5x >=0`

`<=> x<=0`

`c)-3x+7 >=0`

`<=>  -3x >=-7`

`<=> x<=7/3`

`d)3x+7 >=0`

`<=> 3x>=-7`

`<=> x >=-7/3`

`e) x/3 >=0`

`=> x >=0`

`f) -5x >=0`

`=> x <=0`

`f) 4-x >=0`

`=> x <=4`

`h) 1+x^2 >=0`

mà `1+x^2 > 0 AA x`

`=> x in NN^#`

i) `(-5)/(x^2 +6) >=0`

Do -5 <0

`=> x^2 +6 <0 (vô-lí)`

`=> biểu thức ko có nghĩa

`A = -25x^2 +30x -2 = -(25x^2 -30x +2)`

`= -[(5x)^2 - 2*5x*3 +3^2 +2-3^2]`

`=-[(5x-3)^2 -7] = 7-(5x-3)^2`

Do `-(5x-3)^2 <= 0 AA x`

`=> 7- (5x-3)^2 <0 AA x `

hay `A<0 AA x (đpcm)`

`B =9x^2 +6x = (3x)^2 + 2*3x*1 +1 -1)`

`=(3x +1)^2 -1`

Do `(3x+1)^2 >=0 AA x`

`=> (3x+1)^2 -1 >=-1 AA x`

hay `B>=-1`

Dấu ''='' xảy ra khi và chỉ khi `3x+1=0 =>x =-1/3`

Vậy GTNN của `B=-1` khi `x=-1/3`