`a)` Ta có:

`A=-6x^3+x^2+3x+2`

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

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

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

Vì: `(2x-1)(-3x^2-x+1)\vdots(2x-1)`

Do đó để `A\vdots(2x-1)` thì: `3\vdots `(2x-1)`

`->2x-1\in Ư(3)={1;-1;3;-3}`

`->2x\in{2;0;4;-2}`

`->x\in{1;0;2;-1}`

`b)A=-5x^3+3x^2+15x-6`

`->27A=-135x^3+81x^2+405x-162`

`=(-135x^3-45x^2)+(126x^2+42x)+(363x+121)-282`

`=-45x^2(3x+1)+42x(3x+1)+121(3x+1)-282`

`=(-45x^2+42x+121)(3x+1)-283`

Vì: `(-45x^2+42x+121)(3x+1)\vdots(3x+1)`

`->` Để `27A\vdotsB` thì: `283\vdots(3x+1)`

`3x+1\in Ư(283)={+-1;+-283}`

Vì: `3x+1` chia `3` dư `1` do đó:

`->3x+1\in{1;283}`

`->3x\in{0;282}`

`->x\in{0;94}`

Vậy: `...`

Bài 7:

`a)(x-1)(y+2)=7`

Vì: `x,y` nguyên `->x-1,y+2\in Ư(7)={1;-1;7;-7}`

`TH1:x-1=1`

`->x=1+1=2`

Suy ra: `y+2=7`

`->y=7-2=5`

`TH2:x-1=-1`

`->x=-1+1=0`

Suy ra: `y+2=-7`

`->y=-7-2=-9`

`TH3:x-1=7`

`->x=7+1=8`

Suy ra: `y+2=1`

`->y=1-2=-1`

`TH4:x-1=-7`

`->x=-7+1=-6`

Suy ra: `y+2=-1`

`->y=-1-2=-3`

Vậy: `....`

`b)x(y-3)=-12`

Vì: `x,y\inZ` suy ra: `x,y-3\in Ư(-12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}`

`TH1:x=-1`

Suy ra: `y-3=12`

`->y=12+3=15`

`TH2:x=1`

Suy ra: `y-3=-12`

`y=-12+3=-9`

`TH3:x=-2`

Suy ra: `y-3=6`

`->y=6+3=9`

`TH4:x=2`

Suy ra: `y-3=-6`

`->y=-6+3=-3`

`TH5:x=-3`

Suy ra: `y-3=4`

`->y=4+3=7`

`TH6:x=3`

Suy ra: `y-3=-4`

`->y=-4+3=-1`

`TH7:x=-4`

Suy ra: `y-3=3`

`->y=3+3=6`

`TH8:x=4`

Suy ra: `y-3=-3`

`->y=-3+3=0`

`TH9:x=6`

Suy ra: `y-3=-2`

`->y=-2+3=1`

`TH10:x=-6`

Suy ra: `y-3=2`

`->y=2+3=5`

`TH11:x=12`

Suy ra: `y-3=-1`

`->y=-1+3=2`

`TH12:x=-12`

Suy ra: `y-3=1`

`->y=1+3=4`

Vậy: `... `

`c)xy-3x-y=0`

`x(y-3)-y=0`

`x(y-3)-(y-3)-3=0`

`(x-1)(y-3)=3`

Vì: `x,y\inZ` do đó: `x-1,y-3\in Ư(3)={1;-1;3;-3}`

`TH1:x-1=1`

`->x=1+1=2`

Suy ra: `y-3=3`

`->y=3+3=6`

`TH2:x-1=-1`

`->x=-1+1=0`

Suy ra: `y-3=-3`

`->y=-3+3=0`

`TH3:x-1=3`

`->x=3+1=4`

Suy ra: `y-3=1`

`->y=1+3=4`

`TH4:x-1=-3`

Suy ra: `y-3=-1`

`->y=-1+3=2`

Vậy: `...`

`d)xy+2x+2y=-16`

`x(y+2)+(2y+4)=-12`

`x(y+2)+2(y+2)=-12`

`(x+2)(y+2)=-12`

Vì: `x,y\inZ` do đó: `x+2,y+2\in Ư(12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}`

`TH1:x+2=1`

`->x=1-2=-1`

Suy ra: `y+2=12`

`->y=10`

`TH2:x+2=-1`

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

Suy ra: `y+2=-12`

`->y=-14`

`TH3:x+2=2`

`->x=2-2=0`

Suy ra: `y+2=6`

`->y=6-2=4`

`TH4:x+2=-2`

`->x=-2-2=-4`

Suy ra: `y+2=-6`

`->y=-8`

`TH5:x+2=3`

`->x=3-2=1`

Suy ra: `y+2=4`

`->y=2`

`TH6:x+2=-3`

`->x=-3-2=-5`

Suy ra: `y+2=-4`

`->y=-6`

`TH7:x+2=-4`

`->x=-4-2=-6`

Suy ra: `y+2=-3`

`->y=-5`

`TH8:x+2=4`

`->x=2`

Suy ra: `y+2=3`

`->y=3-2=1`

`TH9:x+2=6`

`->x=6-2=4`

Suy ra: `y+2=2`

`->y=0`

`TH10:x+2=-6`

`->x=-6-2=-8`

Suy ra: `y+2=-2`

`->y=-4`

`TH11:x+2=12`

`->x=12-2=10`

Suy ra: `y+2=1`

`->y=1-2=-1`

`TH12:x+2=-12`

`->x=-12-2=-14`

Suy ra: `y+2=-1`

`->y=-1-2=-3`

Vậy: `...`

`(x+1/4-1/3):(21*1/6-1/4)=7/46`

`(x+3/12-4/12):(7*1/2-1/4)=7/46`

`(x-1/12):(7/2-1/4)=7/46`

`(x-1/12):(14/4-1/4)=7/46`

`(x-1/12):13/4=7/46`

`x-1/12=7/46*13/4`

`x-1/12= 91/184`

`x=91/184+1/12`

`x= 319/552`

Vậy: `x=319/552`

Bạn ơi P vs M đâu vậy bạn

`a)` ĐKXĐ của `Q` là:

`{(x>=0),(2-\sqrt{x}\ne0):}`

`{(x>=0),(\sqrt{x}\ne2):}`

`{(x>=0),(x\ne4):}`

`b)Q=((8-x\sqrt{x})/(2-\sqrt{x})+2\sqrt{x})((2-\sqrt{x})/(2+\sqrt{x}))^2`

`=((2^3-(\sqrt{x})^3)/(2-\sqrt{x})+2\sqrt{x})((2-\sqrt{x})/(2+\sqrt{x}))^2`

`=(((2-\sqrt{x})(x+2\sqrt{x}+4))/(2-\sqrt{x})+2\sqrt{x})(2-\sqrt{x})^2/(2+\sqrt{x})^2`

`=(x+2\sqrt{x}+4+2\sqrt{x})*(2-\sqrt{x})^2/(2+\sqrt{x})^2`

`=(x+4\sqrt{x}+4)*(2-\sqrt{x})^2/(\sqrt{x}+2)^2`

`=(\sqrt{x}+2)^2*(2-\sqrt{x})^2/(\sqrt{x}+2)^2`

`=(2-\sqrt{x})^2`

`=x-4\sqrt{x}+4`

`b)x=9` (thỏa đk của `Q`)

Thay `x=9` vào `Q` ta được:

`Q=9-4*\sqrt{9}+4`

`=9-4*3+4`

`=13-12`

`=1`

Vậy: `Q=1` khi `x=9`

`M=(x-5)(x+5)`

`=x^2-5^2`

`=x^2-25`

`P=(x-2)(x^2+2x+4)`

`=(x-2)(x^2+2*x+2^2)`

`=x^3-2^3`

`=x^3-8`

`M+P`

`=(x^2-25)+(x^3-8)`

`=x^2-25+x^3-8`

`=x^3+x^2-33`

`M-1`

`=x^2-25-1`

`=x^2-26`

`P-M`

`=(x^3-8)-(x^2-25)`

`=x^3-8-x^2+25`

`=x^3-x^2+17`

`P-2M`

`=(x^3-8)-2(x^2-25)`

`=x^3-8-2x^2+50`

`=x^3-2x^2+42`

`2M-2P`

`=2(x^2-25)-2(x^3-8)`

`=2x^2-50-2x^3+16`

`=-2x^3-2x^2-34`

lỗi latex quá nnhiều, ko so với đk

`a)\sqrt{4x^2-4x+1}=3`

ĐK: `4x^2-4x+1>=0->(2x-1)^2>=0` (luôn đúng)

Phương trình đã cho tương đương:

`\sqrt{(2x-1)^2}=3`

`|2x-1|=3`

`TH1:x>=1/2`

`->2x-1=3`

`2x=3+1`

`2x=4`

`x=4/2`

`x=2(N)`

`TH2:x<1/2`

`-(2x-1)=3`

`1-2x=3`

`2x=1-3`

`2x=-2`

`x=-2/2`

`x=-1(N)`

Vậy: `S={2;-1}`

`b)3(\sqrt{x}+2)+5=4\sqrt{4x}+1(x>=0)`

`3(\sqrt{x}+2)+5=4*2\sqrt{x}+1`

`3\sqrt{x}+6+5=8\sqrt{x}+1`

`8\sqrt{x}-3\sqrt{x}=11-1`

`5\sqrt{x}=10`

`\sqrt{x}=10/5=2`

`x=2^2=4(N)`

Vậy: `x=4`

`c)\sqrt{1-3x}<2`

ĐK: `1-3x>=0`

`x<=1/3`

Bất phương trình tương đương:

`1-3x<2^2`

`1-3x<4`

`3x>1-4`

`3x> -3`

`x> -3/3`

`x> -1`

Vậy: `-1<x<=1/3`

`P=(a\sqrt{a})(\sqrt{a}-1)+1/(1-\sqrt{a})`

`a)` ĐKXĐ: `{(a>=0),(\sqrt{a}-1\ne0):}`

`{(a>=0),(\sqrt{a}\ne1):}`

`{(a>=0),(a\ne1):}`

`b)P=(a\sqrt{a})/(\sqrt{a}-1)+1/(1-\sqrt{a})`

`=(a\sqrt{a})/(\sqrt{a}-1)-1/(\sqrt{a}-1)`

`=(a\sqrt{a}-1)/(\sqrt{a}-1)`

`=((\sqrt{a}-1)(a+\sqrt{a}+1))/(\sqrt{a}-1)`

`=a+\sqrt{a}+1`

`a=9/4` (thỏa đkxđ)

Thay `a=9/4` vào `P` ta được:

`P=9/4+\sqrt{9/4}+1=9/4+3/2+1`

`=9/4+6/4+4/4=19/4`

Vậy: `P=19/4` khi `a=9/4`.

`bb(1))`

`a)A=(\sqrt{x}+5)/(2\sqrt{x}-1)(x>=0,x\ne1/4)`

Vì: `A` nguyên `->2A` nguyên

`->2A=(2\sqrt{x}+10)/(2\sqrt{x}-1)`

`=((2\sqrt{x}-1)+11)/(2\sqrt{x}-1)`

`=1+11/(2\sqrt{x}-1)`

Để `2A` nguyên thì: `11\vdots(2\sqrt{x}-1)`

`2\sqrt{x}-1\in Ư(11)`

`2\sqrt{x}-1\in{1;-1;11;-11}`

Mà: `2\sqrt{x}-1>=-1\AAx`

Suy ra: `2\sqrt{x}-1\in{-1;1;11}`

`2\sqrt{x}\in{0;2;12}`

`\sqrt{x}\in{0;1;6}`

`x\in{0;1;36}`

Vậy: `x\in{0;1;36}`

`b)B=(3\sqrt{x}+1)/(\sqrt{x}-2)(x>=0;x\ne4)`

`=((3\sqrt{x}-6)+7)/(\sqrt{x}-2)`

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

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

Để `B` nguyên thì: `7\vdots(\sqrt{x}-2)`

`\sqrt{x}-2\in Ư(7)={1;-1;7;-7}`

Mà: `\sqrt{x}-2>=-2\AAx`

`->\sqrt{x}-2\in{1;-1;7}`

`\sqrt{x}\in{3;1;9}`

`x\in{9;1;81}`

Vậy: `x\in{9;1;81}`

`c)C=(3\sqrt{x}+2)/(2\sqrt{x}+1)(x>=0)`

Vì: `C` nguyên `->2C` nguyên

`->2C=(6\sqrt{x}+4)/(2\sqrt{x}+1)`

`=(3(2\sqrt{x}+1)+1)/(2\sqrt{x}+1)`

`=3+1/(2\sqrt{x}+1)`

Để `2C` nguyên thì: `1\vdots(2\sqrt{x}+1)`

`2\sqrt{x}+1\in Ư(1)={1;-1}`

Mà: `2\sqrt{x}+1>=1\AAx`

`->2\sqrt{x}+1=1`

`2\sqrt{x}=0`

`\sqrt{x}=0`

`x=0`

Vậy: `x=0`

`d)D=(2\sqrt{x}+3)/(\sqrt{x}-5)(x>=0;x\ne25)`

`=(2(\sqrt{x}-5)+13)/(\sqrt{x}-5)`

`=2+13/(\sqrt{x}-5)`

Để `D` nguyên thì: `13\vdots(\sqrt{x}-5)`

`\sqrt{x}-5\in Ư(13)={1;-1;13;-13}`

Mà: `\sqrt{x}-5>=-5\AAx`

`->\sqrt{x}-5\in{1;-1;13}`

`\sqrt{x}\in{6;4;18}`

`x\in{36;16;324}`

Vậy: `x\in{36;16;324}`