`A=(-6x+9)/(7x-5)`

Vì: `A` nguyên `->7A` nguyên

`->7*(-6x+9)/(7x-5)=(-42x+63)/(7x-5)`

`=((-42x+30)+33)/(7x-5)`

`=(-6(7x-5)+33)/(7x-5)=-6+33/(7x-5)`

Để `7A` nguyên thì: `33/(7x-5)\inZ`

`->33vdots7x-5`

`->7x-5\in Ư(33)={1;-1;3;-3;11;-11;33;-33}`

`->7x\in{6;4;8;2;16;-6;38;-28}`

`->x\in{6/7;4/7;8/7;2/7;16/7;-6/7;38/7;-4}`

Mà `x\inZ->x=-4`

Vậy: `x=-4`

Bài 3:

`a)A=x^4-3x^3+3x^2-ax+b`

`=(x^4-3x^3+4x^2)+(3x^2-9x+12)+(-a+9)x+(b-12)`

`=x^2(x^2-3x+4)+3(x^2-3x+4)+(-a+9)x+(b-12)`

`=(x^2+3)(x^2-3x+4)+(-a+9)x+(b-12)`

Vì: `(x^2-3x+4)(x^2+3)\vdots(x^2-3x+4)`

Suy ra: `A\vdotsB` khi `-a+9=0` và `b-12=0`

`a=9` và `b=12`

Vậy: `(a;b)=(9;12)`

`b)A=6x^4-7x^3+ax^2+b`

`=(6x^4-6x^3+6x^2)+(-x^3+x^2-x)+(-x^2+x-1)+(a-6)x^2+(b-1)`

`=6x^2(x^2-x+1)-x(x^2-x+1)-(x^2-x+1)+(a-6)x^2+(b-1)`

`=(6x^2-x-1)(x^2-x+1)+(a-6)x^2+(b-1)`

Vì: `(6x^2-x-1)(x^2-x+1)\vdots(x^2-x+1)`

Nên để `A\vdotsB` thì: `a-6=0` và `b-1=0`

`a=6` và `b=1`

Vậy: `(a;b)=(6;1)`

`bb2`

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

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

`=\sqrt{(2\sqrt{3}+2)/(12-4)-(3(2\sqrt{3}-2))/(12-4)}`

`=\sqrt{(2\sqrt{3}+2)/8-(6\sqrt{3}-6)/8}`

`=\sqrt{(2\sqrt{3}+2-6\sqrt{3}+6)/8}`

`=\sqrt{(8-4\sqrt{3})/8}`

`=\sqrt{(4-2\sqrt{3})/4}`

`=\sqrt{((\sqrt{3})^2-2*\sqrt{3}*1+1^2)/4}`

`=\sqrt{(\sqrt{3}-1)^2/4}`

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

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

`->2x=\sqrt{3}-1`

`->2x+1=\sqrt{3}`

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

`->4x^2+4x+1=3`

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

`P=(4(x+1)x^2024-2x^2023+2x+3)/(2x^2+3x)`

`=((4x+4)x^2024-2x^2023+2x+3)/(2x^2+3x)`

`=(4x^2025+4x^2024-2x^2023+2x+3)/(2x^2+3x)`

`=(2x^2023*(2x^2+2x-1)+2x+3)/(2x^2+3x)`

Thay `2x^2+2x-1=0` vào `A` ta được:

`A=(2x^2023*0+2x+3)/(2x^2+3x)`

`=(2x+3)/(x(2x+3))`

`=1/x`

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

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

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

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

`=\sqrt{3}+1`

Vậy: `A=\sqrt{3}+1`

Bài 64:

`x=(\root[3]{10+6\sqrt{3}}(\sqrt{3}-1))/(\sqrt{6+2\sqrt{5}}-\sqrt{5})`

`=(\root[3]{(\sqrt{3})^3+3*(\sqrt{3})^2*1+3*\sqrt{3}*1^2+1^3}*(\sqrt{3}-1))/(\sqrt{(\sqrt{5})^2+2*\sqrt{5}*1+1^2}-\sqrt{5})`

`=(\root[3]{(\sqrt{3}+1)^3}*(\sqrt{3}-1))/(\sqrt{(\sqrt{5}+1)^2}-\sqrt{5})`

`=((\sqrt{3}+1)(\sqrt{3}-1))/(\sqrt{5}+1-\sqrt{5})`

`=(3-1)/1`

`=2/1`

`=2`

Thay `x=2` vào `P` ta được:

`P=(12x^2+4x-55)^2017`

`=(12*2^2+4*2-55)^2017`

`=12*4+8-55)^2017`

`=(48+8-55)^2017`

`=1^2017`

`=1`

Vậy: `P=1`

Câu 2:

`x=3+\sqrt{2}`

`->x-3=\sqrt{2}`

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

`->x^2-6x+9=2`

`->x^2-6x+9-2=0`

`->x^2-6x+7=0`

`M=\sqrt{x^2-6x+16}/(x^3-5x^2+x-1)`

`=\sqrt{x^2-6x+16}/((x^3-6x^2+7x)+(x^2-6x+7)-8)`

`=\sqrt{x^2-6x+16}/(x(x^2-6x+7)+(x^2-6x+7)-8)`

`=\sqrt{(x^2-6x+7+9}/((x^2-6x+7)(x+1)-8)`

Thay `x^2-6x+7=0` vào `M` ta được:

`M=\sqrt{0+9}/(0*(x+1)-8)=\sqrt{9}/(-8)=-3/8`

Vậy: `M=-3/8`

anh ơi câu e sai rồi anh `3*35=105`

`b)(2x+1)(2x-3)-(4x+1)(x+2)=8`

`[2x(2x-3)+(2x-3)]-[4x(x+2)+(x+2)]=8`

`(4x^2-6x+2x-3)-(4x^2+8x+x+2)=8`

`(4x^2-4x-3)-(4x^2+9x+2)=8`

`4x^2-4x-3-4x^2-9x-2=8`

`-13x-5=8`

`-13x=8+5`

`-13x=13`

`x=13/(-13)`

`x=-1`

Vậy: `x=-1`

`e)3(x-7)(x+5)-(x-1)(3x+2)=-13`

`3[x(x+5)-7(x+5)]-[x(3x+2)-(3x+2)]=-13`

`3(x^2+5x-7x-35)-(3x^2+2x-3x-2)=-13`

`3(x^2-2x-35)-(3x^2-x-2)=-13`

`3x^2-6x-105-3x^2+x+2=-13`

`-5x-103=-13`

`5x=-103+13`

`5x=-90`

`x=-90/5`

`x=-18`

Vậy: `x=-18`

`5(x-3)(x-7)-(5x+1)(x-2)=25`

`5[x(x-7)-3(x-7)]-[5x(x-2)+(x-2)]=25`

`5(x^2-7x-3x+21)-(5x^2-10x+x-2)=25`

`5(x^2-10x+21)-(5x^2-9x-2)=25`

`5x^2-50x+105-5x^2+9x+2=25`

`-41x+107=25`

`41x=107-25`

`41x=82`

`x=82/41`

`x=2`

Vậy: `x=2`

=========================

`3(x-7)(x+5)-(x-1)(3x+2)=-13`

`3[x(x+5)-7(x+5)]-[x(3x+2)-(3x+2)]=-13`

`3(x^2+5x-7x-35)-(3x^2+2x-3x-2)=-13`

`3(x^2-2x-35)-(3x^2-x-2)=-13`

`3x^2-6x-105-3x^2+x+2=-13`

`-5x-103=-13`

`5x=-103+13`

`5x=-90`

`x=-90/5`

`x=-18`

Vậy: `x=-18`

`a)Q=((x-2\sqrt{x})/(x-4)-(x-x\sqrt{x}-6)/(x+\sqrt{x}-2)+(\sqrt{x}+1)/(1-\sqrt{x}))*(x+39)/(x+3\sqrt{x}-10)`

`=((\sqrt{x}(\sqrt{x}-2))/((\sqrt{x}+2)(\sqrt{x}-2))-(x-x\sqrt{x}-6)/((\sqrt{x}+2)(\sqrt{x}-1))-(\sqrt{x}+1)/(\sqrt{x}-1))*(x+39)/(x+3\sqrt{x}-10)`

`=(\sqrt{x}/(\sqrt{x}+2)-(x-x\sqrt{x}-6)/((\sqrt{x}+2)(\sqrt{x}-1))-(\sqrt{x}+1)/(\sqrt{x}-1))*(x+39)/(x+3\sqrt{x}-10)`

`=(\sqrt{x}(\sqrt{x}-1)-(x-x\sqrt{x}-6)-(\sqrt{x}+1)(\sqrt{x}+2))/((\sqrt{x}-1)(\sqrt{x}+2))*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=(x-\sqrt{x}-x+x\sqrt{x}+6-(x+3\sqrt{x}+2))/((\sqrt{x}-1)(\sqrt{x}+2))*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=(x\sqrt{x}-\sqrt{x}+6-x-3\sqrt{x}-2)/((\sqrt{x}-1)(\sqrt{x}+2)*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=(x\sqrt{x}-x-4\sqrt{x}+4)/((\sqrt{x}-1)(\sqrt{x}+2))*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=(x(\sqrt{x}-1)-4(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+2))*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=((x-4)(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+2))*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=((\sqrt{x}-2)(\sqrt{x}+2))/(\sqrt{x}+2)*(x+39)/((\sqrt{x}-2)(\sqrt{x}+5))`

`=(x+39)/(\sqrt{x}+5)`

`b)Q=(x-25+64)/(\sqrt{x}+5)`

`=(x-25)/(\sqrt{x}+5)+64/(\sqrt{x}+5)`

`=(\sqrt{x}-5)+64/(\sqrt{x}+5)`

`=(\sqrt{x}+5)+64/(\sqrt{x}+5)-10`

Vì: `\sqrt{x}+5,64/(\sqrt{x}+5)>0`

Suy ra áp dụng BĐT Cauchy ta được:

`Q>=2\sqrt{(\sqrt{x}+5)*64/(\sqrt{x}+5)}-10=2\sqrt{64}-10=2*8-10=6`

Dấu "=" xảy ra: `\sqrt{x}+5=64/(\sqrt{x}+5)`

`->(\sqrt{x}+5)^2=64`

`->\sqrt{x}+5=8`

`->\sqrt{x}=3`

`->x=9(N)`

Vậy: `...`