ĐK: `x-3>=0` suy ra: `x>=3`

Do đó: `x-2>0` hay: `|x-2|=x-2`

Ta được: `|9-2(x-2)|=x-3`

`|9-2x+4|=x-3`

`|13-2x|=x-3`

`TH1:3<=x<=13/2` ta được:

`13-2x=x-3`

`x+2x=13+3`

`3x=16`

`x=16/3(tm)`

`TH2:x>=13/2` ta được:

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

`2x-13=x-3`

`2x-x=-3+13`

`x=10(tm)`

Vậy: `x\in{16/3;10}`

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

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

`8x^3-12x^2+6x-1-8x^3=4x+1-12x^2-3x`

`-12x^2+6x-1=-12x^2+x+1`

`-12x^2+6x-1+12x=x+1`

`6x-1=x+1`

`6x-x=1+1`

`5x=2`

`x=2/5`

Vậy: `x=2/5`

`\sqrt{3x+1}+\sqrt{x+1}=8(x>=-1/3)`

`(\sqrt{3x+1}+\sqrt{x+1})^2=8^2`

`3x+1+x+1+2\sqrt{(3x+1)(x+1)}=64`

`4x+2+2\sqrt{(3x+1)(x+1)}=64`

`2x+1+\sqrt{(3x+1)(x+1)}=32`

`\sqrt{(3x+1)(x+1)}=31-2x(x>=31/2)`

`(3x+1)(x+1)=(31-2x)^2`

`3x^2+3x+x+1=4x^2-124x+961`

`x^2-128x+960=0`

`(x^2-8x)+(-120x+960)=0`

`x(x-8)-120(x-8)=0`

`(x-120)(x-8)=0`

`x-120=0` hoặc `x-8=0`

`x=120(L)` hoặc `x=8(tm)`

Vậy: `x=8`

Ta có:

`x+1/x=5`

`(x+1/x)^2=5^2`

`x^2+2*x*1/x+(1/x)^2=25`

`x^2+2*1+1/x^2=25`

`x^2+1/x^2+2=25`

`x^2+1/x^2=25-2`

`x^2+1/x^2=23`

Vậy: `x^2+1/x^2=23`

-----------

Áp dụng HĐT: `(A+B)^2=A^2+2AB+B^2`

`|-2x-6|=|2x+3|`

`TH1:-2x-6=2x+3`

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

`4x=-9`

`x=-9/4`

`TH2:-2x-6=-(2x-3)`

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

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

`0=9` (vô lý)

`->` Không có `x` thỏa mãn

Vậy: `x=-9/4`

`a)-3n+1 \vdots(-n-1)`

`->(-3n-3)+4\vdots(-n-1)`

`->3(-n-1)+4\vdots(-n-1)`

`->4\vdots(-n-1)`

`->-n-1\in Ư(4)={1;-1;2;-2;4;-4}`

`->-n\in{2;0;3;-1;5;-3}`

`->n\in{-2;0;-3;1;-5;3}`

Vậy: `...`

`b)6n+2\vdotsn-3`

`->(6n-18)+20\vdots(n-3)`

`->6(n-3)+20\vdots(n-3)`

`->20\vdots(n-3)`

`->n-3\in Ư(20)={1;-1;2;-2;4;-4;5;-5;10;-10;20;-20}`

`->n\in{4;2;5;1;7;3;8;-2;13;-7;23;-17}`

Vậy: `...`

`\sqrt{3x+1}+\sqrt{x+1}=8(x>=-1/3)`

`(\sqrt{3x+1}+\sqrt{x+1})^2=8^2`

`3x+1+x+1+2\sqrt{(3x+1)(x+1)}=64`

`4x+2+2\sqrt{(3x+1)(x+1)}=64`

`2x+1+\sqrt{(3x+1)(x+1)}=32`

`\sqrt{(3x+1)(x+1)}=31-2x(x<=31/2)`

`(3x+1)(x+1)=(31-2x)^2`

`3x^2+4x+1=4x^2-124x+961`

`x^2-128x+960=0`

`(x-120)(x-8)=0`

`x-120=0` hoặc `x-8=0`

`x=120(L)` hoặc `x=8(N)`

Vậy: `x=8`

em ko ghi đúng đề cơ mà

Bài 3:

ĐK: `m>=0`

`a)` Đồ thị hàm số đi qua `A(1;0)` ta thay `x=1,y=0` vào suy ra:

`0=(3\sqrt{m}-m)*1-2\sqrt{m}`

`3\sqrt{m}-m-2\sqrt{m}=0`

`\sqrt{m}-m=0`

`\sqrt{m}(1-\sqrt{m})=0`

`\sqrt{m}=0` hoặc `\sqrt{m}=1`

`m=0(N)` hoặc `m=1(N)`

`b)` Đồ thị hàm số song với với `y=-2x`

Suy ra: `3\sqrt{m}-m=-2` và `-2\sqrt{m}\ne0`

`->m-3\sqrt{m}-2=0` và `\sqrt{m}\ne0`

Đặt: `\sqrt{m}=k>=0` ta được:

`k^2-3k-2=0` và `k\ne0`

`k^2-3k+9/4=2+9/4` và `k\ne0`

`(k-3/2)^2=17/4`

`k=3/2+\sqrt{17}/2(N)` hoặc `k=3/2-\sqrt{17}/2(L)` và `k\ne0`

`->\sqrt{m}=3/2+\sqrt{17}/2`

`->m=(3/2+\sqrt{17}/2)^2(N)`

Vậy: `...`

`((x+1)/(\sqrt{x}+1)+1/(x+\sqrt{x})-1/\sqrt{x}):\sqrt{x}/(x+2\sqrt{x}+1)>=2017+\sqrt{2017}(x>0)`

`((x+1)/(\sqrt{x}+1)+1/(\sqrt{x}(\sqrt{x}+1))-1/\sqrt{x})*(x+2\sqrt{x}+1)/\sqrt{x}>=2017+\sqrt{2017}`

`((\sqrt{x}(x+1))/(\sqrt{x}(\sqrt{x}+1))+1/(\sqrt{x}(\sqrt{x}+1))-(\sqrt{x}+1)/(\sqrt{x}(\sqrt{x}+1)))*\sqrt{x}/(\sqrt{x}+1)^2>=2017+\sqrt{2017}`

`(\sqrt{x}(x+1)+1-\sqrt{x}-1)/(\sqrt{x}(\sqrt{x}+1))*(\sqrt{x}+1)^2/\sqrt{x}>=2017+\sqrt{2017}`

`(x\sqrt{x}+\sqrt{x}-\sqrt{x})/(\sqrt{x})*(\sqrt{x}+1)/\sqrt{x}>=2017+\sqrt{2017}`

`(x\sqrt{x})/x*(\sqrt{x}+1)>=2017+\sqrt{2017}`

`\sqrt{x}(\sqrt{x}+1)>=2017+\sqrt{2017}`

`x-2017+\sqrt{x}-\sqrt{2017}>=0`

`(\sqrt{x}-\sqrt{2017})(\sqrt{x}+\sqrt{2017)+(\sqrt{x}-\sqrt{2017})>=0`

`(\sqrt{x}-\sqrt{2017})(\sqrt{x}+\sqrt{2017}+1)>=0`

Vì: `\sqrt{x}+\sqrt{2017}+1>0`

Suy ra: `\sqrt{x}-\sqrt{2017}>=0`

`\sqrt{x}>=\sqrt{2017}`

`x>=2017`

Vậy: `x>=2017`