`\sqrt{x^2+6x+9}=\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}` (ĐK: `x\inR)`

`\sqrt{x^2+2*x*3+3^2}=\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}`

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

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

`|x+3|=|3+\sqrt{3}+3-\sqrt{3}`

`|x+3|=6`

`TH1:x+3=6`

`x=6-3`

`x=3`

`TH2:x+3=-6`

`x=-6-3`

`x=-9`

Vậy: `S={3;-9}`

`->` Chọn `bbD`

`c)` Ta có:

`\root[3]{17\sqrt{5}-38}`

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

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

`=\sqrt{5}-2`

`\sqrt{14-6\sqrt{5}}`

`=\sqrt{3^2-2*3*\sqrt{5}+(\sqrt{5})^2}`

`=\sqrt{(3-\sqrt{5})^2}`

`=|3-\sqrt{5}|`

`=3-\sqrt{5}`

Suy ra: `x=\root[3]{17\sqrt{5}-38}/(\sqrt{5}+\sqrt{14-6\sqrt{5})*(\sqrt{5}+2)`

`=(\sqrt{5}-2)/(\sqrt{5}+3-\sqrt{5})*(\sqrt{5}+2)`

`=(\sqrt{5}-2)/3*(\sqrt{5}+2)`

`=(5-2^2)/3`

`=1/3`

Thay `x=1/3` vào `T` ta được:

`T=(3*(1/3)^3+8*(1/3)^2-2)^2020`

`=((1/3)^2+8*(1/3)^2-2)^2020`

`=[(1/3)^2(1+8)-2]^2020`

`=[(1/3)^2*9-2]^2020`

`=(1-2)^2020`

`=(-1)^2020`

`=-1`

`N=\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}`

`=\sqrt{5-4\sqrt{5}+4}+\sqrt{5+4\sqrt{5}+4}`

`=\sqrt{(\sqrt{5})^2-2*\sqrt{5}*2+2^2}+\sqrt{(\sqrt{5})^2+2*\sqrt{5}*2+2^2}`

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

`=|\sqrt{5}-2|+|\sqrt{5}+2|`

`=(\sqrt{5}-2)+(\sqrt{5}+2)` (vì `\sqrt{5}>\sqrt{4}=2->\sqrt{5}-2>0)`

`=\sqrt{5}-2+\sqrt{5}+2`

`=2\sqrt{5}`

Vậy: `N=2\sqrt{5}`

`->` Chọn `bbD`

`-(-2x+3)^2-(5x-3)^2`

`=-(2x-3)^2-(5x-3)^2`

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

`=-[(4x^2-12x+9)+(25x^2-30x+9)]`

`=-(4x^2-12x+9+25x^2-30x+9)`

`=-(29x^2-42x+18)`

`=-29x^2+42x-18`

Vậy: `-(-2x+3)^2-(5x-3)^2=-29x^2+42x-18`

`M=\sqrt{a/b}+\sqrt{ab}-a\sqrt{1/(ab)}(a,b>0)`

`=\sqrt{a/b}+\sqrt{ab}-\sqrt{a^2*1/(ab)}`

`=\sqrt{a/b}+\sqrt{ab}-\sqrt{a*1/b}`

`=\sqrt{a/b}+\sqrt{ab}-\sqrt{a/b}`

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

`=0+\sqrt{ab}`

`=\sqrt{ab}`

Vậy `M=\sqrt{ab}` với `a>0,b>0`

`->` Chọn `bbB`

Ta có:

`a^2+b^2+1>=ab+a+b`

`2(a^2+b^2+1)>=2(ab+a+b)`

`2(a^2+b^2+1)-2(ab+a+b)>=0`

`2a^2+2b^2+2-2ab-2a-2b>=0`

`(a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)>=0`

`(a-b)^2+(a-1)^2+(b-1)^2>=0`

Vì: `(a-b)^2>=0\AAa,b`

`(a-1)^2>=0\AAa`

`(b-1)^2>=0\AAb`

Suy ra: `(a-b)^2+(a-1)^2+(b-1)^2>=0` (đúng)

Dấu "=" xảy ra khi: `{(a-b=0),(a-1=0),(b-1=0):}`

`->a=b=1`

Vậy: `...`

`a)(n+2)\vdots(n+1)`

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

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

`->n+1\in Ư(1)={1;-1}`

`->n\in{0;-2}`

`b)(n+3)\vdots(n+2)`

`->(n+2+1)\vdots(n+2)`

`->1\vdots(n+2)`

`->(n+2)\in Ư(1)={1;-1}`

`->n\in{-1;-3}`

`c)n\vdots(n+1)`

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

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

`->n+1\in Ư(-1)={1;-1}`

`->n\in{0;-2}`

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

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

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

`->n+3\in Ư(-4)={1;-1;2;-2;4;-4}`

`->n\in{-2;-4;-1;-5;1;-7}`

Ta có:

`(a^2+b^2+c^2)/3>=((a+b+c)/3)^2`

`(a^2+b^2+c^2)/3>=(a+b+c)^2/9`

`3(a^2+b^2+c^2)>=(a+b+c)^2`

`3(a^2+b^2+c^2)>=a^2+b^2+c^2+2ab+2bc+2ca`

`3(a^2+b^2+c^2)-(a^2+b^2+c^2)>=2ab+2bc+2ca`

`2(a^2+b^2+c^2)>=2ab+2bc+2ca`

`2a^2+2b^2+2c^2-2ab-2bc-2ca>=0`

`(a^2-2ab+b^2)+(a^2-2ca+c^2)+(b^2-2bc+c^2)>=0`

`(a-b)^2+(a-c)^2+(b-c)^2>=0`

`(a-b)^2>=0`

`(a-c)^2>=0`

`(b-c)^2>=0`

Suy ra: `(a-b)^2+(b-c)^2+(c-a)^2>=0` (đúng)

Dấu "=" xảy ra: `{(a-b=0),(b-c=0),(c-a=0):}`

`->a=b=c`

`a)-2/5+x=-1/3+3/4`

`-2/5+x=5/12`

`x=2/5+5/12`

`x=1/6`

`b)-2 2/3+5/3:x=1/3-0,75`

`-8/3+5/3:x=1/3-3/4`

`-8/3+5/3:x=-5/12`

`5/3:x=8/3-5/12`

`5/3:x=9/4`

`x=5/3:9/4`

`x=20/27`

`c)(2x+(-2)/3)(x-0,3)=0`

`2x+(-2)/3=0` hoặc `x-0,3=0`

`2x=2/3` hoặc `x=0,3`

`x=2/3:2` hoặc `x=0,3`

`x=1/3` hoặc `x=0,3`

`d)(x+2/3)^2=1/36`

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

`TH1:x+2/3=1/6`

`x=1/6-2/3`

`x=-1/2`

`TH2:x+2/3=-1/6`

`x=-1/6-2/3=-6/6`

`e)(2x-3/2)^3=27/8`

`(2x-3/2)^3=(3/2)^3`

`2x-3/2=3/2`

`2x=3/2+3/2`

`2x=3`

`x=3/2`

`(\sqrt{x}-(3\sqrt{x})/(1-\sqrt{x}))*((\sqrt{x}-1)/(x\sqrt{x}+4x+4\sqrt{x}))(x>0,x\ne1)`

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

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

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

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

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

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

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