1: \(A=\dfrac{\left(x+1\right)^3}{\left(x+1\right)^2}=x+1\)

\(B=\dfrac{\left(x+1\right)\cdot\left(x^2-x+1\right)}{x+1}=x^2-x+1\)

2: A=B

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

=>x^2-2x=0

=>x=0 hoặc x=2

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`1)\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}`

`2)`

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

`<=>\sqrt(x-2)^2}=1`

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

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

`b)\sqrt{x^2-3x}-\sqrt{x-3}=0`              `ĐK: x >= 3`

`<=>\sqrt{x}\sqrt{x-3}-\sqrt{x-3}=0`

`<=>\sqrt{x-3}(\sqrt{x}-1)=0`

`<=>[(\sqrt{x-3}=0),(\sqrt{x}-1=0):}`

`<=>[(x-3=0),(\sqrt{x}=1):}<=>[(x=3(t//m)),(x=1(ko t//m)):}`

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{7}{2}\end{matrix}\right.\)

\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{37}{4}\)

\(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\dfrac{153}{8}\)

\(C=x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2=\dfrac{977}{16}\)

\(D=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\dfrac{\sqrt{65}}{2}\)

\(E=\left(2x_1+x_2\right)\left(2x_2+x_1\right)=2\left(x_1^2+x_2^2\right)+5x_1x_2=1\)

`a,` Đthang đi qua `A(3, 12)`.

`-> x = 3, y = 12 in y`.

`<=> 12 = 9a.`

`<=> a = 12/9 = 4/3.`

`b,` Đthang đi qua `B(-2;3)`.

`=> x = -2, y = 3 in y`.

`<=> 3=4a`.

`<=> a = 3/4`.

`3x^2+10x+3=0`

Ptr có: `\Delta'=5^2-3.3=16 > 0`

   `=>` Ptr có `2` nghiệm pb

 `=>` Áp dụng Viét có: `{(x_1+x_2=[-b]/a=-10/3),(x_1 .x_2=c/a=1):}`

~~~~~~~~~~~~~

`A=x_1 ^2+x_2 ^2`

`A=(x_1+x_2)^2-2x_1 .x_2`

`A=(-10/3)^2-2.1=82/9`

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`B=x_1 ^3+x_2 ^3`

`B=(x_1+x_2)(x_1 ^2-x_1 .x_2+x_2 ^2)`

`B=(x_1+x_2)[(x_1+x_2)^2 -3x_1 .x_2]`

`B=(-10/3).[(-10/3)^2-3.1]=-730/27`

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`C=x_1 ^4+x_2 ^4`

`C=(x_1 ^2+x_2 ^2)^2 -2x_1 ^2 .x_2 ^2`

`C=[(x_1+x_2)^2-2x_1 .x_2]^2-2(x_1 .x_2)^2`

`C=[(-10/3)^2-2.1]^2-2. 1^2=6562/81`

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`D=|x_1-x_2|`

`D=\sqrt{(x_1-x_2)^2}`

`D=\sqrt{(x_1+x_2)^2-4x_1.x_2}`

`D=\sqrt{(-10/3)^2-4.1}=8/3`

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`E=(2x_1+x_2)(2x_2+x_1)`

`E=4x_1 .x_2+2x_1 ^2+2x_2 ^2+x_1 .x_2`

`E=5x_1 . x_2+2(x_1+x_2)^2-4x_1 .x_2`

`E=x_1 .x_2+2(x_1+x_2)^2`

`E=1+2(-10/3)^2=209/9`

`a)` Thay `x=-3` vào ptr có:

   `(-3)^2-6.(-3)+2m+1=0`

`<=>9+18+2m+1=0`

`<=>m=-14`

`b)` Ptr có: `\Delta'=(-3)^2-(2m+1)=9-2m-1=8-2m`

Ptr có `2` nghiệm phân biệt `<=>\Delta' > 0`

    `<=>8-2m > 0<=>m < 4`

`c)` Ptr có nghiệm kép `<=>\Delta' =0`

           `<=>8-2m=0<=>m=4`

\(A=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}>=1>0\)

=>A>|A|

Ta có: A= \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)= \(1+\dfrac{1}{\sqrt{x}+1}\)

Vì x ≥0⇒\(\sqrt{x}\) ≥0⇒\(\sqrt{x}+1 \)≥ 1 ⇒ \(1+\dfrac{1}{\sqrt{x}+1}\)≥ 2

hay A≥ 2>0

Khi đó ta có: A=|A|

Vậy A=|A|