Câu 2:

\(A\left(x\right)=x^2+3x+1\)

\(B\left(x\right)=2x^2-2x-3\)

a) Tính A(x) là sao em?

b) \(A\left(x\right)+B\left(x\right)=\left(x^2+3x+1\right)+\left(2x^2-2x-3\right)\)

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

\(=\left(x^2+2x^2\right)+\left(3x-2x\right)+\left(1-3\right)\)

\(=3x^2+x-2\)

Câu 1:

\(M\left(x\right)=x^3+3x-2x-x^3+2\)

\(=\left(x^3-x^3\right)+\left(3x-2x\right)+2\)

\(=x+2\)

Bậc của M(x) là 1

Câu 3

a) \(3x\left(x^2-x+1\right)\)

\(=3x.x^2-3x.x+3x.1\)

\(=3x^3-3x^2+3x\)

b) \(\left(2x^2+x-1\right):\left(x+1\right)\)

\(=\left(2x^2+2x-x-1\right):\left(x+1\right)\)

\(=\left[\left(2x^2+2x\right)-\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left[2x\left(x+1\right)-\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left(x+1\right)\left(2x-1\right):\left(x+1\right)\)

\(=2x-1\)

`3b)\sqrt{25x-25}-15/2\sqrt{(x-2)/9}=6+3/2\sqrt{x-1}`

ĐK:`x>=1`

`pt<=>sqrt{25(x-1)}-15/2*1/3sqrt{x-1}-3/2sqrt{x-1}=6`

`<=>5sqrt{x-1}-5/2sqrt{x-1}-3/2sqrt{x-1}=6`

`<=>5sqrt{x-1}-4sqrt{x-1}=6`

`<=>sqrt{x-1}=6`

`<=>x-1=36`

`<=>x=37(tmddk)`

Vậy `S={37}`

3b) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\dfrac{3}{2}\sqrt{x-1}\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\sqrt{\dfrac{1}{9}.\left(x-1\right)}-\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}.\dfrac{1}{3}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow\sqrt{x-1}=6\Rightarrow x-1=36\Rightarrow x=37\)

Bài 3: 

c) ĐKXĐ: \(x\ge0\)

Ta có: \(\sqrt{4+5\sqrt{x}}=3\)

\(\Leftrightarrow5\sqrt{x}+4=9\)

\(\Leftrightarrow5\sqrt{x}=5\)

\(\Leftrightarrow x=1\)

Vậy: S={1}

d) Ta có: \(x-\sqrt{4x-20}=20\)

\(\Leftrightarrow\sqrt{4x-20}=x-20\)

\(\Leftrightarrow\left(x-20\right)^2=4x-20\)

\(\Leftrightarrow x^2-40x+400-4x+20=0\)

\(\Leftrightarrow x^2-44x+420=0\)

\(\Delta=\left(-44\right)^2-4\cdot1\cdot420=256\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{44-16}{2}=\dfrac{28}{2}=14\\x_2=\dfrac{44+16}{2}=\dfrac{60}{2}=30\end{matrix}\right.\)

Vậy: S={14;30}

\(a,=\dfrac{2}{3}\cdot\dfrac{21}{4}+\dfrac{5}{8}=\dfrac{7}{2}+\dfrac{5}{8}=\dfrac{33}{8}\\ b,=\left(\dfrac{1}{27}\cdot27\right)^{2020}\cdot27=1^{2020}\cdot27=27\\ c,=\dfrac{2^{30}\cdot2^{19}}{2^{48}}=2\)

Câu 1: 2y + 8/3 =y +3/2

            2y-y= 3/2 - 8/3

                y =-7/6

 Câu 2: 12/6=2 

Suy ra  a/5=b/3=c/17=d/9=2

  vậy a=10,b=6,c=34,d=18