`2/7` của `-42` là:
`-42 . 2/7 = -12`

ĐKXĐ: `2x^2 +x-2 >= 0`

\(\sqrt{2x^2+x-2}=0\\ \Leftrightarrow\left(\sqrt{2x^2+x-2}\right)^2=0^2\\ \Leftrightarrow2x^2+x-2=0\)

Ta có: 

\(\Delta=1^2-4.2.\left(-2\right)\\ =1+16\\ =17>0\)

​Suy ra phương trình luôn có 2 nghiệm phân biệt​

\(\Rightarrow x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{17}}{2.2}=\dfrac{-1+\sqrt{17}}{4}\\ x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{17}}{2.2}=\dfrac{-1-\sqrt{17}}{4}\)

Thử lại thấy cả 2 giá trị của `x` đều thỏa mãn

Vậy phương trình có tập nghiệm \(S=\left\{\dfrac{-1+\sqrt{17}}{4};\dfrac{-1-\sqrt{17}}{4}\right\}\)

`ĐKXĐ: {(x >= 0),(2x^2 +x -2 >= 0):} `

\(\sqrt{2x^2+x-2}=x\\ \Leftrightarrow2x^2+2x-2=x^2\\ \Leftrightarrow x^2+x-2=0\\ \Leftrightarrow x^2-x+2x-2=0\\ \Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)

Thử lại `x=1` thỏa mãn ĐKXĐ

Vậy pt có tập nghiệm `S={1}`

\(\sqrt{2x^2+x-2}=0\\ \Leftrightarrow\left(\sqrt{2x^2+x-2}\right)^2=0^2\\ \Leftrightarrow2x^2+x-2=0\)

Ta có: 

\(\Delta=1^2-4.2.\left(-2\right)\\ =1+16\\ =17>0\)

Suy ra phương trình luôn có 2 nghiệm phân biệt

\(\Rightarrow x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{17}}{2.2}=\dfrac{-1+\sqrt{17}}{4}\\ x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{17}}{2.2}=\dfrac{-1-\sqrt{17}}{4}\)

Vậy phương trình có tập nghiệm \(S=\left\{\dfrac{-1+\sqrt{17}}{4};\dfrac{-1-\sqrt{17}}{4}\right\}\)

\(\left(9\dfrac{1}{4}+12\dfrac{1}{2}\right)\times2\\ =\left(\dfrac{37}{4}+\dfrac{25}{2}\right)\times2\\ =\left(\dfrac{37}{4}+\dfrac{50}{4}\right)\times2\\ =\dfrac{87}{4}\times2\\ =\dfrac{87}{2}\\ \left(12\dfrac{3}{4}+16\dfrac{1}{2}\right)\times2\\ =\left(\dfrac{51}{4}+\dfrac{33}{2}\right)\times2\\ =\left(\dfrac{51}{4}+\dfrac{66}{4}\right)\times2\\ =\dfrac{117}{4}\times2\\ =\dfrac{117}{2}\)

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

`<=> |x| = 3/2`

`<=> x= ` \(\pm\dfrac{3}{2}\)

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ \Leftrightarrow2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\right)=2.\dfrac{49}{99}\\ \Leftrightarrow\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=1-\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\)

\(\Leftrightarrow2x=98\\ \Leftrightarrow x=49\)

\(M=x^2+11xy-y^2\\ =\left(\dfrac{5}{2}\right)^2+11.\dfrac{5}{2}.\dfrac{-4}{3}-\left(-\dfrac{4}{3}\right)^2\\ =\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}\\ =-\dfrac{1159}{36}\)

`M+(5x^2 -2xy)=6x^2 +9xy -y^2`

`=> M=6x^2+9xy-y^2-(5x^2 -2xy)`

`=> M=6x^2+9xy-y^2-5x^2 +2xy`

`=> M=x^2+11xy-y^2`

Ta có:
\(\left(2x-5\right)^{2018}\ge0\forall x;\left(3y+4\right)^{2020}\ge0\forall y\\ \Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho `(2x-5)^2018 + (3y+4)^2020 <= 0`

`=> `(2x-5)^2018 + (3y+4)^2020 =0`

Dấu "=" xảy ra `<=> {(2x-5=0),(3y+4=0):} <=> {(x= 5/2),(y = -4/3):}`

Thay `x= 5/2 ; y = -4/3` vào `M` ta có:
\(M=6x^2+11xy-y^2\\ =6.\left(\dfrac{5}{2}\right)^2+11.\dfrac{5}{2}.\dfrac{-4}{3}-\left(-\dfrac{4}{3}\right)^2\\ =6.\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}\\ =\dfrac{75}{2}-\dfrac{346}{9}\\ =-\dfrac{17}{18}\)