\(a,2\left(x-3\right)< x+7\\ \Leftrightarrow2x-6-x-7< 0\\ \Leftrightarrow x-13< 0\\ \Leftrightarrow x< 13\\ b,x-\dfrac{x+2}{6}\ge\dfrac{x-1}{3}+5+\dfrac{2x}{5}\\ \Leftrightarrow30x-5\left(x+2\right)\ge10\left(x-1\right)+30.5+6.2x\\ \Leftrightarrow30x-5x-10\ge10x-10+150+12x\\ \Leftrightarrow30x-5x-10-10x+10-150-12x\ge0\\ \Leftrightarrow3x-150\ge0\\ \Leftrightarrow3x\ge150\\ \Leftrightarrow x\ge50\)

\(a,\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =\left(6x^2+14x+9x+21\right)-\left(6x^2-10x+33x-55\right)\\ =\left(6x^2+23x+21\right)-\left(6x^2+23x-55\right)\\ =6x^2+23x+21-6x^2-23x+55\\ =76\)

Vậy ...

\(b,\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)\\ =3x^4-2x^3+x^2+6x^3-4x^2+2x+9x^2-6x+3-4x^3+4x-3x^4-6x^2\\ =3\)

Vậy ...

`a, 2/3 xx x = 2/7`

`x = 2/7 : 2/3`

`x= 2/7 xx 3/2`

`x=3/7`

`b, 2/5 : x= 3/7`

`x= 2/5 : 3/7`

`x= 2/5 xx 7/3`

`x=14/15`

`ĐKXĐ: x ne +- 2`

\(\dfrac{x-2}{x+2}+\dfrac{4x}{x^2-4}=\dfrac{x+2}{x-2}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{4x}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}+\dfrac{4x}{\left(x+2\right)\left(x-2\right)}-\dfrac{x^2+4x+4}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+4+4x-x^2-4x-4}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{-4x}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow-4x=0\\ \Leftrightarrow x=0\)

Vậy pt có tập nghiệm `S={0}`

`a, 5= sqrt25`

Mà `sqrt25 < sqrt27 => 5 < sqrt27`

`b, sqrt{2+1}=sqrt{3}`

Vì `3>sqrt{3}=> 3>sqrt{2+1}`

`c, sqrt{15-1}=sqrt{14}`

`3=sqrt{9}`

Vì `sqrt{9} < sqrt{14} => 3<sqrt{15-1}`

`d, 2+sqrt{6} < 2 +sqrt{9} = 2+3=5`

Gọi độ dài `2` cạnh góc vuông lần lượt là `a,b(a,b>0)`

Theo Bài ra ta có:
`{(x^2 +y^2=13^2),(xy:2=30):}`

`<=>{((x+y)^2 -2xy=169),(xy=60):}`

`<=>{((x+y)^2 -2.60=169),(xy=60):}`

`<=>{((x+y)^2 -120=169),(xy=60):}`

`<=>{((x+y)^2 =289),(xy=60):}`

`<=>{(x+y=17),(xy=60):}`

`<=>{(x=17-y),((17-y)y-60=0):}`

`<=>{(x=17-y),((17y-y^2-60=0):}`

`<=>{(x=17-y),((y^2 -17y+60=0):}`

`<=>{(x=17-y),((y^2 -17y+60=0):}`

​\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=17-5\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=17-12\\y=12\end{matrix}\right.\end{matrix}\right.\)​

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=5\\y=12\end{matrix}\right.\end{matrix}\right.\)

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4+5x^3-2x^2+x-7x^3-35x^2+14x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x-2-x^3-11x\\ =-7x^2-2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-12x-3x+9x^2\right)-\left(3x^2-12x+5x-20\right)\\ =4x-12x^2-3x+9x^3-3x^2+12x-5x+20\\ =9x^3-15x^2+8x+20\)

Ta có:

`8=7+1=x+1`

`B=x^15 -8x^14 + 8x^13 + ... +8x-5`

`=x^15 - (x+1)x^14 + (x+1)x^13 + ...+(x+1)x-5`

`=x^15 - x^15 -x^14 + x^14+ x^13 + ...+x^2+x-5`

`=x-5`

`=7-5`

`=2`