a)2x²-7x-2x²-2x=7

-9x=7

x=-7/9

b)3x²+24x-x²-2x²-2x=2

22x=2

x=1/11

c: \(3x\left(x-7\right)-2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

d: \(7x^2-28=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(2x^3+5x^2-3x=0\Leftrightarrow x\left(2x^2+5x-3\right)=0\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(\dfrac{2x+1}{x-1}< 1\)

\(\dfrac{2x+1}{x-1}-1< 0\)

\(\dfrac{2x+1-x+1}{x-1}< 0\)

\(\dfrac{x+2}{x-1}< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2< 0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2>0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -2\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x>-2\\x< 1\end{matrix}\right.\end{matrix}\right.\) 

Tự kết luận nha

ng ta dg hoc ma vao anh

b: \(\Leftrightarrow8\left(1-3x\right)-2\left(3x+2\right)=140-15\left(2x+1\right)\)

=>8-24x-6x-4=140-30x-15

=>-30x+30x=115-4=111

=>0x=111(vô lý)

Bài 5:

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

=>(6x-1)(x-3)=(2x+5)(3x+2)

=>6x^2-18x-x+3=6x^2+4x+15x+10

=>-19x+3=19x+10

=>-38x=7

=>x=-7/38

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

chụp rõ hơn một xíu đc ko bạn

nhỏ quá bn ạ

a) (a+b)^3 +15a

Bài 3: 

Xét ΔIAB có 

\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)

\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)

hay \(\widehat{DAB}+\widehat{ABC}=230^0\)

Xét tứ giác ABCD có 

\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)

\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)

mà \(\widehat{C}-\widehat{D}=10^0\)

nên \(2\cdot\widehat{C}=160^0\)

\(\Leftrightarrow\widehat{C}=80^0\)

\(\Leftrightarrow\widehat{D}=70^0\)