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

          (2 x - 3) - (x + 2) = ( x - 2)-3(x - 5)

\(\Leftrightarrow\)2x - 3 - x - 2 = x - 2 - 3x + 15

\(\Leftrightarrow\)x - 5 = 13 - 2x

\(\Leftrightarrow\)3x = 18

\(\Leftrightarrow\)x = 6

  Vậy x = 6 là giá trị cần tìm

\(=\)\(\dfrac{13}{x-1}+\dfrac{5}{2(x-1)}-\dfrac{2}{(x-1)}\)\(\)

\(=\)\(\dfrac{26+5-4}{2(x-1)}\)

\(=\)\(\dfrac{27}{2(x-1)}\)

cảm ơn bn nhìu nha

Ta có: 2x + 3y + 5z - 119 = 0

=>  2x + 3y + 5z = 119

 \(\frac{x+2}{3}=\frac{y+3}{5}=\frac{z-4}{7}\Leftrightarrow\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}=\frac{2x+4+3y+9+5z-20}{6+15+35}=\frac{119+4+9-20}{56}=\frac{112}{56}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x+2}{3}=2\\\frac{y+3}{5}=2\\\frac{z-4}{7}=2\end{cases}\Rightarrow}\hept{\begin{cases}x+2=6\\y+3=10\\z-4=14\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=7\\z=18\end{cases}}\)

Vậy...

Bài 1:

a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(3-2x\right)^2=\left(x-2\right)^2\\x< =\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3-x+2\right)\left(2x-3+x-2\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\)

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

b: \(\left|x\right|< 3\)

nên -3<x<3

c: \(\left|x\right|\ge5\)

nên \(\left[{}\begin{matrix}x\ge5\\x\le-5\end{matrix}\right.\)

Bài 2: 

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=7\end{matrix}\right.\)

c) \(2x=3y=5z\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng tính chát dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒\(\left\{{}\begin{matrix}x=5.15=75\\y=5.10=50\\z=5.6=30\end{matrix}\right.\)

Ta có :

\(\left(2x^2-3x+1\right)-\left(2x^2-3x+4\right)=0\)

\(\Leftrightarrow2x^2-3x+1-2x^2+3x-4=0\)

\(\Leftrightarrow-3=0\left(ktm\right)\)

\(\Leftrightarrow x\in\varnothing\)

a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)

\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)

\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)

\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)

\(x=\dfrac{-1}{3}\)

b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)

\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)

\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)

\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)

\(2x=\dfrac{-17}{24}\)

\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)

\(x=\dfrac{-17}{48}\)

c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)

a, 1/4 + 3/4 : x = -2

     3/4 : x = -2 - 1/4 

     3/4 : x = -9/4

             x = 3/4 : -9/4

             x = -1/3