đề câu c thiếu rồi

A,x thuộc{-3,-2,-1,0,1,2,3,4}

B,x thuộc{-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}

câu c mk chịu

du sr ae nha cai cho trong la 6 nha

mình chỉ có thể giúp bạn hai câu đầu thôi thông cảm nha

( ko có dấu gạch nên ghi là "gạch " ) 

gạch x + 1 gạch = 2^3 - 1

gạch x + 1 gạch = 8 - 1

gạch x + 1 gạch = 7

x + 1 = 7 hoặc x + 1 = - 7 

 x = 7 - 1 hoặc x = ( -7 ) - 1

 x = 6 hoặc x = - 8

Vậy  x = 6 hoặc x = - 8

gạch 2 - x gạch = 5

2 - x = 5 hoặc 2 - x = - 5

x = 2 - 5 hoặc x = 2 -  ( -5 ) 

x = ( -3 ) hoặc x = 7

Vậy x = ( -3 ) hoặc x = 7

\(\frac{3}{4}+\frac{1}{4}\cdot x+x-\frac{7}{6}\cdot x=\frac{5}{12}\)

\(\frac{3}{4}+\frac{1}{4}\cdot x-\frac{7}{6}\cdot x+x\cdot1=\frac{5}{12}\)

\(\frac{3}{4}+x\left(\frac{1}{4}-\frac{7}{6}+1\right)=\frac{5}{12}\)

\(\frac{3}{4}+x\cdot\frac{1}{12}=\frac{5}{12}\)

\(x\cdot\frac{1}{12}=\frac{5}{12}-\frac{3}{4}\)

\(x\cdot\frac{1}{12}=\frac{5}{12}-\frac{9}{12}\)

\(x\cdot\frac{1}{12}=\frac{-1}{3}\)

\(x=\frac{-1}{3}\text{ : }\frac{1}{12}\)

\(x=\frac{-1}{3}\cdot12\)

\(x=\frac{-12}{3}\)

\(x=-4\)

\(\text{b, }0,25\cdot x-\frac{2}{3}\cdot x=-1\frac{1}{6}\)

\(\frac{1}{4}\cdot x-\frac{2}{3}\cdot x=\frac{-7}{6}\)

\(x\cdot\left(\frac{1}{4}-\frac{2}{3}\right)=\frac{-7}{6}\)

\(x\cdot\frac{-5}{12}=\frac{-7}{6}\)

\(x=\frac{-7}{6}\text{ : }\frac{-5}{12}\)

\(x=\frac{-7}{6}\cdot\frac{12}{-5}\)

\(x=\frac{-14}{-5}\)

x - 30 : 6 = 8 x 9 

x - 5 = 72

    x  = 72 + 5 

   x   = 77

Vậy x = 77

\(\frac{4}{x+2}+\frac{-3}{x-2}+\frac{12}{x^2-4}.\)

\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-4}{x^2-4}\)

\(\frac{4}{x+2}+\frac{\left(-2\right)}{x-2}+\frac{12}{x^2-4}\)

\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{1}{x+2}\)

Đặt \(g\left(x\right)=f\left(x\right)+h\left(x\right)\left(1\right)\)trong đó \(h\left(x\right)=ax^2+bx+c\left(2\right)\)

Tìm \(a,b,c\)sao cho \(g\left(1\right)=g\left(2\right)=g\left(3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}g\left(1\right)=f\left(1\right)+h\left(1\right)=0\\g\left(2\right)=f\left(2\right)+h\left(2\right)=0\\g\left(3\right)=f\left(3\right)+h\left(3\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}h\left(1\right)=-5\\h\left(2\right)=-11\\h\left(3\right)=-21\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b+c=-5\\4a+2b+c=-11\\9a+3b+c=-21\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a+b+c=-5\\3a+b=-6\\5a+b=-10\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-2\\b=0\\c=-3\end{cases}}\)Thay vào (2) ta được:

\(h\left(x\right)=4x-3\)

Vì \(g\left(1\right)=g\left(2\right)=g\left(3\right)=0\)mà g(x)  bậc 4 có hệ số cao nhất là 1 nên ta có 

 \(g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-x_0\right)\)

Từ \(\left(1\right)\Rightarrow f\left(x\right)=g\left(x\right)-h\left(x\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-x_0\right)+4x-3\)

\(f\left(-1\right)=\left(-1-1\right)\left(-1-2\right)\left(-1-3\right)\left(-1-x_0\right)+4.\left(-1\right)-3\)

\(=-24\left(-1-x_0\right)-7\)

\(f\left(5\right)=\left(5-1\right)\left(5-2\right)\left(5-3\right)\left(5-x_0\right)+4.5-3\)

\(=24\left(5-x_0\right)+17\)

\(\Rightarrow f\left(-1\right)+f\left(5\right)\)\(=-24\left(-1-x_0\right)-7+24\left(5-x_0\right)+17\)

                                            \(=24+24x_0+120-24x_0+10\)

                                             \(=154\)

Xl nha bài sai sót vì thay a,b,c sai r xl 

\(x^5-x^4+3x^3+3x^2-x+1=0\)

\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^4-2x^3+5x^2-2x+1=0\left(#\right)\end{cases}}\)

\(\Leftrightarrow x=-1\)(vì biểu thức # vô nghiệm) (cái này bạn tự cm)

vậy....