1: \(3x^2+mx+n\)

\(=3x^2+15x+\left(m-15\right)x+5m-75-5m+75+n\)

\(3x^2+mx+n\) chia x+5 dư 27

=>-5m+75+n=27

=>-5m+n=27-75=-48

2: \(x^3+mx+n\)

\(=x^3+x^2-x^2-x+\left(m+1\right)x+m+1+n-m-1\)

\(=\left(x+1\right)\left(x^2-x+m+1\right)+n-m-1\)

\(x^3+mx+n\) chia x+1 dư 7

=>n-m-1=7

=>n=m+1+7=m+8

\(x^3+mx+n\)

\(=x^3-3x^2+3x^2-9x+\left(m+9\right)x+3m+27+n-3m-27\)

\(=\left(x-3\right)\left(x^2+3x+m+9\right)+n-3m-27\)

\(x^3+mx+n\) chia x-3 dư 5

=>n-3m-27=5

=>n=3m+27+5=3m+32

mà n=m+8

nên 3m+32=m+8

=>2m=-24

=>m=-12

n=m+8=-12+8=-4

Đáp án A

Đáp án D

a ) 9 . 3x = 81

          3x = 9

            x = 3

b ) 2x : 4 = 1

     2x      = 4

       x      = 2

c ) 2x - 64 = 2

     2x        = 66

       x        = 33

d ) 2x = 16

       x  = 8

e ) 3 ^ 2 . 3 ^ 4 . 3x = 3 ^ 10

     3 ^ ( 2 + 4 + x )  = 3 ^ 10

=> 2 + 4 + x = 10

                 x = 4

f ) 2x + 4 . 2x = 5 . 2 ^ 5

    2x + 8 x     = 160

    10x            = 160

        x            = 16

\(a.9.3x=81\)

\(3x=81:9\)

\(3x=9\)

\(x=9:3\)

\(x=3\)

Đáp án B

Bài 3:

1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)

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

Vậy.......................

2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)

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

Vậy........................

3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)

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

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

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

Vậy............................

4, 5 tương tự nhé bn!

bài 3

1 (x-1)(x+2)+5x-5=0

=>(x-1)(x+2)+(5x-5)=o

=>(x-1)(x+2)+5(x-1)=0

=>(x-1)(x+2+5)=0

=>(x-1)(x+7)=0

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

vậy x=1 hoặc x=-7

2. (3x+5)(x-3)-6x-10=0

=>(3x+5)(x-3)-(6x+10)=0

=>(3x+5)(x-3)-2(3x+5)=0

=>(3x+5)(x-3-2)=0

=>(3x+5)(x-5)=0

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

\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}\)\(=3\)

\(\Leftrightarrow\frac{x-152}{2011}+\frac{x-2135}{28}+\frac{x-2039}{124}=3\)\(\Leftrightarrow\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=3-1-1-1\)

\(\Leftrightarrow\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

\(\Leftrightarrow\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

        Vì \(\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)>0\)

  \(\Rightarrow x-2163=0\)

\(\Leftrightarrow x=2163\)

          Vậy \(x=2163\)

a) 86 - x < 86 - 4

     86 - x < 82

             x < 86 - 82

             x < 4

b) hình như đề bài sai hay sao á

không phải sai đau