kho qua

\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)

\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)

\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)

Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:

\(E=5x.0+105=105\)

trả lời 

5^x-3 -20 = 105

5^x-3  = 105 + 20

5^x-3 = 125

5^x-3 = 5³

x-3 = 3

x =3+3

x = 6

Vậy x = 6

Ta có: \(\frac{5x}{2}=\frac{6y}{5}=\frac{7z}{3}\) => \(\frac{x}{\frac{2}{5}}=\frac{y}{\frac{5}{6}}=\frac{z}{\frac{3}{7}}\)

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

 \(\frac{x}{\frac{2}{5}}=\frac{y}{\frac{5}{6}}=\frac{z}{\frac{3}{7}}=\frac{y-x-z}{\frac{5}{6}-\frac{2}{5}-\frac{3}{7}}=\frac{\frac{1}{105}}{\frac{1}{210}}=2\)

=> \(\hept{\begin{cases}\frac{x}{\frac{2}{5}}=2\\\frac{y}{\frac{5}{6}}=2\\\frac{z}{\frac{3}{7}}=2\end{cases}}\) => \(\hept{\begin{cases}x=\frac{4}{5}\\y=\frac{5}{3}\\z=\frac{6}{7}\end{cases}}\)

1. xy + 5x + 5y = 92

=> (xy + 5x) + (5y + 25) = 92 + 25

=> x(y + 5) + 5(y + 5) = 117

=> (x + 5)(y + 5) = 117

=> x + 5 \(\in\)Ư(117) = {-1;1;-3;3;-9;9;-13;13;-39;39;-117;117}

Mà x >= 0 => x + 5 >= 5

=> x + 5 \(\in\){9;13;39;117}

Ta có bảng sau:

Vậy; (x;y) \(\in\){(4;8);(8;4)}

khó quá ak! Nhìn rối cả mắt.

các câu còn lại tương tự như bài mình vừa làm

1) 2(3x+5)-6=9 

  2(3x+5)    =9+6

  2(3x+5)    =15

    3x+5     = 15:2

   3x+5    = \(\frac{15}{2}\)

  3x      = \(\frac{15}{2}-5\)

  3x        =\(\frac{5}{2}\)

   x        = \(\frac{5}{2}:3\)

   x       = \(\frac{5}{6}\)

2) 5x+3(4+2x)=25

   5x+12+6x=25

   5x+6x      =-12+25

   11x          =13

      x           =13:11

      x           =\(\frac{13}{11}\)

3) 3(4x+1)+2(x-1)=105

   12x+3+2x-2=105

   12x+2x      = -3+2+105

    14x           =104

       x            = 104:14

       x            = \(\frac{52}{7}\)

4) \(30-[2\left(x-3\right)-2]=14\)

                 \(2\left(x-3\right)-2\)=\(30-14\)

                 \(2\left(x-3\right)-2=16\)

                 \(2\left(x-3\right)=16+2\)

                  \(2\left(x-3\right)=18\)

                  \(x-3=18:2\)

                  \(x-3=9\)

                  \(x=9+3\)

                  \(x=12\)

     k mình nha !!!!!!

\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)

\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)

\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)

\(\Rightarrow x\in\left\{3;6;2\right\}\)

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

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

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

\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-4\)

\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Leftrightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)

\(\Rightarrow200-x=0\)

\(\Leftrightarrow x=200\)

Vậy....