Mỗi câu đúng được 0,75 điểm

a) x × 7 + 85 = 106

             x × 7 = 106 – 85

             x × 7 = 21

                   x = 21 : 7

                   x = 3

b) x : 9 + 72 = 137

              x : 9 = 137 - 72

              x : 9 = 65

                   x = 65 x 9

                   x = 585

con cai nit

a: Thay x=49 vào A, ta được:

\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)

b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)

P>1 khi P-1>0

=>\(\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}>0\)

=>\(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)

\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

a, 5\(x\)(\(x^2\) - 9) = 0

    \(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

    3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0

    (\(x+3\))( 3 - \(x\)) = 0

     \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\){ -1; 10}

a) 5x(x2-9)=0

=> TH1 5x=0

      <=> x= 0

     TH2: 2x-9=0

       <=> 2x=9

       <=> x = \(\dfrac{9}{2}\)

b, 3(x+3) - x²- 3x = 0   

<=> 3x + 9 - x² -3x = 0

<=> - x² +9 = 0

<=> - x² = -9

 <=> x = 3

c, x² -9x -10 = 0

<=> x² -x + 10x -10 = 0

<=> x(x-1)+10(x-1)=0     

 <=> (x-1)(x+10)=0

=> TH1: x-1=0

       <=> x=1

      TH2: x +10=0

      <=>  x=-10

câu a là 34,5 :x = 1,5 x 2 hả bn

bạn chép lại đi bị lệnh chữ mình không làm dc

( 18 - 9x ) x 7 = 819 - 756

( 18 - 9x ) x 7 = 63

  18 - 9x         = 63 : 7

  18 - 9x         = 9

         9x         = 18 - 9

         9x         = 9

           x         = 9 : 9

           x         = 1

( 18 - 9\(x\) ) x 7 = 819 - 756

=>7x18 - 7x 9\(x\)= 63

=>126 - 63\(x\)= 63

=>63\(x\)= 126 - 63

=>63\(x\)= 63

=>\(x\)= 36 : 36 = 1

OK 

(18-9)*7=63

9*7=63

Mỗi phần đúng được (0,75đ)

a) 76 + x = 257

             x = 257 – 76

             x = 181

b) 672 : x = 6

              x = 672 : 6

              x = 112

\(a,\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ b,\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1,5+\dfrac{-3}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\-\dfrac{3}{x}=-1,5=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)

a: \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)=0\)

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

b: \(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{5}+\left(-3\right):x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\left(-3\right):x=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{9}{16}\cdot\dfrac{4}{3}=\dfrac{3}{4}\\x=\left(-3\right):\dfrac{-1}{5}=15\end{matrix}\right.\)