a) x – 258 = 347

               x = 347 + 258

               x = 605

b) x × 9 = 819

           x = 819 : 9

           x = 91

co ai biet ko

Mng oi câu b 1202⁰ nha mng

Bài 1: 

a, 58.32 + 58.68 - 800

= 58.(32 + 68) - 800

= 58.100 - 800

= 5800 - 800

= 5000

b, 12020 + 280 : [55 - (7 - 4)³]

  =  12020 + 280 : [ 55 - 3³ ]

  =  12020 + 280 : [ 28]

  = 12020 + 10

  = 12030 

c, (96 - 19 - 45) - (55 + 96 - 119)

  = 96 - 19 - 45 - 55 - 96 + 119

 = (96 - 96) + (119 - 19) - (45 + 55)

= 0 + 100 - 100

= 0 

Bài 2:

a,7\(x\) - 70  = 105

   7\(x\)         = 105 + 70

   7\(x\)         = 175

     \(x\)          = 175 : 7

    \(x\)           = 25

2:

a: x+201=351

=>x=351-201

=>x=150

b: \(8\cdot5^2-27:25\)

\(=8\cdot25-\dfrac{27}{25}\)

\(=200-1,08=198,92\)

d: \(2023-23:\left[9+2\left(2^3-0,21\right)\right]\)

\(=2023-23:\left[9+16-0,42\right]\)
\(=2023-\dfrac{23}{25-0,42}\)

\(=2023-\dfrac{1150}{1229}=\dfrac{2485117}{1229}\)

b: 2(x-21)=84

=>x-21=84/2=42

=>x=42+21=63

c: 135-4(81-x)=55

=>4(81-x)=135-55=80

=>81-x=20

=>x=61

( 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

a: \(A=\left|3x-9\right|+1.5\ge1.5\forall x\)

Dấu '=' xảy ra khi x=3

b: \(B=\left|x-7\right|-14\ge-14\forall x\)

Dấu '=' xảy ra khi x=7

a, Ta có \(A=\left|3x-9\right|+1,5\)

Ta thấy: \(\left|3x-9\right|\ge0\Rightarrow\left|3x-9\right|+1,5\ge1,5\Rightarrow A\ge1,5\)

Dấy"=" xảy ra \(\Leftrightarrow3x-9=0\Leftrightarrow3x=9\Leftrightarrow x=3\)

Vậy \(A_{min}=1,5\Leftrightarrow x=0\)

b, Ta có \(B=\left|x-7\right|-14\)

Ta thấy: \(\left|x-7\right|\ge0\Rightarrow\left|x-7\right|-14\ge-14\Rightarrow B\ge-14\)

Dấu "=" xảy ra \(\Leftrightarrow x-7=0\Leftrightarrow x=7\)

Vậy \(B_{min}=-14\Leftrightarrow x=7\)

c, Ta có: \(C=-\left|\dfrac{1}{2}x-4\right|+13\Rightarrow C=13-\left|\dfrac{1}{2}x-4\right|\)

Ta thấy: \(\left|\dfrac{1}{2}x-4\right|\ge0\Rightarrow13-\left|\dfrac{1}{2}x-4\right|\le13\Rightarrow C\le13\)

Dấu "=" xảy ra \(\Leftrightarrow\dfrac{1}{2}x-4=0\Leftrightarrow\dfrac{1}{2}x=4\Leftrightarrow x=8\)

Vậy \(C_{max}=13\Leftrightarrow x=8\)

d, Ta có: \(D=-\left|1,5-x\right|-14\Rightarrow D=-14-\left|1,5-x\right|\)

Ta thấy: \(\left|1,5-x\right|\ge0\Rightarrow-14-\left|1,5-x\right|\le-14\Rightarrow D\le-14\)

Dấu "=" xảy ra\(\Leftrightarrow1,5-x=0\Rightarrow x=1,5\)

Vậy \(D_{max}=-14\Leftrightarrow x=1,5\)

Hoctot

a,x/9=5/3 = x=15

b,17/x=85/105 = x = 21

c,6/8=15/x = x =  20

d,x/8=2/3 = x = 6

 e,3/x-7=27/135 =x= 22

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: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)

\(\frac{7.x}{8}-5.\left(x-9\right)=20.x+\frac{1,5}{6}\)

\(\Leftrightarrow7.x-5.\left(x-9\right)=20x.8+\frac{1,5}{6}.8\)

\(\Leftrightarrow7.x-40\left(x-9\right)=160x+2\)

\(\Leftrightarrow7x-40+360=160x+2\)

\(\Leftrightarrow-33x+360=160x+2\)

\(\Leftrightarrow-33x=160x+x-360\)

\(\Leftrightarrow-33x=160x-358\)

\(\Leftrightarrow-33x-160x=160x-358\)

\(\Leftrightarrow-193x=-358\)

\(\Rightarrow x=\frac{358}{193}\)

Khong chac nhe

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

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