x : 2,5 = 11,95 - 7,35

x : 2,5 = 4,6 

x         = 4,6 x 2,5 

x         =     11,5

bang 11.5

giup minh voi

b₁)

\(11.95-x:2,5=7.35\)

\(\Rightarrow x:2=11,95-7,35=4,6\)

\(\Rightarrow x=4,6.2=9,2\)

b₂)

\(360:7,2.0,9=360:72.0,9.10=360:72.9=5.9=45\)

Bài 14: 

a: Ta có: \(P=\dfrac{2\sqrt{x}}{\sqrt{x}+m}+\dfrac{\sqrt{x}}{\sqrt{x}-m}-\dfrac{m^2}{4x-4m^2}\)

\(=\dfrac{8\sqrt{x}\left(\sqrt{x}-m\right)+4\sqrt{x}\left(\sqrt{x}+m\right)-m^2}{4\left(\sqrt{x}+m\right)\left(\sqrt{x}-m\right)}\)

\(=\dfrac{8x-8m\sqrt{x}+4x+4m\sqrt{x}-m^2}{4\left(\sqrt{x}+m\right)\left(\sqrt{x}-m\right)}\)

\(=\dfrac{-m^2-4m\sqrt{x}+12x}{4\left(\sqrt{x}+m\right)\left(\sqrt{x}-m\right)}\)

a) -1816

b) - 700

c)  4800

d) 8

ai làm hộ mik đúng và xog trc mik tick

`(3x)/(2.5)+(3x)/(5.8)+(3x)/(8.11)+(3x)/(11.14)=1/21`
`=>x(3(2.5)+3/(5.8)+3/(8.11)+3/(11.14))=1/21`
`=>x(1/2-1/5+1/5-1/8+1/8-1/11-1/14)=1/21`
`=>x*(1/2-1/14)=1/21`
`=>x*3/7=1/21`
`=>x=1/21:3/7=1/9`
Vậy `x=1/9`

Vậy 

\(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+\dfrac{3x}{8.11}+\dfrac{3x}{11.14}=\dfrac{1}{21}\) 

\(x.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\) 

\(x.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\) 

                       \(x.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\) 

                                      \(x.\dfrac{3}{7}=\dfrac{1}{21}\) 

                                          \(x=\dfrac{1}{21}:\dfrac{3}{7}\) 

                                          \(x=\dfrac{1}{9}\)

3x²-75=0
<=> 3x²=75
<=> x²=25
<=> x=5
2x²-98=0
<=> 2x²=98
<=>   x²=49
<=>   x=7
x²-7x=0
<=> x(x-7)=0
<=> x=0 hoặc x=7
-3x²+5x=0
x(-3x+5)=0
x=0 hoặc -3x+5=0
x=0 hoặc -3x=-5
x=0 hoặc    x=5/3
x²+4x+4=0
(x+2)²=0
x+2=0
x=-2
 

1. 3x² - 75 = 0

<=> 3x² = 75

<=> x² = 25

<=> x = \(\sqrt{25}\)

<=> x = 5

2. x² - 7x = 0

<=> x(x - 7) = 0

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

3. x² - 14x + 13 = 0

<=> x² - 13x - x + 13 = 0

<=> x(x - 13) - (x - 13) = 0

<=> (x - 1)(x - 13) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=13\end{matrix}\right.\)

4. 2x² - 98 = 0

<=> 2x² = 98

<=> x² = 49

<=> x = \(\sqrt{49}\)

<=> x = 7

5. -3x² + 5x = 0

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

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

6. x² - 2x - 80 = 0

<=> x² + 8x - 10x - 80 = 0

<=> x(x + 8) - 10(x + 8) = 0

<=> (x - 10)(x + 8) = 0

<=> \(\left[{}\begin{matrix}x-10=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

7. x² = 81

<=> x² - 9² = 0

<=> (x - 9)(x + 9) = 0

<=> \(\left[{}\begin{matrix}x-9=0\\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

8. x² + 4x + 4 = 0

<=> x² + 2.x.2 + 2² = 0

<=> (x + 2)² = 0

<=> 0 = 0² - (x + 2)²

<=> (0 + x + 2)(0 - x + 2) = 0

<=> (x + 2)(-x + 2) = 0

<=> \(\left[{}\begin{matrix}x+2=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

9. 4x² + 12x + 5 = 0

<=> 4x² + 2x + 10x + 5 = 0

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

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

<=> \(\left[{}\begin{matrix}2x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

x-(17-x)=x-7

x+17+x=x-7

2x-17=x-7

x=-7+17

x=10

Ta có: \(x-\left(17-x\right)=x-7\)

    \(\Leftrightarrow x-17+x=x-7\)

    \(\Leftrightarrow x=10\)

Vậy \(x=10\)

= x - x = x - 17 - 7

= x - x = x - 10

suy ra x = 10

ta có 

ab x aba = abab

=>  số đó chỉ có thể là 10 

vì ta thấy 10 x 10 = 100

            10 x 101 = 1010

Đặt phép tính hàng dọc ta có:

       a b a            vì axb=b => a=1

         a b              vì b+b=b => b=0

__________                        => *=0

      b * b                  Thử lại:  101x10=1010(đúng)

   a b a                     Vậy a=1;b=0

___________

   a b a b

a=1,b=0 bn nha