Ta có: C% = \(\dfrac{S}{S+100}\times100\%\)

<=> C% = \(\dfrac{\dfrac{m_{ct}}{m_{dm}}\times100}{\dfrac{m_{ct}}{m_{dm}}\times100+100}\) x100%= \(\dfrac{\dfrac{100m_{ct}}{m_{dm}}}{\dfrac{100\left(m_{ct}+m_{dm}\right)}{m_{dm}}}\) x100%

= \(\dfrac{100m_{ct}}{m_{dm}}\times\dfrac{m_{dm}}{100m_{dd}}\) x 100% = \(\dfrac{m_{ct}}{m_{dd}}\times100\%\) (Luôn đúng)

=> Đpcm

a) ( a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

b) ( a + b - c )² = a² + b² + c² + 2ab - 2bc - 2ac

c) (a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac

Làm lại câu c. Ghi nhầm đề:

c) Để C = -0,5 => \(\dfrac{1}{2\left(x+1\right)}=-0,5\)

=> \(2\left(x+1\right)=-2\)

=> \(x+1=-1\)

=> x = -2 (TM ĐKXĐ)

Thay x = -2 vào C ta được C = -0,5 (TM)

Vậy nghiệm của PT: S = \(\left\{-2\right\}\)

Haizzzzzzzzzzz!

ĐKXĐ: \(x\ne0;\dfrac{-1}{2};\dfrac{1}{2}\)

\(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right)\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)

=

\(\dfrac{4x\left(x+1\right)+1}{4x^2}.\left[\dfrac{\left(1-2x\right)\left(1+2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{\left(1-2x\right)\left(1+2x\right)}.\dfrac{\left(1-2x\right)^2}{1+2x}\right]\)\(-\dfrac{1}{2x}\)

= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\left(\dfrac{1-4x^2}{\left(2x+1\right)^2}-\dfrac{1-2x}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)

= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\dfrac{2x\left(1-2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)

= \(\dfrac{1-2x}{2x}-\dfrac{1}{2x}=\dfrac{-2x}{2x}=1\)

C_M của HCl 18,25% = \(\dfrac{C\%\times10D}{M}\) = \(\dfrac{18,25\times10\times1,2}{36,5}\) = 6M

Tương tự: C_M của HCl 13% = \(\dfrac{13\times10\times1,123}{36,5}\) = 4M

=> n_HCl (1) = 6V₁

n_HCl (2) = 4V₂

=> C_M dd HCl 4,5M = \(\dfrac{n_1+n_2}{V_1+V_2}\) = \(\dfrac{6V_1+4V_2}{V_1+V_2}\) = 4,5M

=> 6V₁ + 4V₂ = 4,5V₁ + 4,5V₂

=> 1,5V₁ = 0,5V₂

=> \(\dfrac{V_1}{V_2}=\dfrac{0,5}{1,5}=\dfrac{1}{3}\)