sai đề rồi phải ko bạn ơi mình giải mãi mà cứ thấy kết quả lá số thập phân

\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\) (1)

\(\frac{y}{5}=\frac{z}{3}\Rightarrow\frac{y}{10}=\frac{z}{6}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

Và x + y + z = 46

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y+z}{15+10+6}=\frac{46}{31}\)

Ta có:

\(\frac{x}{15}=\frac{46}{31}\Rightarrow x=\frac{46}{31}.15=\frac{690}{31}\)

\(\frac{y}{10}=\frac{46}{31}\Rightarrow y=\frac{46}{31}.10=\frac{460}{31}\)

\(\frac{z}{6}=\frac{46}{31}\Rightarrow z=\frac{46}{31}.6=\frac{276}{31}\)

Vậy \(x=\frac{690}{31};y=\frac{460}{31};z=\frac{276}{31}\)

( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 24

= ( x² + 7x + 10 ) ( x² + 7x + 12 ) - 24

Đặt x² + 7x + 10 = y

Ta có : 

y² + 2y - 24 = ( y - 4 ) ( y + 6 ) = ( x² + 7x + 6 ) ( x² + 7x + 16 )

                    = ( x + 1 ) ( x + 6 ) ( x² + 7x + 16 )

Đặt x²+7x+10=t

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=t\left(t+2\right)-24=t^2+2t-24\)

\(=\left(t^2+2t+1\right)-25=\left(t+1\right)^2-5^2=\left(t-4\right)\left(t+6\right)\)=(x²+7x+6)(x²+7x+16)

=(x²+x+6x+6)(x²+7x+16)=[x(x+1)+6(x+1)](x²+7x+16)=(x+1)(x+6)(x²+7x+16)

\(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt:  \(x^2+7x+10=t\)Khi đó B trở thành:

\(B=t\left(t+2\right)-24\)

\(=t^2+2t-24=\left(t-4\right)\left(t+6\right)\)

đến đây bạn thay trở lại

a,\(\left(x-2\right)\left(x+2\right)=0\)

\(< =>\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)

b,\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)

\(< =>\orbr{\begin{cases}\frac{3x}{4}-\frac{2}{4}=0\\\frac{3x}{12}+\frac{16}{12}=0\end{cases}}\)

\(< =>\orbr{\begin{cases}3x-2=0\\3x+16=0\end{cases}}< =>\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)

\(\left(x-2\right)\left(x+2\right)=0\)

=> \(\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)

=> \(\orbr{\begin{cases}\frac{3}{4}x-\frac{1}{2}=0\\0,25x+\frac{4}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)

x-890=398

=>x=398+890

=>x=1288.

Vậy x=1288

x - 890 = 398

x = 398 + 890

x = 1288

x-890=398

x      =398+890

x     =1288

5^x+1-5^x=500

(5^x-5^x)+1=500

0.5^x=499

Sai đề

5^x + 1 - 5^x = 500

=> 5^x - (-1) - 5^x = 500

=> 5^x - 5^x - (-1) = 500

=> 0 - (-1) = 500

=> 1 = 500

=> Sai đề

x×2+x×3=4

x×(2+3)=4

x×5=4

4 khong chia het cho 5 

x.2+x.3=4

x.(2+3) =4

x.5       =4

   x       =4:5

   x       =0,8

=> x=0,8

\(x\times2+x\times3=4\)

\(x\times\left(2+3\right)=4\)

\(x\times5=4\)

\(x=4\div5\)

\(x=\frac{4}{5}\)

CHÚC BẠN HỌC GIỎI

\(\frac{4}{x+2}+\frac{-3}{x-2}+\frac{12}{x^2-4}.\)

\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-4}{x^2-4}\)

\(\frac{4}{x+2}+\frac{\left(-2\right)}{x-2}+\frac{12}{x^2-4}\)

\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{1}{x+2}\)