3.x +2 : 2.x -1
ban nao giai ho mk vs
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\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\) (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)
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,\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)
\(< =>\orbr{\begin{cases}\frac{3x}{4}-\frac{2}{4}=0\\\frac{3x}{12}+\frac{16}{12}=0\end{cases}}\)
\(< =>\orbr{\begin{cases}3x-2=0\\3x+16=0\end{cases}}< =>\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)
\(\left(x-2\right)\left(x+2\right)=0\)
=> \(\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)
=> \(\orbr{\begin{cases}\frac{3}{4}x-\frac{1}{2}=0\\0,25x+\frac{4}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)
5x + 1 - 5x = 500
=> 5x - (-1) - 5x = 500
=> 5x - 5x - (-1) = 500
=> 0 - (-1) = 500
=> 1 = 500
=> Sai đề
\(\frac{4}{x+2}+\frac{-3}{x-2}+\frac{12}{x^2-4}.\)
\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-4}{x^2-4}\)
\(\frac{4}{x+2}+\frac{\left(-2\right)}{x-2}+\frac{12}{x^2-4}\)
\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{1}{x+2}\)
3.x có nghĩa là 3 nhân x hả bn????
Trên mạng có câu hỏi tt ă, áp dụng