\(\frac{x}{5}-\frac{2}{y}=\frac{2}{15}\)
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a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)
\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)
\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)
\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)
\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)
Vậy x = \(\frac{1}{3}\)
b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)
\(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)
\(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)
\(x.\frac{5}{6}=\frac{-2}{15}\)
\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)
Vậy x = \(\frac{-4}{25}\)
c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)
\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)
\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)
\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)
\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)
Vậy x = \(\frac{-4}{25}\)
Ủng hộ tớ nha m.n
\(\frac{x}{5}+-\frac{2}{y}=\frac{2}{15}\)
\(\Leftrightarrow\frac{2}{y}=\frac{x}{5}-\frac{2}{15}\)
\(\Leftrightarrow\frac{2}{y}=\frac{3x}{15}-\frac{2}{15}=\frac{3x-2}{15}\)
\(\Leftrightarrow2\cdot15=y\left(3x-2\right)\)
\(\Leftrightarrow30=y\left(3x-2\right)\)
Từ trên ta có được \(y\inƯ\left(30\right)\)
\(\left(3x-2\right)\inƯ\left(30\right)\)
mà \(Ư\left(30\right)=\left\{\pm1;\pm2;\pm3;\pm5;\pm6;\pm10;\pm15;\pm30\right\}\)
* Đến đây bạn lập bảng các trường hợp để tìm ra x,y thỏa mãn ạ *
=> Từ bảng ta tìm được x = 1; y = 30
a)\(x-\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}+\frac{3}{5}=\frac{6}{5}\)
b)\(|x|-\frac{4}{5}=\frac{2}{3}\\ \Rightarrow|x|=\frac{2}{3}+\frac{4}{5}=\frac{22}{15}\\ \Rightarrow|x|=\frac{22}{15}\\ \Rightarrow x=\frac{22}{15}\)
c)\(\frac{x}{-5}=\frac{24}{15}\\ \Rightarrow x=\frac{-5\cdot24}{15}=-8\)
d)\(\frac{x}{4}=\frac{y}{5} và x-y=21\)
Theo tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{x}{4}=\frac{y}{5}=\frac{x-y}{4-5}=\frac{21}{-1}=-21\)
Do đó :
\(\frac{x}{4}=-21\Rightarrow x=-84\)
\(\frac{y}{5}=-21\Rightarrow y=-105\)
\(x-\frac{3}{5}=\frac{3}{5}\)
\(x=\frac{3}{5}+\frac{3}{5}\)
\(x=\frac{6}{5}\)
\(\left|x\right|-\frac{4}{5}=\frac{2}{5}\)
\(\left|x\right|=\frac{2}{5}+\frac{4}{5}\)
\(\left|x\right|=\frac{6}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-\frac{6}{5}\end{cases}}\)
\(\frac{x}{-5}=\frac{24}{15}\)
\(\Rightarrow x.15=\left(-5\right).24\)
\(\Rightarrow x.15=-120\)
\(\Rightarrow x=-120:15\)
\(\Rightarrow x=-8\)
Đặt: \(\left\{{}\begin{matrix}a=\sqrt{x}+1\\b=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{a}-\frac{1-b}{b}=\frac{22}{15}\\\frac{3}{a}+\frac{5+b}{b}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{a}-\frac{1}{b}+1=\frac{22}{15}\\\frac{3}{a}+\frac{5}{b}+1=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{a}-\frac{1}{b}=\frac{7}{15}\\\frac{3}{a}+\frac{5}{b}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{6}{a}-\frac{3}{b}=\frac{7}{5}\\\frac{6}{a}+\frac{10}{b}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{6}{a}-\frac{3}{b}=\frac{7}{5}\\\frac{13}{b}=\frac{13}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3=\sqrt{x}+1\\5=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5-x=1\end{matrix}\right.\)
Vậy pt có \(n_0\) \(S=\left\{4;1\right\}\)
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
Xin lỗi, câu trên mình trả lời nhầm, cho mình trả lời lại nha!
Để mẫu số 5 và y quy đòng lên thành mẫu số 15 thì y phải bằng 3
Ta có:
\(\frac{x}{5}-\frac{2}{y}=\frac{x}{5}-\frac{2}{3}=\frac{x\cdot3}{15}-\frac{10}{15}=\frac{2}{15}\) => x . 3 - 10 = 2 => x . 3 = 2 + 10 => x . 3 = 12 => x = 12 : 3 => x = 4
Vậy x = 4, y = 3
\(\frac{4}{5}-\frac{2}{3}=\frac{2}{15}\)
\(\Rightarrow x=4;y=3\)