\(a^2\left(a+1\right)-b^2\left(b-1\right)-3ab\left(a-b+\frac{2}{3}\right)\)

\(=a^3+a^2-b^3+b^2-3a^2b+3ab^2-2ab\)

\(=\left(a^3-3a^2b+3ab^2-b^2\right)+\left(a^2-2ab+b^2\right)\)

\(=\left(a-b\right)^3+\left(a-b\right)^2\)(*)

Thay a - b = 7 vào (*), ta có:

(*) \(\Leftrightarrow7^3+7^2=392\)

Gọi tử số là x

Mẫu số là 15 - x

Theo đề ra, ta có phương trình:

\(\frac{x-5}{15-x+2}=\frac{1}{5}\)

\(\Leftrightarrow\frac{x-5}{17-x}=\frac{1}{5}\)

\(\Leftrightarrow5\left(x-5\right)=17-x\)

\(\Leftrightarrow5x-25=17-x\)

\(\Leftrightarrow5x+x=17+25\)

\(\Leftrightarrow6x=42\)

\(\Leftrightarrow x=7\)

Vậy tử số là 7, mẫu số là 15 - 7 = 8 => Phân số ban đầu là \(\frac{7}{8}\)

mong ban hay nhan co dau di, khong dau tui nhin ngua mat la tui con tra loi dai dai do

Phiền bạn đừng đăng linh tinh !!!

Học tốt !!!

hình tự vẽ nhé 

ok banj

ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\x-2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}}\)

\(\frac{1-x}{x+1}+3=\frac{2x+3}{x-2}\)

\(\frac{\left(1-x\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}\)

\(2x^2-8=2x^2+5x+3\)

\(2x^2-8-2x^2-5x-3=0\)

\(-11-5x=0\)

\(5x=-11\)

\(x=-\frac{11}{5}\)Theo ĐKXĐ => tm 

Bài làm

a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\)

\(P=\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)

\(P=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)

\(P=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x+2}\)

\(P=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{x+1}\)

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

b) Thay \(x=\frac{1}{2}\)vào P ta được:

\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}\)

\(P=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{2}{2}}\)

\(P=\frac{3}{2}:\frac{-1}{2}\)

\(P=\frac{3}{2}.\left(-2\right)\)

\(P=-3\)

Vậy giá trị của \(P=-3\) tại \(x=\frac{1}{2}\)

a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)

\(\Leftrightarrow P=\left(\frac{x}{x-2}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)

\(\Leftrightarrow P=\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)

\(\Leftrightarrow P=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)

\(\Leftrightarrow P=\frac{x^2+2x+1}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x+2}{x+1}\)

\(\Leftrightarrow P=\frac{\left(x+1\right)^2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{x+1}{x-2}\)

Vậy \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)

b) Ta có \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)

Thay x=\(\frac{1}{2}\left(tm\right)\)vào P ta có:

\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{4}{2}}=\frac{\frac{3}{2}}{\frac{-3}{2}}=\frac{3}{2}:\frac{-3}{2}=-1\)

Vậy \(P=-1\)khi x=\(\frac{1}{2}\)

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