toàn cấp 2 ha

Bài làm

\(xy^2+2xy+x=32y\)

\(\Leftrightarrow x\left(y^2+2y+1\right)=32y\)

\(\Leftrightarrow x=\frac{32y}{y^2+2y+1}\)

\(\Leftrightarrow\frac{32y}{\left(y+1\right)^2}\)

\(\Leftrightarrow x=\frac{32y}{y+1}-\frac{32y}{\left(y+1\right)^2}\)

Để x là số nguyên dương thì 

\(\left(y+1\right)^2\inƯ_{\left(32\right)}\)và\(\left(y+1\right)^2\)là số chính phương 

\(\Rightarrow\left(y+1^2\right)=\left\{1;4;16\right\}\)

\(\Leftrightarrow y+1=\left\{1;2;4\right\}\)

\(\Leftrightarrow y=\left\{0;1;3\right\}\)

Vì y là số nguyên dương 

Nên: \(\hept{\begin{cases}y=1\Rightarrow x=8\\y=3\Rightarrow x=6\end{cases}}\)

Vậy   x = 8; y = 1

hoặc x = 6; y = 3

# Chúc bạn học tốt #

Bạn có thể giải thích rõ dòng: 4 và 5 không. Mình thấy nó chưa được chính xác.

\(A=x^2+y^2-2\left(x-y\right)\)

\(A=x^2+y^2-2x+2y\)

\(A=\left(x^2-2x+1\right)+\left(y^2+2y+1\right)-2\)

\(A=\left(x-1\right)^2+\left(y+1\right)^2-2\)

Vì \(\left(x-1\right)^2\ge0;\left(y+1\right)^2\ge0\)\(\Rightarrow A\ge-2\)

Dấu ''='' xảy ra khi: \(\hept{\begin{cases}x-1=0\\y+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

Vậy GTNN của A là A=-2 khi x=1 và y=-1

\(B=x\left(x-3\right)\left(x+1\right)\left(x+4\right)\)

\(B=\left[x\left(x+1\right)\right]\left[\left(x-3\right)\left(x+4\right)\right]\)

\(B=\left(x^2+x\right)\left(x^2+x-12\right)\)

Đặt \(x^2+x=a\)ta được;

\(B=a\left(a-12\right)=a^2-12a=\left(a^2-2.a.6+36\right)-36\)\(=\left(a-6\right)^2-36\)

Vì \(\left(a-6\right)^2\ge0\)\(\Rightarrow\left(a-6\right)^2-36\ge-36\)

Dấu ''='' xảy ra khi \(a-6=0\Rightarrow a=6\Rightarrow x^2+x-6=0\)\(\Rightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)

\(\Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\)\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy GTNN của B là B=-36 khi x=-3 hoặc x=2

\(a^4-2a^3+a^2\)

\(=a^2\left(a^2-2a+1\right)\)

\(=a^2\left(a-1\right)^2\ge0\)

Vậy \(a^4-2a^3+a^2\ge0\)

\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\left(ĐK:x\ne\pm1\right)\)

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

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

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

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

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

\(=\frac{2\sqrt{x}-6}{x-1}\)

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

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

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

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

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

\(=\frac{2x\sqrt{x}-6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

Bài lm có sai sót chỗ nào thì mong bn sửa lại........

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

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

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

\(=\frac{x+3\sqrt{x}-x+4\sqrt{x}-3}{\sqrt{x}+3}\)

\(=\frac{7\sqrt{x}-3}{\sqrt{x}+3}\)

\(\frac{x-3\sqrt{x}}{\sqrt{x}-3}-\frac{x+4\sqrt{x}+3}{\sqrt{x}+3}\)

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

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

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

\(=\sqrt{x}-\sqrt{x}-1=-1\)