rút gọn bt :A=m^2(m^2-n).(m^3-n^6).(m+n^2).Rất cảm ơn nào giúp mình câu này(tích phải bằng 0)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ĐKXĐ: x \(\ne\pm3\)
b. M = \(\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
= \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\) = \(\frac{9+6x+x^2}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)
c. M = 0 hay \(\frac{x+3}{x-3}=0\) => x + 3 = 0 <=> x = -3 (Loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(m+2\right)\left(n+3\right)=7\)
\(\Rightarrow m+2,n+3\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
Do \(m,n\in N\) nên không có m và n thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(N=1:\left(\frac{x+2}{\sqrt{x^3}-1}+\frac{\sqrt{x}+1}{x+1+\sqrt{x}}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(N=1:\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}-\frac{\left(x+1+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{x+2+x-1-x-1-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{\sqrt{x}}{\left(x+1+\sqrt{x}\right)}\right)\)
\(N=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
y b
chia 2 ve cho can 2
tc
\(\sqrt{x}+1+\frac{1}{\sqrt{x}}\)
tc \(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)(bdt cosi)
\(\sqrt{x}+1+\frac{1}{\sqrt{x}}\ge3\)
=> dpcm
may mk loi font chu thong cam viet ko co dau