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?Amanda?, Phạm Lan Hương, Phạm Thị Diệu Huyền, Vũ Minh Tuấn, Nguyễn Ngọc Lộc , @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @Trần Thanh Phương
giúp e với ạ! Cần trước 5h chiều nay! Cảm ơn mn nhiều!
Tranh thủ làm 1, 2 bài rồi ăn cơm:
1/ Đặt \(m=n-2008>0\)
\(\Rightarrow2^{2008}\left(369+2^m\right)\) là số chính phương
\(\Rightarrow369+2^m\) là số chính phương
m lẻ thì số trên chia 3 dư 2 nên ko là số chính phương
\(\Rightarrow m=2k\Rightarrow369=x^2-\left(2^k\right)^2=\left(x-2^k\right)\left(x+2^k\right)\)
b/
\(2\left(a^2+b^2\right)\left(a+b-2\right)=a^4+b^4\) \(\left(a+b>2\right)\)
\(\Rightarrow2\left(a^2+b^2\right)\left(a+b-2\right)\ge\frac{1}{2}\left(a^2+b^2\right)^2\)
\(\Rightarrow a^2+b^2\le4\left(a+b-2\right)\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2\le0\Rightarrow a=b=2\)
\(\Rightarrow x=y=4\)
\(A=\frac{a}{ab+c\left(a+b+c\right)}+\frac{b}{bc+a\left(a+b+c\right)}+\frac{c}{ca+b\left(a+b+c\right)}\)
\(=\frac{a}{\left(b+c\right)\left(a+c\right)}+\frac{b}{\left(a+b\right)\left(a+c\right)}+\frac{c}{\left(a+b\right)\left(c+b\right)}\)
Áp dụng bđt AM-GM ta có
\(A=\frac{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\ge27.\frac{a^2+b^2+c^2+ab+bc+ca}{8\left(a+b+c\right)^3}\)\(=\frac{a^2+b^2+c^2+ab+bc+ca}{8}\)
\(=\frac{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}{8}\)\(\ge\frac{9-\frac{\left(a+b+c\right)^2}{3}}{8}=\frac{9-3}{8}=\frac{3}{4}\)
Dấu "=" xảy ra khi a=b=c=1
a/ Một cách đơn giản hơn:
\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
\(P=\frac{x-\frac{1}{2}+y-\frac{1}{2}}{y^2}+\frac{y-\frac{1}{2}+z-\frac{1}{2}}{z^2}+\frac{z-\frac{1}{2}+x-\frac{1}{2}}{x^2}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(P=\left(x-\frac{1}{2}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\left(y-\frac{1}{2}\right)\left(\frac{1}{y^2}+\frac{1}{z^2}\right)+\left(z-\frac{1}{2}\right)\left(\frac{1}{x^2}+\frac{1}{z^2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(P\ge\frac{2}{xy}\left(x-\frac{1}{2}\right)+\frac{2}{yz}\left(y-\frac{1}{2}\right)+\frac{2}{zx}\left(z-\frac{1}{2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(P\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\)
\(P\ge\sqrt{3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}-1=\sqrt{3}-1\)
\(P_{min}=\sqrt{3}-1\) khi \(x=y=z=\sqrt{3}\)
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
a/ Đơn giản là dùng phép thế:
\(x+2y+x+y+z=0\Rightarrow x+2y=0\Rightarrow x=-2y\)
\(x+y+z=0\Rightarrow z=-\left(x+y\right)=-\left(-2y+y\right)=y\)
Thế vào pt cuối:
\(\left(1-2y\right)^2+\left(y+2\right)^2+\left(y+3\right)^2=26\)
Vậy là xong
b/ Sử dụng hệ số bất định:
\(\left\{{}\begin{matrix}a\left(\frac{x}{3}+\frac{y}{12}-\frac{z}{4}\right)=a\\b\left(\frac{x}{10}+\frac{y}{5}+\frac{z}{3}\right)=b\end{matrix}\right.\)
\(\Rightarrow\left(\frac{a}{3}+\frac{b}{10}\right)x+\left(\frac{a}{12}+\frac{b}{5}\right)y+\left(\frac{-a}{4}+\frac{b}{3}\right)z=a+b\) (1)
Ta cần a;b sao cho \(\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}=-\frac{a}{4}+\frac{b}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}\\\frac{a}{3}+\frac{b}{10}=-\frac{a}{4}+\frac{b}{3}\end{matrix}\right.\) \(\Rightarrow\frac{a}{2}=\frac{b}{5}\)
Chọn \(\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\) thay vào (1):
\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
1,ĐK: \(x,y\ne-2\)
HPT<=> \(\left\{{}\begin{matrix}x\left(x+2\right)+y\left(y+2\right)=\left(x+2\right)\left(y+2\right)\left(1\right)\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x^2\left(x+2\right)^2+2xy\left(x+2\right)\left(y+2\right)+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)
=> \(2xy\left(x+2\right)\left(y+2\right)=0\)
<=>\(2xy=0\) (do x+2 và y+2 \(\ne0\))
<=> \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Tại x=0 thay vào (1) có: \(y\left(y+2\right)=2\left(y+2\right)\) <=> y= \(\pm2\) => y=2 (vì y khác -2)
Tại y=0 thay vào (1) có: \(x\left(x+2\right)=2\left(x+2\right)\) => x=2
Vậy HPT có 2 nghiệm duy nhất (2,0),(0,2)
2, ĐK: \(y\ne-1\)
HPT <=> \(\left\{{}\begin{matrix}x^2=2\left(x+3\right)\left(y+1\right)\left(1\right)\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)
=> \(\frac{6\left(3+x\right)\left(y+1\right)}{y+1}=4-x\)
<=> 6(x+3)=4-x
<=> \(14=-7x\)
<=> \(x=-2\) thay vào (1) có \(4=2\left(y+1\right)\)
<=>y=1\(\)( tm)
Vậy hpt có một nghiệm duy nhất (-2,1)
3,\(\left\{{}\begin{matrix}x^2-y=y^2-x\left(1\right)\\x^2-x=y+3\left(2\right)\end{matrix}\right.\)
PT (1) <=> \(\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)
<=> (x-y)(x+y+1)=0
<=>\(\left[{}\begin{matrix}x=y\\y=-x-1\end{matrix}\right.\)
Tại x=y thay vào (2) có \(y^2-y=y+3\) <=> \(y^2-2y-3=0\) <=> (y-3)(y+1)=0 <=> \(\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Tại y=-1-x thay vào (2) có: \(x^2-x=-1-x+3\) <=> \(x^2=2\) <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}y=-1-\sqrt{2}\\y=-1+\sqrt{2}\end{matrix}\right.\)
Vậy hpt có 4 nghiệm (3,3),(-1,-1), ( \(\sqrt{2},-1-\sqrt{2}\)),( \(-\sqrt{2},-1+\sqrt{2}\))
4,\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\left(2\right)\end{matrix}\right.\)(đk:\(x\ne0,y\ne0\))
<=> \(\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=\frac{9}{2}\\\left(y+\frac{1}{y}\right)\left(x+\frac{1}{x}\right)=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)
Có \(\left\{{}\begin{matrix}u+v=\frac{9}{2}\\uv=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\v\left(\frac{9}{2}-v\right)=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left(v-\frac{5}{2}\right)\left(v-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left[{}\begin{matrix}v=\frac{5}{2}\\v=2\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\\\left[{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
Tại \(\left\{{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=\frac{5}{2}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)\left(y-\frac{1}{2}\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Tại \(\left\{{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\y+\frac{1}{y}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-2\right)\left(x-\frac{1}{2}\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt có 4 nghiệm (1,2),( \(1,\frac{1}{2}\)) ,( 2,1),(\(\frac{1}{2},1\)).
10.
\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y+1\right)=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\y=2x+1\end{matrix}\right.\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=y^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=y^2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=x^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=\left(2x+1\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\3x\left(x+1\right)=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2x+1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x=-1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
3 g) \(xyz=x+y+z+2\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=\Sigma_{cyc}\left(x+1\right)\left(y+1\right)\)
\(\Rightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\) .Đặt \(\frac{1}{x+1}=a;\frac{1}{y+1}=b;\frac{1}{z+1}=c\Rightarrow x=\frac{1-a}{a}=\frac{b+c}{a};y=\frac{c+a}{b};z=\frac{a+b}{c}\) vì a + b + c = 1.
Khi đó \(P=\Sigma_{cyc}\frac{1}{\sqrt{\frac{\left(b+c\right)^2}{a^2}+2}}=\Sigma_{cyc}\frac{a}{\sqrt{2a^2+\left(b+c\right)^2}}\)
\(=\sqrt{\frac{2}{9}+\frac{4}{9}}.\Sigma_{cyc}\frac{a}{\sqrt{\left[\left(\sqrt{\frac{2}{9}}\right)^2+\left(\sqrt{\frac{4}{9}}\right)^2\right]\left[2a^2+\left(b+c\right)^2\right]}}\)
\(\le\sqrt{\frac{2}{3}}\Sigma_{cyc}\frac{a}{\sqrt{\left[\frac{2}{3}a+\frac{2}{3}b+\frac{2}{3}c\right]^2}}=\frac{\sqrt{6}}{2}\left(a+b+c\right)=\frac{\sqrt{6}}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=2\)
3c) Nhìn quen quen, chả biết có lời giải ở đâu hay chưa nhưng vẫn làm:D (Em ko quan tâm nha!)
\(P=3-\Sigma_{cyc}\frac{2xy^2}{xy^2+xy^2+1}\ge3-\Sigma_{cyc}\frac{2xy^2}{3\sqrt[3]{\left(xy^2\right)^2}}=3-\frac{2}{3}\Sigma_{cyc}\sqrt[3]{\left(xy^2\right)}\)
\(\ge3-\frac{2}{3}\Sigma_{cyc}\frac{x+y+y}{3}=3-\frac{2}{3}\left(x+y+z\right)=3-2=1\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Câu 2:
Từ điều kiện bài này có thể đặt ẩn phụ và AM-GM ra luôn kết quả, nhưng hơi rắc rối khi người ta hỏi từ đâu mà có cách đặt ẩn phụ như vậy, do đó ta giải trâu :D
\(x^2+y^2+z^2+xyz=4\)
\(\Leftrightarrow\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{4}+2\left(\frac{x}{2}.\frac{y}{z}.\frac{z}{2}\right)=1\)
\(\Leftrightarrow\frac{xy}{2z}.\frac{xz}{2y}+\frac{xy}{2z}.\frac{yz}{2x}+\frac{yz}{2x}.\frac{xz}{2y}+2\left(\frac{xy}{2z}.\frac{yz}{2x}.\frac{xy}{2y}\right)=1\)
Đặt \(\left(\frac{xy}{2z};\frac{zx}{2y};\frac{yz}{2x}\right)=\left(m;n;p\right)\Rightarrow mn+np+pn+2mnp=1\)
\(\Leftrightarrow2\left(n+1\right)\left(m+1\right)\left(p+1\right)=\left(n+1\right)\left(m+1\right)+\left(n+1\right)\left(p+1\right)+\left(m+1\right)\left(p+1\right)\)
\(\Leftrightarrow\frac{1}{n+1}+\frac{1}{m+1}+\frac{1}{p+1}=2\)
\(\Leftrightarrow1=\frac{n}{n+1}+\frac{m}{m+1}+\frac{p}{p+1}\ge\frac{\left(\sqrt{n}+\sqrt{m}+\sqrt{p}\right)^2}{m+n+p+3}\)
\(\Leftrightarrow m+m+p+2\left(\sqrt{mn}+\sqrt{np}+\sqrt{mp}\right)\le m+n+p+3\)
\(\Leftrightarrow\sqrt{mn}+\sqrt{np}+\sqrt{mp}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\le\frac{3}{2}\Leftrightarrow x+y+z\le3\)
Câu 1:
\(2xyz=1-\left(x+y+z\right)+xy+yz+zx\)
\(\Rightarrow xy+yz+zx=2xyz+\left(x+y+z\right)-1\)
\(VT=x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\)
\(=\left(x+y+z\right)^2-2\left(x+y+z\right)-4xyz+2\)
\(VT\ge\left(x+y+z\right)^2-2\left(x+y+z\right)-\frac{4}{27}\left(x+y+z\right)^3+2\)
\(VT\ge\frac{4}{27}\left[\frac{15}{4}-\left(x+y+z\right)\right]\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}\)
(Do \(0< x;y;z< 1\Rightarrow x+y+z< 3< \frac{15}{4}\))
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
Vũ Minh Tuấn, buithianhtho, Băng Băng 2k6, Akai Haruma, Nguyễn Thành Trương, No choice teen, Nguyễn Thanh Hằng, HISINOMA KINIMADO, Bùi Thị Vân, Arakawa Whiter, @tth_new, @Nguyễn Việt Lâm, @Nguyễn Thị Ngọc Thơ
mn giúp e vs ạ! Thanks nhiều!
Nghỉ tết r học chăm thế