a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)

b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)

c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)

d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)

e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)

Giải pt: { máy tính cho ra x=-1 , x=4 }

\(\left(x+1\right)\sqrt{16x+17}=8x^2-15x-23\) (1)

ĐK: \(16x+17\ge0\Leftrightarrow x\ge-\dfrac{17}{16}\)

(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(N\right)\\\left\{{}\begin{matrix}16x+17=\left(x-\dfrac{23}{8}\right)^2\\x\ge\dfrac{23}{8}\end{matrix}\right.\end{matrix}\right.\)(2)

(2) \(\Leftrightarrow16x+17=x^2-\dfrac{23}{4}x+\dfrac{529}{64}\Leftrightarrow x^2-\dfrac{87}{4}-\dfrac{559}{64}=0\) (Xấu quéc!! Pt này không có nghiệm = 4---> sai ở đâu vậy ạ??)

Cảm ơn trước nak ^^!

Câu I:

1.\(\dfrac{x}{4}=\dfrac{y}{7}\Rightarrow x=4k;y=7k\)

\(\Rightarrow xy=4k.7k=28k^2=112\)

\(\Leftrightarrow k=\pm2\)

*Với k=-2\(\Rightarrow x=-8;y=-14\)

*Với k=2\(\Rightarrow x=8;y=14\)

Vậy (x;y)=(-8;-14);(8;14).

2.Giả sử \(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\) với a,b,c khác 0

Đặt a=3k;b=5k;c=15k

\(\Rightarrow\dfrac{ab+ac}{2}=\dfrac{a\left(b+c\right)}{2}=\dfrac{3k.20k}{2}=30k^2\)

\(\dfrac{bc+ba}{3}=\dfrac{b\left(a+c\right)}{3}=\dfrac{5k.18k}{3}=30k^2\)

\(\dfrac{ca+cb}{4}=\dfrac{c\left(a+b\right)}{4}=\dfrac{15k.8k}{4}=30k^2\)

\(\Rightarrow\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}=\dfrac{ca+cb}{4}=30k^2\)

Vậy \(\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}=\dfrac{ca+cb}{4}\) thì \(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\)

3. Có : \(P=\left|2013-x\right|+\left|2014-x\right|\)\(=\left|2013-x\right|+\left|x-1014\right|\)\(\ge\left|2013-x+x-2014\right|=\left|-1\right|=1\)

Vậy P_min=1\(\Leftrightarrow\left(2013-x\right)\left(x-2014\right)\ge0\)

\(\Leftrightarrow-x^2+4027x-4054182\ge0\)

\(\Leftrightarrow2013\le x\le2014\)

Câu III:

2.Có:\(A=\dfrac{x_1^6}{x_2^6}+\dfrac{x_2^6}{x_1^6}\)\(=\dfrac{x_1^{12}+x_2^{12}}{x_1^6x_2^6}\)

Theo hệ thức Vi-et:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2}{2}=1\\x_1x_2=\dfrac{-1}{2}\end{matrix}\right.\)

Có: \(x_1^{12}+x_2^{12}=\left(x_1^6+x^6_2\right)^2-2x_1^6x_2^6\)\(=\left[\left(x_1^3+x_2^3\right)^2-2x_1^3x_2^3\right]^2-2x_1^6x_2^6\)

\(=\left\{\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]^2-2x_1^3x_2^3\right\}^2-2x_1^6x_2^6\)

\(=\left\{\left[1-3.\dfrac{-1}{2}.1\right]^2-2.\left(\dfrac{-1}{2}\right)^3\right\}^2-2.\dfrac{1}{2^6}\)

\(=\left\{\dfrac{25}{4}+\dfrac{1}{4}\right\}^2-\dfrac{1}{32}\)=\(\dfrac{1351}{32}\)

\(\Rightarrow A=\dfrac{\dfrac{1351}{32}}{\dfrac{1}{64}}\)\(=2702\)

Câu II:

1. b)\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)\(\left(x\ne-2;-4;-6;-8\right)\)

\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

\(\Leftrightarrow\left(\dfrac{2}{x+2}-1\right)+\left(\dfrac{8}{x+8}-1\right)=\left(\dfrac{4}{x+4}-1\right)+\left(\dfrac{6}{x+6}-1\right)\)

\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{x}{x+8}=\dfrac{x}{x+4}+\dfrac{x}{x+6}\)

\(\Leftrightarrow x\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)

Với \(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\left(\dfrac{1}{x+4}+\dfrac{1}{x+6}\right)=0\)

\(\Leftrightarrow\left(2x+10\right)\left(\dfrac{1}{\left(x+2\right)\left(x+8\right)}-\dfrac{1}{\left(x+4\right)\left(x+6\right)}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\left(TM\right)\\\dfrac{1}{\left(x+2\right)\left(x+8\right)}-\dfrac{1}{\left(x+4\right)\left(x+6\right)}=0\end{matrix}\right.\)

Với \(\frac{1}{\left(x+2\right)\left(x+8\right)}-\frac{1}{\left(x+4\right)\left(x+6\right)}=0\)

Cho \(\left\{{}\begin{matrix}x,y,z>0\\x+2y+3z=3\end{matrix}\right.\)

Tìm MaxP biết:

\(P=\dfrac{88y^3-x^3}{2xy+16y^2}+\dfrac{297z^3-8y^3}{6yz+36z^2}+\dfrac{11x^3-27z^3}{3xz+4x^2}\)

Đặt 2y=a, 3z=b \(\Rightarrow x+a+b=3\)

\(\Rightarrow P=\dfrac{11a^3-x^3}{ax+4a^2}+\dfrac{11b^3-a^3}{ab+4b^2}+\dfrac{11x^3-b^3}{bx+4x^2}\)

Ta chứng minh bđt sau:

\(\dfrac{11a^3-x^3}{ax+4a^2}\le3a-x\Leftrightarrow11a^3-x^3\le\left(3a-x\right)\left(ax+4a^2\right)\Leftrightarrow11a^3-x^3\le12a^3+3a^2x-ax^2-4a^2x\Leftrightarrow a^3-a^2x-ax^2+x^3\ge0\Leftrightarrow a^2\left(a-x\right)-x^2\left(a-x\right)\ge0\Leftrightarrow\left(a-x\right)^2\left(a+x\right)\ge0\left(luondung\right)\)tương tự:

\(\dfrac{11x^3-b^3}{bx+4x^2}\le3x-b,\dfrac{11b^3-a^3}{ab+4b^2}\le3b-a\)

\(\Rightarrow P\le3\left(x+a+b\right)-\left(a+b+x\right)=2\left(a+b+x\right)=2.3=6\)

\(MaxP=6\Leftrightarrow x=1,y=\dfrac{1}{2},z=\dfrac{1}{3}\)

KẾT QUẢ CUỘC THI TOÁN DO DƯƠNG PHAN KHÁNH DƯƠNG TỔ CHỨC .

Giải nhất : Ngô Tấn Đạt . Phần thưởng : Thẻ cào 100k + 30GP

Giải nhì : Hoàng Thảo Linh và Diệp Băng Dao . Phần thưởng : Thẻ cào 50k + 20GP

Giải ba : Truy kích và Luân Đào . Phần thưởng : 15GP

Nhờ thầy @phynit trao giải cho những bạn trên ạ . Cảm ơn các bạn dã ủng hộ cuộc thi của mình . GOOD LUCK !

ĐÁP ÁN VÒNG 3 : " CUỘC THI TOÁN DO DƯƠNG PHAN KHÁNH DƯƠNG TỔ CHỨC "

Câu 1 :

a ) ĐKXĐ : \(x\ge0\) , \(x\ne25\) , \(x\ne9\)

b )

\(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\dfrac{-5}{\sqrt{x}+5}:\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\)

\(=\dfrac{-5}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}+5}\)

\(=\dfrac{-5}{\sqrt{x}+5}\times\dfrac{\sqrt{x}+5}{-\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5}{\sqrt{x}+3}\)

c )

Để biểu thức A nhận giá trị nguyên thì \(5\) phải chia hết cho \(\sqrt{x}+3\)

Ta có : \(Ư\left(5\right)=\left(-5;-1;1;5\right)\) . Mà \(\sqrt{x}+3\ge3\) .

\(\Rightarrow\sqrt{x}+3=5\Rightarrow\sqrt{x}=2\Rightarrow x=4\left(N\right)\)

Vậy \(x=4\) thì biểu thức A nhận giá trị nguyên .

d )

Ta có :

\(B=\dfrac{A\left(x+16\right)}{5}=\dfrac{5\left(x+16\right)}{\dfrac{\sqrt{x}+3}{5}}=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

Theo BĐT Cô - Si cho hai số không âm ta có :

\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\ge2\sqrt{\sqrt{x}+3\times\dfrac{25}{\sqrt{x}+3}}=2\sqrt{25}=10\)

\(\Rightarrow\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge10-6=4\)

Dấu \("="\) xảy ra khi \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}+3=5\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Vậy GTNN của \(B\) là 4 khi \(x=4\)

Câu 2 :

a ) \(\left(x^2-x+1\right)\left(x^2+4x+1\right)=6x^2\)

\(\Leftrightarrow x^4+4x^3+x^2-x^3-4x^2-x+x^2+4x+1-6x^2=0\)

\(\Leftrightarrow x^4+3x^3-8x^2+3x+1=0\)

Xét : 0 không phải là nghiệm của phương trình trên .

\(\Leftrightarrow x^2+3x-8+\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}\right)+\left(3x+\dfrac{3}{x}\right)-8=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-10=0\)

Đặt \(x+\dfrac{1}{x}=t\) . Phương trình trở thành :

\(t^2+3t-10=0\)

\(\Delta=9+40=49>0\)

\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{-3+\sqrt{49}}{2}=2\\t_2=\dfrac{-3-\sqrt{49}}{2}=-5\end{matrix}\right.\)

Với \(t_1=2\) :

\(\Leftrightarrow x+\dfrac{1}{x}=2\)

\(\Leftrightarrow\) \(\dfrac{x^2}{x}+\dfrac{1}{x}=\dfrac{2x}{x}\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

Với \(t=-5\) :

\(\Leftrightarrow x+\dfrac{1}{x}=-5\)

\(\Leftrightarrow\) \(\dfrac{x^2}{x}+\dfrac{1}{x}=\dfrac{-5x}{x}\)

\(\Leftrightarrow x^2+5x+1=0\)

\(\Delta=25-4=21>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{21}}{2}\\x_2=\dfrac{-5-\sqrt{21}}{2}\end{matrix}\right.\)

Vậy \(S=\left\{1;\dfrac{-5+\sqrt{21}}{2};\dfrac{-5-\sqrt{21}}{2}\right\}\)

b ) \(3x^2+2x=2\sqrt{x^2+x}+1-x\)

\(\Leftrightarrow3\left(x^2+x\right)-2\sqrt{x^2+x}-1=0\)

\(\Leftrightarrow3\left(x^2+x\right)-3\sqrt{x^2+x}+\sqrt{x^2+x}-1=0\)

\(\Leftrightarrow3\sqrt{x^2+x}\left(\sqrt{x^2+x}-1\right)+\left(\sqrt{x^2+x}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2+x}-1\right)\left(3\sqrt{x^2+x}+1=0\right)\)

\(\) \(\Leftrightarrow\left(\sqrt{x^2+x}-1\right)=0\) . Vì \(3\sqrt{x^2+x}+1>0\)

\(\Leftrightarrow x^2+x-1=0\)

\(\Delta=1+4=5>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-1+\sqrt{5}}{2}\\x_2=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy ..............................

c )

\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\) ( ĐK : \(x\ge-1\) )

\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}-2x-\sqrt{\left(x+1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\left(\sqrt{x+3}-2x\right)\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x}+1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x+1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy......................

d ) \(x^2+9x+20=2\sqrt{3x+10}\) ( ĐK : \(x\ge-\dfrac{10}{3}\) )

\(\Leftrightarrow\left(x^2+6x+9\right)+\left(3x+10-2\sqrt{3x+10}+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{3x+10}=1\end{matrix}\right.\Leftrightarrow x=-3\)

Vậy...............................

Câu 3 :

a )

\(VT=\dfrac{\sqrt{\dfrac{abc+4}{a}-4\sqrt{\dfrac{bc}{a}}}}{\sqrt{abc}-2}\)

\(=\dfrac{\sqrt{\dfrac{abc+4}{a}-\dfrac{4\sqrt{abc}}{a}}}{\sqrt{abc}-2}\)

\(=\dfrac{\sqrt{\dfrac{abc+4-4\sqrt{abc}}{a}}}{\sqrt{abc}-2}\)

\(=\dfrac{\sqrt{\dfrac{\left(\sqrt{abc}-2\right)^2}{a}}}{\sqrt{abc}-2}\)

\(=\dfrac{\dfrac{\sqrt{abc}-2}{\sqrt{a}}}{\sqrt{abc}-2}=\dfrac{1}{\sqrt{a}}\left(đpcm\right)\)

b )

Nếu trong \(a+bc;b+ca;c+ab\) không có số nào lớn hơn 1 thì giá trị của mỗi số hạng củaVT ít nhất là \(\dfrac{1}{3}\)

Nếu trong \(a+bc;b+ca;c+ab\) có một số lớn hơn 1 khi đó : \(c=\dfrac{1-ab}{a+b}\)và \(a+b< 1\)

Theo BĐT Cô - Si dưới dạng engel ta có :

\(\dfrac{1}{2a+2bc+1}+\dfrac{1}{2b+2ca+1}\ge\dfrac{4}{2a+2b+2bc+2ca+2}=\dfrac{2}{a+b+2-ab}\)

Khi đó ta cần chứng minh :

\(\dfrac{2}{2+a+b-ab}+\dfrac{1}{2c+2ab+1}\ge1\)

Hay :\(\dfrac{2}{a+b-ab+2}+\dfrac{a+b}{a+b-2ab+2ab\left(a+b\right)+2}\ge1\)

Ta có :

\(VT=\dfrac{4+4\left(a+b\right)-4ab+3ab\left(a+b\right)+\left(a+b\right)^2}{\left(2+a+b-ab\right)\left(2+a+b-2ab+2ab\left(a+b\right)\right)}\)

Đặt \(S=a+b< 1;P=ab\) . Ta cần chứng minh :

\(\dfrac{4+4S-4P+3SP+S^2}{4S-6P+3SP+S^2+2S^2P-2P^2+2SP^2+4}\ge1\)

\(\Leftrightarrow2P\ge2S^2P-2P^2+2S^2P\)

\(\Leftrightarrow2P\left(1-S\right)\left(P+S+1\right)\ge0\) ( Đúng vì \(S< 1\) )

Dấu \("="\) xảy ra khi \(\left(a;b;c\right)=\left(0;1;1\right)\) và hoàn vị .

Câu 4 :

a )

Tứ giác ADHE có : \(\widehat{A}=\widehat{D}=\widehat{E}=90^0\)

\(\Rightarrow ADHE\) là hình chữ nhật .

\(\Rightarrow\widehat{AED}=\widehat{HAE}\)

Ta lại có : \(\widehat{HAE}=\widehat{ABC}\) ( Cùng phụ với góc C )

\(\Rightarrow\widehat{AED}=\widehat{ABC}\)

Xét \(\Delta AED\) và \(\Delta ABC\) ta có :

\(\left\{{}\begin{matrix}\widehat{A}:Chung\\\widehat{AED}=\widehat{ABC}\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta AED\sim\Delta ABC\left(g-g\right)\)

b )

Ta có : \(\left\{{}\begin{matrix}S_{ADE}=\dfrac{1}{2}S_{ADHE}\\S_{ABC}=2S_{ADHE}\end{matrix}\right.\Rightarrow S_{ADE}=\dfrac{1}{4}S_{ABC}\Rightarrow\) \(\dfrac{S_{ADE}}{S_{ABC}}=\dfrac{1}{4}\)

Mặt khác : \(\Delta ADE\sim\Delta ABC\) ( Câu a )

\(\Rightarrow\) \(\dfrac{S_{ADE}}{S_{ABC}}=\left(\dfrac{DE}{BC}\right)^2=\dfrac{1}{4}\)

\(\Rightarrow\) \(\dfrac{DE}{BC}=\dfrac{1}{2}\Rightarrow DE=\dfrac{1}{2}BC\)

Gọi M là trung điểm của BC .

\(\Delta ABC\) vuông tại A . \(\Rightarrow AM=\dfrac{1}{2}BC\)

\(\Rightarrow DE=AM\)

Mà \(AH=DE\) ( Do ADHE là hình chữ nhật )

\(\Rightarrow AM=AH\) ( Đường trung tuyến cũng là đường cao )

\(\Rightarrow\Delta ABC\) vuông cân tại A ( đpcm )

Câu 5 :

Ta có :

\(\left\{{}\begin{matrix}2011+y^2=y^2+xy+yz+zx=\left(x+y\right)\left(y+z\right)\\2011+z^2=z^2+xy+yz+zx=\left(x+z\right)\left(y+z\right)\\2011+x^2=x^2+xy+yz+zx=\left(x+y\right)\left(x+z\right)\end{matrix}\right.\)

\(\Rightarrow Q=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=2\left(xy+yz+zx\right)=2.2011=4022\)

Đây là đề bài:

Kiểm tra hộ mik lời giải, nếu có cách khác các bn góp ý cho mik nha, thnks nhiều!

Có \(P=\dfrac{2}{x^2+y^2}+\dfrac{35}{xy}+2xy\\ \Leftrightarrow P=\left(\dfrac{2}{x^2+y^2}+\dfrac{1}{xy}\right)+\dfrac{2}{xy}+\left(\dfrac{32}{xy}+2xy\right)\)

Xét nhóm 1: Áp dụng BĐT\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\left(1\right)\ge2\left(\dfrac{4}{\left(x+y\right)^2}\right)\ge2\left(\dfrac{4}{4^2}\right)=\dfrac{1}{2}\Rightarrow Min\left(1\right)=\dfrac{1}{2}\Leftrightarrow x=y\\\)

Xét nhóm 2: Vì \(x+y\le4\Rightarrow2\sqrt{xy}\le4\Rightarrow xy\le4\Rightarrow\dfrac{1}{xy}\ge\dfrac{1}{4}\Rightarrow Min\left(2\right)=\dfrac{1}{2}\Leftrightarrow xy=4\\ \)

Xét nhóm 3:Áp dụng BĐT Cô-si ta được:\(\dfrac{32}{xy}+2xy\ge2\sqrt{\dfrac{32}{xy}\cdot2xy}=16\Rightarrow Min\left(3\right)=16\Leftrightarrow x=y\\ \)

Từ các NX trên\(\Rightarrow MinP=\dfrac{1}{2}+\dfrac{1}{2}+16=17\left(ĐK:\right)x=y;xy=4hayx=y=2\)

(1) giải pt quy về \(ax^2+bx+c=0\)

1) \(x^2=3x\)                        2) \(x^2-3x=4\) 

3) \(x^4-5x^2+6=0\)                      4) \(x^3=9x\)

5) \(\left(x+2\right)\left(x-3\right)=x^2-4\)              6) \(\dfrac{x+11}{x^2-1}-\dfrac{x-1}{x+1}=\dfrac{2\left(x+7\right)}{x+1}\)

giúp mk vs mk cần gấp

a,\(\left|3x-4\right|+\left|3y+5\right|=0\) b,\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\) c,\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)

d,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\) e,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)

Giups mình giải bài tìm x,y,z này nhé!!! Cảm ơn nhiều ạ!!!