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:

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

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

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

(x – 2014)^3 + (x + 2012)^3 = 8(x – 1)^3

\(\Leftrightarrow\left(x-2014\right)^3+\left(x+2012\right)^3=\left(2x-2\right)^3\)(1)

Đặt \(\hept{\begin{cases}x-2014=a\\x+2012=b\\2x-2=c\end{cases}}\)thay vào pt (1) ta được: 

\(a^3+b^3=c^3\)

\(\Leftrightarrow a^3+b^3-c^3=0\)

\(\Leftrightarrow\left(a+b\right)^3-c^3-ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-ab\left(a+b\right)=0\)(2)

Thay \(a=x-2014;b=x+2012;c=2x-2\)hay \(a+b-c=0\)vào (2) ta được:

\(\left(x-2014\right)\left(x+2012\right)\left(2x-2\right)=0\)

... nốt 

Tìm x,biết:

a/\(x+5x^2=0\Leftrightarrow......\)
b/\(x+1=\left(x+1\right)^2\Leftrightarrow..........\)
c/\(x^3+x=0\Leftrightarrow.......\)
d/\(5x\left(x-2\right)-\left(2-x\right)=0\)
e/\(x\left(2x-1\right)+\frac{1}{3}-\frac{2}{3}x=0\Leftrightarrow........\)
g/\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow.....\)
h/\(x^2-3x=0\Leftrightarrow.....\)
i/\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow.....\)

Cho đề \(\hept{\begin{cases}2y^2-x^2=1\\2\left(x^3-y\right)=y^3-x\end{cases}\Leftrightarrow}\)\(\hept{\begin{cases}2\left(y^2+1\right)-\left(x^2+1\right)=2\\x\left(2x^2+1\right)-y\left(y^2+2\right)=0\end{cases}}\)

đặt \(a=y^2+1,b=x^2+1\)

\(\Leftrightarrow\hept{\begin{cases}2a-b=2\\x\left(2b-1\right)-y\left(a+1\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}b=2a-2\\x\left(4a-5\right)-ya-y=0\end{cases}}}\Leftrightarrow\hept{\begin{cases}b=2a-2\\a=\frac{5x+y}{4x-y}\end{cases}\Leftrightarrow\hept{\begin{cases}b=\frac{2x+4y}{4x-y}\\a=\frac{5x+y}{4x-y}\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}y^2+1=\frac{5x+y}{4x-y}\left(1\right)\\x^2+1=\frac{2x+4y}{4x-y}\left(2\right)\end{cases}}\)

pt(1)-pt(2),ta dc:\(\left(x-y\right)\left(\frac{3}{4x-y}+x+y\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=y\left(3\right)\\\frac{3}{4x-y}+x+y=0\left(4\right)\end{cases}}\)

CM:PT (4) vô nghiệm giúp mình nha!Và xem lại nếu mình có lm sai hay thiếu đk j đó hãy chỉ giúp mình nha!!!Hoặc pt(4) có nghiệm thì hãy giải giúp mình luôn nha!Thanks

Khi thử đổi biến chứng minh Iran 96 và cái kết.... Mà chả biết lúc đổi biến có tính sai chỗ nào ko mà kết quả nó nhìn khủng khiếp quá:(

Cho a, b, c là các số không âm thỏa mãn không có 2 số nào đồng thời bằng 0. Chứng minh rằng:

\(\left(ab+bc+ca\right)\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\right)\ge\frac{9}{4}\)

Đặt \(\left(a+b+c;ab+bc+ca;abc\right)=\left(3u;3v^2;w^3\right)\)

Cần chứng minh

\(\left(ab+bc+ca\right)\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\right)\ge\frac{9}{4}\)

\(\Leftrightarrow v^2\left(\left(3v^2+a^2\right)^2+\left(3v^2+b^2\right)^2+\left(3v^2+c^2\right)^2\right)\ge3\left(9uv^2-w^3\right)\)

\(\Leftrightarrow v^2\left(27v^4+6v^2\left(a^2+b^2+c^2\right)+a^4+b^4+c^4\right)\ge3\left(9uv^2-w^3\right)\)

\(\Leftrightarrow v^2\left(27v^4+6v^2\left(9u^2-6v^2\right)+a^4+b^4+c^4\right)\ge3\left(9uv^2-w^3\right)\)

\(\Leftrightarrow v^2\left(27v^4+6v^2\left(9u^2-6v^2\right)+81u^4-108u^2v^2+18v^4+12uw^3\right)\ge3\left(9uv^2-w^3\right)\)

\(\Leftrightarrow135u^4v^2-144u^2v^4+12uv^2w^3-27uv^2+45v^6+3w^3\ge0\)

1.     \(\frac{x^3-10x^2+25x}{x^2-5x}\)\(=0\)  ( đkxđ: \(x\ne0;5\))

<=> \(\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0\)<=>   \(x-5=0\)<=> vô n_o

_{2.       \(A=\)\(\frac{2x^2-2}{x^3-x^2-4x+4}\)\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)}   ( a, đkxđ: \(x\ne1;\pm2\))

b, \(A=0\)<=> \(\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0\)<=> \(x=-1\)( TM) . Vậy \(A=0\Leftrightarrow x=-1\)

3.    \(B=\frac{3x^2-12}{\left(x-3\right)\left(x^2+4x+4\right)}\)\(=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)^2}\) ( a,  đkxđ: \(x\ne3,-2\))

\(b,B=0\Leftrightarrow\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}=0\Leftrightarrow x=2\left(tm\right)\). Vậy \(B=0\Leftrightarrow x=2\)

có  \(ab=\frac{3}{5}\Leftrightarrow a=\frac{3}{5b}\)

có  \(bc=\frac{4}{5}\Rightarrow c=\frac{4}{5b}\)

mà \(ac=\frac{4}{5}\Leftrightarrow\frac{3}{5b}.\frac{4}{5b}=\frac{3}{4}\Leftrightarrow\frac{12}{25.b^2}=\frac{3}{4}\Leftrightarrow12.4=3.25.b^2\)

\(\Leftrightarrow\frac{48}{75}=b^2\Leftrightarrow b^2=\frac{16}{25}\Leftrightarrow b=\pm\frac{4}{5}\)

với \(b=\frac{4}{5}\)thì \(a=3:\left(5.\frac{4}{5}\right)=3:4=\frac{3}{4}\)

\(c=4:\left(5.\frac{4}{5}\right)=4:4=1\)

với \(b=-\frac{4}{5}\)thì  \(a=3:\left(5.\frac{-4}{5}\right)=3:-4=-\frac{3}{4}\)

\(c=4:\left(5.\frac{-4}{5}\right)=4:-4=-1\)

vậy \(\left(a;b;c\right)\in\left\{\left(\frac{4}{5};\frac{3}{4};1\right);\left(\frac{-4}{5};\frac{-3}{4};-1\right)\right\}\)

Dự đoán đẳng thức xảy ra khi x = y = z = 1.

Đặt x = 1 + a ; y = 1 + b , ( a , b \(\in\) R ). Từ giả thiết suy ra z = 1 - a - b.

Ta có: 
\(x^2+y^2+z^2+xy+yz+zx\)\(=\left(1+a\right)^2+\left(1+b\right)^2+\left(1-a-b\right)^2+\left(1+a\right)\left(1+b\right)+\left(1+b\right)\left(1-a-b\right)+\left(1-a-b\right)\left(1+a\right)=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+6\ge6.\)

Đẳng thức xảy ra khi và chỉ khi.

\(b=0;a+\frac{b}{2}=0\Leftrightarrow a=0;b=0\Leftrightarrow x=y=z=1.\)

Naruto nhận nè bạn.