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|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
\(\Leftrightarrow16-3\left(x+1\right)< 24+2\left(x-1\right)\)
=>16-3x-3<24+2x-2
=>-3x+13<2x+22
=>-5x<9
hay x>-9/5
\(x^3+y^3=\left(x^2+y^2\right)\sqrt{x^2-xy+y^2}\)
\(\Leftrightarrow\left(x^3+y^3\right)^2=\left(x^2+y^2\right)^2.\left(x^2-xy+y^2\right)\)
\(\Leftrightarrow\left(x+y\right)^2.\left(x^2-xy+y^2\right)^2=\left(x^2+y^2\right)^2.\left(x^2-xy+y^2\right)\)
\(\Leftrightarrow\left(x+y\right)^2.\left(x^2-xy+y^2\right)=\left(x^2+y^2\right)^2\)
\(\Leftrightarrow\left(x^3+y^3\right)\left(x+y\right)=\left(x^2+y^2\right)^2\)
\(\Leftrightarrow x^4+x^3y+xy^3+y^4=x^4+y^4+2x^2y^2\)
\(\Leftrightarrow x^3y+xy^3-2x^2y^2=0\)
\(\Leftrightarrow xy\left(x^2-2xy+y^2\right)=0\)
\(\Leftrightarrow\sqrt{4x-3}.\left(x-y\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{4x-3}=0\\\left(x-y\right)^2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}4x-3=0\\x-y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\x=y\end{cases}}\)
Xét trường hợp:
Với x=3/4
=>\(x=\frac{3}{4}\Leftrightarrow y.\frac{3}{4}=0\Leftrightarrow y=0\)
Với: \(x=y\)
Có: \(xy=\sqrt{4x-3}\Leftrightarrow x^2y^2=4x-3\Leftrightarrow x^4-4x+3=0\Leftrightarrow x\left(x^3-1\right)-3\left(x-1\right)=0\)\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^2+2x+3\right)=0\)( vì x^2+2x+3 luôn dương. Tự c/m nhé )
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)\(\Leftrightarrow x=y=1\)
KL:.................................
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{8}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+3\left(\dfrac{1}{8}-\dfrac{1}{y}\right)=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{8}-\dfrac{3}{x}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{1}{x}=\dfrac{1}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\x=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{12}\\x=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=24\end{matrix}\right.\)