\(Tacó:\) \(x+y=3\)

\(\dfrac{x}{2}+\dfrac{x}{2}+y=3\)

Áp dụng BĐT Cô si cho 3 số, ta có:

\(3=\left(\dfrac{x}{2}+\dfrac{x}{2}+y\right)\ge3.^3\sqrt{\dfrac{x}{2}.\dfrac{x}{2}.y}=3.^3\sqrt{\dfrac{x^2y}{4}}\)

⇒\(1\ge^3\sqrt{\dfrac{x^2y}{4}}\)

⇒\(1\ge\dfrac{x^2y}{4}\Leftrightarrow x^2y\le4\)

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\(\left(1+\frac{1}{x}\right).\left(1+\frac{1}{y}\right).\left(1+\frac{1}{z}\right)=2\)

Giả sử \(x\ge y\ge z>0\)

\(\Rightarrow\frac{1}{x}\le\frac{1}{y}\le\frac{1}{z}\)

\(\Rightarrow1+\frac{1}{x}\le1+\frac{1}{y}\le1+\frac{1}{z}\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\le \left(1+\frac{1}{z}\right)^3\)

\(\Rightarrow2\le\left(1+\frac{1}{z}\right)^3\)

\(\Rightarrow1+\frac{1}{z}\ge\sqrt[3]{2}\)
\(\Rightarrow\frac{1}{z}\ge\sqrt[3]{2}-1\)

\(\Rightarrow z\le\frac{1}{\sqrt[3]{2}-1}< 4\)

Mà z thuộc N^* \(\Rightarrow z\in\left\{1;2;3\right\}\)

TH1 : \(z=1\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{1}\right)=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=1\)

Ta có : \(1+\frac{1}{x}>1;1+\frac{1}{y}>1\)\(\Rightarrow\left(\frac{1}{x}+1\right)\left(1+\frac{1}{y}\right)>1\left(lọai\right)\)

TH2 : \(z=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{2}\right)=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)

Ta có : \(\left(1+\frac{1}{y}\right)^2\ge\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)

\(\Rightarrow1+\frac{1}{y}\ge\sqrt{\frac{4}{3}}\)

\(\Rightarrow\frac{1}{y}\ge\frac{2\sqrt{3}}{3}-1\)

\(\Rightarrow y\le\frac{1}{\frac{2\sqrt{3}}{3}-1}< 7\)

\(\Rightarrow y\in\left\{1;2;3;4;5;6\right\}\)

Nếu y = 1 \(\Rightarrow\left(1+1\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = -3 ( loại )

Nếu y = 2 \(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = -9 ( loại )

Nếu y = 3 \(\Rightarrow\left(1+\frac{1}{3}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > \(x\in\varnothing\)

Nếu y = 4 \(\Rightarrow\left(1+\frac{1}{4}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 15 ( tm )

Nếu y = 5 \(\Rightarrow\left(1+\frac{1}{5}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 9 ( tm )

Nếu y = 6 \(\Rightarrow\left(1+\frac{1}{6}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 7 ( tm )

TH3 : z =3 thì bạn làm tương tự nhé

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ĐKXĐ : \(x\ne-2,x\ne2\)

\(\frac{x+1}{x-2}+\frac{5}{2+x}=\frac{12}{x^2-4}+1\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\frac{5\left(x-2\right)}{x^2-4}=\frac{12}{x^2-4}+\frac{x^2-4}{x^2-4}\)

\(\Leftrightarrow x^2+3x+2-5x+10=8+x^2\)

\(\Leftrightarrow12-2x=8\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy phương trình vô nghiệm

\(\frac{x+1}{x-2}+\frac{5}{2+x}=\frac{12}{x^2-4}+1\)

ĐKXĐ: \(x\ne\pm2\)

\(\Leftrightarrow\frac{x+1}{x-2}+\frac{-5}{x-2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{12}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x^2+2x+x+2-5x-10=12+x^2+2x-2x-4\)

\(\Leftrightarrow x^2-2x-8=8+x^2\)

\(\Leftrightarrow x^2-2x-8-8-x^2=0\)

\(\Leftrightarrow-2x-16=0\)

\(\Leftrightarrow-2x=0+16\)

\(\Leftrightarrow-2x=16\)

\(\Leftrightarrow x=16:\left(-2\right)\)

\(\Leftrightarrow x=-8\)(TMĐKXĐ)

Vậy S = \(\left\{-8\right\}\)

a, Ta có \(\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=1\end{cases}}\)

Thay x = 1 ta được \(A=\frac{1+1}{2-1}=2\)

b, đk x khác -2 ; -1 

\(B=\frac{2x}{x+1}+\frac{3}{x-2}-\frac{2x^2+1}{x^2-x-2}\)

\(=\frac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}=-\frac{1}{x+1}\)

c, \(P=AB=\frac{-x\left(x+1\right)}{\left(x+1\right)\left(2-x\right)}=\frac{x}{x-2}=\frac{x-2+2}{x-2}=1+\frac{2}{x-2}\)

`Answer:`

\(\frac{x-2}{x+2}-\frac{x^2+2}{x^2+2x}=\frac{3}{x}\) ĐKXĐ: \(x\ne0;x\ne-2\)

\(\Leftrightarrow\frac{x\left(x-2\right)}{x\left(x+2\right)}-\frac{x^2+2}{x\left(x+2\right)}=\frac{3\left(x+2\right)}{x\left(x+2\right)}\)

\(\Rightarrow x\left(x-2\right)-\left(x^2+2\right)=3\left(x+2\right)\)

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

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

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\frac{8}{5}\)

tui ko bt

Ko bé ơi

Ghi đúng môn học nhé

a. PTHH: `HCl+NaOH->NaCl+H_2O`

                   0,6     0,6        0,6               mol

b. `300ml=0,3l`

`n_{HCl}=C_M . V=2.0,3=0,6mol`

\(\rightarrow n_{NaOH}=n_{HCl}=0,6mol\)

\(\rightarrow m_{NaOH}=n.M=0,6.40=24g\)

\(\rightarrow m_{dd_{NaOH}}=\frac{m.100\%}{C\%}=\frac{24.100\%}{20\%}=120g\)

c. Theo phương trình \(n_{NaCl}=n_{HCl}=0,6mol\)

\(\rightarrow m_{NaCl}=n.M=0,6.58,5=35,1g\)

ta chuyển vế ,đc PT sau:

\(x^2+\left(m-7\right)x+\frac{12}{x}-m^2+3m=0\)

\(\Leftrightarrow x^2+mx-7x+\frac{12}{x}-m^2+3m=0\left(1\right)\)

thay x=3 vào PT (1) ta có:

\(\left(1\right)\Leftrightarrow9+3m-3.7+\frac{12}{3}-m^2+3m=0\)

\(\Leftrightarrow-m^2+6m-8=0\)

\\(\Leftrightarrow m^2-6m+8=0\)

\(\Leftrightarrow m^2-6m+9-1=0\)

\(\Leftrightarrow\left(m-3\right)^2-1=0\)

\(\Leftrightarrow\left(m-4\right).\left(m-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}m-4=0\\m-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=4\\m=2\end{cases}}}\)