Đặt \(\hept{\begin{cases}b+c+d=x>0\\a+c+d=y>0\\a+b+d=z>0\end{cases}}\)và \(a+b+c=t>0\)

\(\Rightarrow\hept{\begin{cases}a=\frac{y+z+t-2x}{3}\\b=\frac{x+z+t-2y}{3}\\c=\frac{x+y+t-2z}{3}\end{cases}}\)và \(d=\frac{x+y+z-2t}{3}\)

Từ đó ta có:\(Q=\frac{y+z+t-2x}{3x}+\frac{x+z+t-2y}{3y}+\frac{x+y+t-2z}{3z}+\frac{x+y+z-2t}{3t}\)

\(=\frac{y}{3x}+\frac{z}{3x}+\frac{t}{3x}-\frac{2}{3}+\frac{x}{3y}+\frac{z}{3y}+\frac{t}{3y}-\frac{2}{3}+\frac{x}{3z}+\frac{y}{3z}+\frac{t}{3z}-\frac{2}{3}+\frac{x}{3t}+\frac{y}{3t}+\frac{z}{3t}-\frac{2}{3}\)

\(=\left(\frac{y}{3x}+\frac{x}{3y}\right)+\left(\frac{z}{3x}+\frac{x}{3z}\right)+\left(\frac{t}{3x}+\frac{x}{3t}\right)+\left(\frac{z}{3y}+\frac{y}{3z}\right)+\left(\frac{t}{3y}+\frac{y}{3t}\right)+\left(\frac{t}{3z}+\frac{z}{3t}\right)-\frac{8}{3}\)

Áp dụng BĐT AM-GM ta được:

\(\frac{y}{3x}+\frac{x}{3y}\ge2\sqrt{\frac{y}{3x}.\frac{x}{3y}}=\frac{2}{3}\)

CMTT \(\Rightarrow Q\ge\frac{4}{3}\)

Dấu"="xảy ra \(\Leftrightarrow a=b=c=d\)

a) P = \(\left(\frac{x}{x-1}+\frac{1}{x^2-x}\right):\left(\frac{1}{x+1}+\frac{2}{x^2+1}\right)\)

=> P = \(\left(\frac{x^2}{\left(x-1\right)x}+\frac{1}{x\left(x-1\right)}\right):\left(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2}{\left(x+1\right)\left(x-1\right)}\right)\)

=> P = \(\left(\frac{x^2+1}{x\left(x-1\right)}\right):\left(\frac{x-1+2}{\left(x+1\right)\left(x-1\right)}\right)\)

=> P = \(\frac{x^2+1}{x\left(x-1\right)}:\frac{x+1}{\left(x+1\right)\left(x-1\right)}\)

=> P = \(\frac{x^2+1}{x\left(x-1\right)}\cdot\left(x-1\right)\)

=> P = \(\frac{x^2+1}{x}\)

b) ĐKXĐ: x \(\ne\)0; x \(\ne\)\(\pm\)1

Để P > -1

=> \(\frac{x^2+1}{x}>-1\)

=> \(\frac{x^2+1}{x}+1>0\)

=> \(\frac{x^2+1+x}{x}>0\)

Do x² + x + 1 > 0 \(\forall\)x (vì x² + x + 1 = x² + x + 1/4 + 3/4 = (x + 1/2)² + 3/4 > 0 : giải thích)

=> x > 0

Vậy để P > -1 <=> x > 0 và x \(\ne\)1

a)

\(P=\left(\frac{x}{x-1}+\frac{1}{x^2-x}\right):\left(\frac{1}{x+1}+\frac{1}{x^2+1}\right)\)

\(P=\left(\frac{x}{x-1}+\frac{1}{x\left(x-1\right)}\right):\left(\frac{1}{x+1}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right)\)

\(P=\left(\frac{x^2}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}\right):\left(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right)\)

\(P=\frac{x^2+1}{x\left(x-1\right)}:\frac{x-1}{\left(x+1\right)\left(x-1\right)}\)

\(P=\frac{x^2+1}{x\left(x-1\right)}:\frac{1}{x+1}\)

?????????????????? Đề 

tự làm nốt k hiểu đề cho sai à 

giúp em với mọi người ơi:<<<<<

tinh chu vi hinh tu giacco do dai cac cac canh lan luot la 35cm,30cm,25cm,24cm

CHU VI LÀ 35+30+25+24=114

Ngại làm quá nên chỉ lấy hình , vào TKHĐ là thấy :

Điều kiện: Tinh bột phải chín, trong khi ăn phải nhai kĩ

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