khó vc

Từ giả thiết:

\(a^2=2\left(b^2+c^2\right)\ge\left(b+c\right)^2\Rightarrow\left(\dfrac{a}{b+c}\right)^2\ge1\Rightarrow\dfrac{a}{b+c}\ge1\)

\(P=\dfrac{a}{b+c}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\ge\dfrac{a}{b+c}+\dfrac{\left(b+c\right)^2}{a\left(b+c\right)+2bc}\ge\dfrac{a}{b+c}+\dfrac{\left(b+c\right)^2}{a\left(b+c\right)+\dfrac{1}{2}\left(b+c\right)^2}\)

\(P\ge\dfrac{a}{b+c}+\dfrac{1}{\dfrac{a}{b+c}+\dfrac{1}{2}}\)

Đặt \(\dfrac{a}{b+c}=x\ge1\)

\(\Rightarrow P\ge x+\dfrac{1}{x+\dfrac{1}{2}}=\dfrac{4}{9}\left(x+\dfrac{1}{2}\right)+\dfrac{1}{x+\dfrac{1}{2}}+\dfrac{5}{9}x-\dfrac{2}{9}\)

\(P\ge2\sqrt{\dfrac{4}{9}\left(x+\dfrac{1}{2}\right).\dfrac{1}{\left(x+\dfrac{1}{2}\right)}}+\dfrac{5}{9}.1-\dfrac{2}{9}=\dfrac{5}{3}\)

\(P_{min}=\dfrac{5}{3}\) khi \(x=1\) hay \(a=2b=2c\)

Mình sẽ giúp  bạn cho

mình giúp cho

1) \(2^{x+2}-96=2^x\)\(\Leftrightarrow2^{x+2}-2^x=96\)\(\Leftrightarrow2^x\left(2^2-1\right)=96\)

\(\Leftrightarrow3.2^x=96\)\(\Leftrightarrow2^x=32=2^5\)\(\Leftrightarrow x=5\)

Vậy \(x=5\)

2) \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

\(\Rightarrow a=b\), \(b=c\), \(c=a\)\(\Rightarrow a=b=c\)

Câu 1:

\(2^{x+2}-96=2^x\)

\(\Leftrightarrow2^{x+2}-2^x=96\)(chuyển vế nha bạn)

\(\Leftrightarrow2^x.\left(2^2-1\right)=96\)

\(\Leftrightarrow2^x.3=96\Rightarrow2^x=32=\left(+-6\right)^2\)

\(\Rightarrow x=2\)

Câu 2:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow a=b.1=b\)và \(b=c.1=c\)và \(c=a.1=a\)

\(\Rightarrow a=b=c\)

5x + 2 chia hết cho 9 - 2x
=> 2(5x + 2) = 10x + 4 chia hết cho 9 - 2x
=> 10x + 4 + 5(9 - 2x) = 10x + 4 + 45 - 10x = 49 chia hết cho 9 - 2x
=> 9 - 2x thuộc Ư(49) = {1, 7, 49}
=> 2x thuộc {8, 2, -40}
=> x thuộc {1, 4, -20}
Vậy x thuộc {1, 4, -20}

Học tốt nhé!