Câu hỏi của nguyen phuong thao - Toán lớp 7 - Học toán với OnlineMath

\(\dfrac{a+4b-c}{c}=\dfrac{c+4a-b}{b}=\dfrac{b+4c-a}{a}\\ \Rightarrow\dfrac{a+4b+c}{c}=\dfrac{c+4a+b}{b}=\dfrac{b+4c+a}{a}\\ =\dfrac{6\left(a+b+c\right)}{a+b+c}=6\\ \Rightarrow\dfrac{1}{a+b+c}=6\\ \Rightarrow a+b+c=\dfrac{1}{6}\)

Tới này dễ rồi

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a) (x + y + z)² \(\le3\left(x^2+y^2+z^2\right)\)(1) 

<=> \(x^2+y^2+z^2+2xy+2yz+2zx\le3x^2+3y^2+3z^2\)

<=> \(2x^2+2y^2+2z^2-2xy-2xz-2yz\ge0\)

<=> (x - y)² + (y - z)² + (z - x)² \(\ge0\) (đúng) 

=> (1) đúng "=" khi x = y = z

b) \(A=1\sqrt{4a+1}+1.\sqrt{4b+1}+1.\sqrt{4c+1}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4a+1+4b+1+4c+1\right)}\)

\(=\sqrt{3.\left[4\left(a+b+c\right)+3\right]}=\sqrt{21}\left(\text{vì }a+b+c=1\right)\)

"=" xảy ra <=> \(\dfrac{1}{\sqrt{4a+1}}=\dfrac{1}{\sqrt{4b+1}}=\dfrac{1}{\sqrt{4c+1}};a+b+c=1\)

<=> a = b = c = 1/3 

\(M=\frac{\left(a+1\right)^2+2a}{a\left(a+1\right)}+\frac{\left(b+1\right)^2+2b}{b\left(b+1\right)}+\frac{\left(c+1\right)^2+2c}{c\left(c+1\right)}\)

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

\(M=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)

\(M\ge3+\frac{9}{a+b+c}+2\left(\frac{9}{a+b+c+3}\right)\ge3+3+3=9\)

Dấu "=" xảy ra khi a=b=c=1

\(P^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\)

\(\le\left(1^2+1^2+1^2\right)\left(4a+3+4b+3+3c+3\right)\)

\(=63\)

\(\Rightarrow P\le\sqrt{63}=3\sqrt{7}\).

Dấu \(=\)khi \(\hept{\begin{cases}4a+3=4b+3=4c+3\\a+b+c=3\end{cases}}\Leftrightarrow a=b=c=1\).

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r