câu hỏi ko tl cx thấy xàm xàm xàm xmà

1. a) Ta có: M  = |x + 15/19| \(\ge\)0 \(\forall\)x

Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19

Vậy MinM = 0 <=> x = -15/19

b) Ta có: N = |x  - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x

Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7

Vậy MinN = -1/2 <=> x = 4/7

2a) Ta có: P = -|5/3 - x|  \(\le\)0 \(\forall\)x

Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3

Vậy MaxP = 0 <=> x = 5/3

b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x

Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10

Vậy MaxQ = 9 <=> x = 1/10

3a) Ta có:

|x - y - 5| + 2007.(y - 3)²⁰⁰⁴ = 0

<=> \(\hept{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=y+5\\y=3\end{cases}}\)

<=> \(\hept{\begin{cases}x=8\\y=3\end{cases}}\)

b) Ta có :

(x + y)²⁰¹⁶ + 2007.|y - 1| = 0

<=> \(\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-y\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

c) (x - 1)² + (y + 3)² = 0

<=>  \(\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

\(A=\left(a+b+1\right)\left(a^2+b^2\right)+\frac{4}{a+b}+1-1\ge\left(a+b+1\right)2\sqrt{\left(ab\right)^2}+\frac{\left(2+1\right)^2}{a+b+1}-1\)

\(=2\left(a+b+1\right)+\frac{9}{a+b+1}-1\ge2\sqrt{ab}+1+2\sqrt{\frac{9\left(a+b+1\right)}{a+b+1}}-1\ge2+6=8\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}a^2=b^2\left(1\right)\\\frac{2}{a+b}=1\left(2\right)\\a+b+1=\frac{9}{a+b+1}\left(3\right)\end{cases}}\)

pt \(\left(1\right)\)\(\Leftrightarrow\)\(a=b\) ( vì a, b > 0 ) 

pt \(\left(2\right)\)\(\Leftrightarrow\)\(a=b=1\)

pt \(\left(3\right)\)\(\Leftrightarrow\)\(\left(a+b+1\right)^2=9\)\(\Leftrightarrow\)\(a+b+1=3\) ( đúng vì \(a=b=1\) ) 

Vậy GTNN của \(A\) là \(8\) khi \(a=b=1\)

Chúc bạn học tốt ~ 

Ta co:

\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=\left(\frac{1}{a}-2\right)^2+\left(\frac{1}{b}-2\right)^2+6\left(\frac{1}{a}+\frac{1}{b}\right)-6\ge\frac{24}{a+b}-6=18\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Ta có : \(ab+bc+ca=2abc\)

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

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có :

\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{12}\)

Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)

Ta có : \(ab+bc+ca=2abc\)

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

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z^2\right)}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)

Please !!!!!