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

\(\frac{1}{a}+\frac{1}{b}\ge\sqrt{2}\)

\(< =>MIN=\sqrt{2}\)

dấu "=" xảy ra khi \(\frac{1}{a}=\frac{1}{b}\)

ÁP DỤNG BĐT \(\frac{1}{A}+\frac{1}{B}\ge\frac{4}{A+B}\), ĐẲNG THỨC XẢY RA <=> A = B TA CÓ :

\(\frac{1}{A}+\frac{1}{B}\ge\frac{4}{A+B}=\frac{4}{2\sqrt{2}}=\sqrt{2}\). ĐẲNG THỨC XẢY RA <=> A = B = √2

VẬY GTNN CỦA BIỂU THỨC = √2 <=> A = B = √2

Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\), ta có:

 \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+1+b+1+c+1}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

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

\(=\frac{3^2}{2\left(a+b+c\right)}=\frac{9}{2.4,5}=\frac{9}{9}=1\)

dấu "=" xảy ra khi \(\frac{1}{a+b}=\frac{1}{b+c}=\frac{1}{c+a}=\frac{1}{3}\)

hmm thật ra bạn ghi nhầm đề đúng hog , mk sửa lại nhá:(

Cho hàm số \(y=f\left(x\right)=\frac{1}{2}x+5\)

Tính f(0);    f(1);    f(2);   f(3);     f(-2);     f(-10)

\(f\left(0\right)=\frac{1}{2}.0+5=5\)

\(f\left(2\right)=\frac{1}{2}.2+5=6\)

\(f\left(3\right)=\frac{1}{2}.3+5=6,5\)

\(f\left(-2\right)=\frac{1}{2}.\left(-2\right)+5=4\)

\(f\left(-10\right)=\frac{1}{2}.\left(-10\right)+5=0\)

\(b,\sqrt{x-1}+\sqrt{4x-4}+\frac{1}{5}\sqrt{25x-25}=24\)

\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}+\frac{1}{5}\sqrt{25\left(x-1\right)}=24\)

\(\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=24\)

\(\sqrt{x-1}\left(1+2+1\right)=24\)

\(4\sqrt{x-1}=24\)

\(\sqrt{x-1}=6\)

\(x-1=36\)

\(x=37\left(TM\right)\)