\(\dfrac{1}{9}=\dfrac{1}{27}\)

Làm gì có x nhỉ?

x đâu

b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)

\(\Rightarrow50x\ge0\Rightarrow x\ge0\)

Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)

Thay (1) vào đề bài:

\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)

\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)

\(\Rightarrow49x+\dfrac{16}{99}=50x\)

\(\Rightarrow x=\dfrac{16}{99}\)

Vậy \(x=\dfrac{16}{99}.\)

\(P=\dfrac{1}{a\left(2b+2c-1\right)}+\dfrac{1}{b\left(2c+2a-1\right)}+\dfrac{1}{c\left(2a+2b-1\right)}\)

\(P=\dfrac{1}{a\left[2b+2c-\left(a+b+c\right)\right]}+\dfrac{1}{b\left[2c+2a-\left(a+b+c\right)\right]}+\dfrac{1}{c\left[2a+2b-\left(a+b+c\right)\right]}\)

\(P=\dfrac{1}{a\left(b+c-a\right)}+\dfrac{1}{b\left(c+a-b\right)}+\dfrac{1}{c\left(a+b-c\right)}\)

\(P=\dfrac{1}{ab+ac-a^2}+\dfrac{1}{bc+ab-b^2}+\dfrac{1}{ca+bc-c^2}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow P\ge\dfrac{\left(1+1+1\right)^2}{-a^2-b^2-c^2+2ab+2bc+2ca}=\dfrac{9}{-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]}\) ( 1 )

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow a^2+b^2+c^2-2\left(ab+bc+ca\right)\ge-\left(ab+bc+ca\right)\)

\(\Rightarrow-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]\le ab+bc+ca\)

\(\Rightarrow\dfrac{9}{-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]}\ge\dfrac{9}{ab+bc+ca}\)

Từ ( 1 )

\(\Rightarrow P\ge\dfrac{9}{ab+bc+ca}\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow1\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{1}{3}\ge ab+bc+ca\)

\(\Rightarrow27\le\dfrac{9}{ab+bc+ca}\)

\(\Rightarrow P\ge27\)

Vậy \(P_{min}=27\)

a: \(B=\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)

\(=\dfrac{3x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+2}\)

\(=\dfrac{3}{x-3}\)

b: |2x+1|=5

=>2x+1=5 hoặc 2x+1=-5

=>x=-3(loại) hoặc x=2(nhận)

Khi x=2 thì \(B=\dfrac{3}{2-3}=-3\)

c: Để B=-3/5 thì x-3=-5

=>x=-2(loại)

d: Để B<0 thì x-3<0

=>x<3

4:

\(1.7-\left|x-2016\right|< =1.7\)

Dấu = xảy ra khi x=2016

7:

=>-25<=2x-7<=25

=>-18<=2x<=32

=>-9<=x<=16

=>Số phần tử thỏa mãn là 16 phần tử

đề sai