I often call Nhuan Trach Primary School is my second home.
School yard, through not very large but very spacious and clean. High school entrance and spacious, the two sides is the flower corridor leads straight to the playground and the classroom.

Xet \(P-\frac{1}{3}=\frac{x^2-x+1}{x^2+x+1}-\frac{1}{3}=\frac{3x^2-3x+3-\left(x^2+x+1\right)}{x^2+x+1}=\frac{2x^2-4x+2}{x^2+x+1}\)

 =\(\frac{2\left(x^2-2x+1\right)}{x^2+x+1}=\frac{2\left(x-1\right)^2}{x^2+x+1}\ge0\) (do \(x^2+x+1>0\forall x\) )

Suy ra \(P\ge\frac{1}{3}\)

Dau = xay ra khi \(x-1=0\Leftrightarrow x=1\)

Ta CM 1 số BĐT phụ sau :

\(\left(a+b\right)^2\ge4ab\Leftrightarrow a^2+2ab-4ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\left(true\right)\)

và \(x^2+x+1=x^2+2x+1-x\ge\left(x+1\right)^2-\frac{\left(x+1\right)^2}{4}=\frac{3\left(x+1\right)^2}{4}\)

\(\Rightarrow P=\frac{x^2-x+1}{x^2+x+1}=1-\frac{2x}{x^2+x+1}\)

\(\ge1-\frac{\frac{\left(x+1\right)^2}{2}}{x^2+x+1}\ge1-\frac{\frac{\left(x+1\right)^2}{2}}{\frac{3\left(x+1\right)^2}{4}}=1-\frac{2}{3}=\frac{1}{3}\)

Dấu "=" xảy ra khi \(x+1=0\Leftrightarrow x=-1\)

\(\frac{1^{2n-1}}{2}=\frac{1}{8}\) 

\(1^{2n-1}=1\cdot2:8\)  

\(1^{2n-1}=\frac{1}{4}\)            ( vô lí vì \(1^{2n-1}=1\forall n\)

Vậy không có n thỏa mãn 

\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4.\left(1^{2n-1}\right)}{8}=\frac{1}{8}\)

\(\Leftrightarrow1^{2n-1}=\frac{1}{4}\)

\(\Leftrightarrow1^{2n}=\frac{1}{4}\)

\(\Leftrightarrow1^n.1^2=\frac{1}{4}\)

\(\Leftrightarrow n=-4\)

People don't use this road very often.

==> This road didn't use to use.....................

People don't use this road very often

-People didn't use this road very often

k nhé!Học tốt!

\(\frac{-32}{-2^n}=4\)

\(\Leftrightarrow-2^n=-8\)

\(\Leftrightarrow n=3\)

\(\frac{-32}{-2^n}=4\)   

\(\frac{32}{2^n}=4\)   

\(\frac{2^5}{2^n}=2^2\)    

\(2^{5-n}=2^2\)   

5 - n = 2

n  =3

\(\left(2x+1\right)^3=125\)

\(\Leftrightarrow2x+1=15\)

\(\Leftrightarrow2x=14\Leftrightarrow x=7\)

\(\left(2x+1\right)^3=125\) 

\(\left(2x+1\right)^3=5^3\) 

\(2x+1=5\) 

\(x=2\)

a) \(\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\)

=> \(\frac{12}{13}x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)

=> \(x=2:\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)

b) \(x:\frac{13}{3}=-2,5\)

=> \(x:\frac{13}{3}=-\frac{5}{2}\)

=> \(x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)

c) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

=> \(\frac{4x-3}{12}=-\frac{10}{12}\)

=> 4x - 3 = -10

=> 4x = -10 + 3 = -7

=> x = -7/4

Bài 2 :

\(A=a\cdot\frac{1}{3}+a\cdot\frac{1}{4}-a\cdot\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\cdot\frac{5}{12}\)

Thay a = -3/5 vào biểu thức ta có : \(A=\left(-\frac{3}{5}\right)\cdot\frac{5}{12}=\frac{-3}{12}=\frac{-1}{4}\)

\(B=b\cdot\frac{5}{6}+b\cdot\frac{3}{4}-b\cdot\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\cdot\frac{13}{12}\)

Thay b = 12/13 vào ta được kết quả là 1

a ) \(\frac{25}{9}-\frac{12}{13}\cdot x=\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}\cdot x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)

\(\Rightarrow x=2\div\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)

Vậy ...

b ) \(x\div\frac{13}{3}=-\frac{5}{2}\)

\(\Rightarrow x\div\frac{13}{3}=-\frac{5}{2}\)

\(\Rightarrow x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)

Vậy ..

c ) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

\(\Rightarrow\frac{4x-3}{12}=-\frac{10}{12}\)

\(\Rightarrow4x-3=-10\)

\(\Rightarrow4x=-10+3=-7\)

\(\Rightarrow x=-\frac{7}{4}\)

Vậy ....

cái đó là toán lớp 7 mất rồi