\(A=x-\sqrt{x}\)   \(\left(ĐKXĐ:x\ge0\right)\)

\(A=x-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\) 

\(A=\left(x-\frac{1}{2}\right)^2\) \(-\frac{1}{4}\) 

Có \(\left(x-\frac{1}{2^2}\right)\ge0\forall x\ge0\) 

     \(\left(x-\frac{1}{2}\right)^2\) -    1/4  >= \(\frac{-1}{4}\)mọi x>=0

   Dấu = sảy ra \(\Leftrightarrow\) x- \(\frac{1}{2}\) = 0

                        \(\Leftrightarrow\) x = 1 / 2  (  t/m  ) 

 vậy A đạt GTNN là -1/4 tại x = 1/2

Tớ nhầm nhé \(x\) từ dòng thứ 3 xuống pahir thay =\(\sqrt{x}\)

A=\(x-\sqrt{x}\)  ( ĐKXĐ x >=0 )

   = \(x-2.\sqrt{x}.\frac{1}{2}\) \(+\frac{1}{4}-\frac{1}{4}\) 

   = \(\left(\sqrt{x}-\frac{1}{2}\right)^2\) -  \(\frac{1}{4}\) 

Có \(\left(\sqrt{x}-\frac{1}{2}\right)^2\) \(\ge0\forall x\ge0\)

  \(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2\) \(-\frac{1}{4}\ge\frac{-1}{4}\forall x\) >= 0

dấu = sảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}\)= 0 \(\Leftrightarrow\) x = \(\frac{1}{4}\) ( t/m )

Vậy A đạt GTNN là \(\frac{-1}{4}\) tại x= \(\frac{1}{4}\)

Vừa nãy tớ làm sai 

Làm sai thì thôi nha!

ĐK: x>=0 \(A=x-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

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

đẳng thức xảy ra khi x = 1/4

A=\(x-\sqrt{x}=x-2\times\sqrt{x}\times\frac{1}{2}+\frac{1}{4}-\frac{1}{4}=(\sqrt{x}-\frac{1}{2})^2+\frac{1}{4}\ge\frac{1}{4}\)

GTNN A=\(\frac{1}{4}\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

mk viết nhầm GTNN A=\(\frac{-1}{4}\) mới đúng

11111111110000000000000000000000000

1111111111000000000000000000000000

=1.111111111x10²⁴

Sắp xếp lại theo thứ tự giảm dần của biến:

\(P\left(x\right)=-5x^3-\frac{1}{3}+8x^4+x=8x^4-5x^3+x-\frac{1}{3}\)

\(Q\left(x\right)=x^2-5x-2x^3+x^4-\frac{2}{3}=x^4-2x^3+x^2-5x-\frac{2}{3}\)

\(\Rightarrow P\left(x\right)+Q\left(x\right)=\left(8x^4-5x^3+x-\frac{1}{3}\right)+\left(x^4-2x^3+x^2-5x-\frac{2}{3}\right)\)

\(=\left(8x^4+x^4\right)-\left(5x^3+2x^3\right)+x^2-\left(5x-x\right)-\left(\frac{1}{3}+\frac{2}{3}\right)\)

\(=9x^4-7x^3+x^2-4x-1\)

Câu b tương tự

1+2=3

kb nha🙎 🙎 🙎 🙎 🙎

giúp mik zứi😭😭😭

a; 60 \(\times\) (\(\dfrac{7}{12}\) + \(\dfrac{4}{15}\)) 

 =  60 \(\times\) \(\dfrac{7}{12}\) + 60 \(\times\) \(\dfrac{4}{15}\)

= 35 + 16

= 51

b; \(\dfrac{1}{2}\) \(\times\)\(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\) \(\times\) \(\dfrac{5}{6}\) \(\times\) \(\dfrac{7}{8}\) \(\times\) \(\dfrac{8}{9}\)

= \(\dfrac{2\times3\times4\times5\times6\times7\times8}{2\times3\times4\times5\times6\times7\times8}\) \(\times\) \(\dfrac{1}{9}\)

= 1 \(\times\) \(\dfrac{1}{9}\)

= \(\dfrac{1}{9}\)

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}ĐK:x\ne1;-7\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2+7x-2x-14=x^2-x+4x-4\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow x^2+5x-14-x^2-3x+4=0\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)

\(2xy-x-y=3\)

\(\Rightarrow4xy-2x-2y=6\)

\(\Rightarrow2x\left(2y-1\right)-\left(2y-1\right)=7\)

\(\Rightarrow\left(2y-1\right)\left(2x-1\right)=7=1\left(-7\right)=\left(-1\right)7=\left(-7\right)1=7\left(-1\right)\)

Làm nốt

\(\frac{y}{3}-\frac{1}{x}=\frac{1}{3}\)

\(\Rightarrow\frac{xy-3}{3x}=\frac{1}{3}\)

\(\Rightarrow3xy-9=3x\)

\(\Rightarrow3xy-3x=9\)

\(\Rightarrow3x\left(y-1\right)=9=3\cdot3=\left(-3\right)\left(-3\right)=9\cdot1=\left(-9\right)\left(-1\right)=\left(-1\right)\left(-9\right)=1\cdot9\)

Làm nốt.

\(xy-x+2y=3\)

\(\Rightarrow x\left(y-1\right)+2\left(y-1\right)=1\)

\(\Rightarrow\left(y-1\right)\left(x+2\right)=1=1\cdot1=\left(-1\right)\left(-1\right)\)

Tới đây bí=((