\(6\left(\frac{-2}{3}\right)+12\left(\frac{-2}{3}\right)^2+18\left(\frac{-2}{3}\right)^2\)

\(=\frac{-2}{3}\times6\times\left[1+2\left(\frac{-2}{3}\right)+3\left(\frac{-2}{3}\right)\right]\)

\(=-4\times\left[1+\left(\frac{-2}{3}\right)\left(2+3\right)\right]\)

\(=-4\times\left(1+\frac{-10}{3}\right)\)

\(=-4+\frac{40}{3}\)

\(=-\frac{12}{3}+\frac{40}{3}\)

\(=\frac{28}{3}\)

\(18\left(\frac{-2}{3}\right)^3\) MÀ BẠN

CM: a) Xét t/giác ABE và t/giác DBE

có :  AB = BD (gt)

    \(\widehat{A}=\widehat{D_1}=90^0\) (gt)

   BE : chung

=> t/giác ABE = t/giác DBE (ch - cgv)

=> \(\widehat{B_1}=\widehat{B_2}\) (2 góc t/ứng)

=> BE là tia p/giác của \(\widehat{ABC}\)

b) Xét t/giác ABC có \(\widehat{A}=90^0\) => \(\widehat{B}+\widehat{C}=90^0\)

Xét t/giác DEC có \(\widehat{D_2}=90^0\) => \(\widehat{E_1}+\widehat{C}=90^0\)

=> \(\widehat{B}=\widehat{E_1}\) 

 mà \(\widehat{B_1}=\widehat{B_2}=\frac{\widehat{B}}{2}\) (cmt)

          => \(\frac{\widehat{E_1}}{2}=\widehat{B_1}\) =>  \(\widehat{B_1}=\frac{1}{2}\widehat{E_1}\) hay \(\widehat{ABE}=\frac{1}{2}\widehat{CED}\)

A = x^2 + 17/x^2 + 7

= x^2 + 7 + 10/x^2 + 7

= 1 + 10/x^2 +  7

để A đạt giá trị lớn nhất thì 10/x^2 + 7 lớn nhất

=> x^2 + 7 nhỏ nhất

mà x^2 > 0 => x^2 + 7 > 7 

=> x^2 + 7 = 7

=> x^2 = 0

=> x = 0

vậy_

là 2,22

https://h.vn/hoi-dap/question/242583.html

\(|x-1|+|x-2|\)=1

<=> x-1 + x -2 = 1 <=> 2x = 4 

<=> x=2 

hk tốt 

|x - 1| + |x - 2| = 1

=> x - 1 + x - 2 =  1

=> 2x - 3 = 1

=> 2x = 4

=> x = 2

vậy_

a) Thiếu đề

b) Áp dụng t/c của dãy tỉ số bằng nhau, ta có :

 \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) => \(\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}=\frac{4x+3y+2z}{4+6+6}=\frac{14}{16}=\frac{7}{8}\)

=> \(\hept{\begin{cases}\frac{x}{1}=\frac{7}{8}\\\frac{y}{2}=\frac{7}{8}\\\frac{z}{3}=\frac{7}{8}\end{cases}}\) => \(\hept{\begin{cases}x=\frac{7}{8}.1=\frac{7}{8}\\y=\frac{7}{8}.2=\frac{7}{4}\\z=\frac{7}{8}.3=\frac{21}{8}\end{cases}}\)

Vậy ...

Sửa lại xíu :

 \(a)\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và \(x-2y+3z=14\)

\(b)\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)và \(4x+3y+2z=36\)

Ta có : \(x=\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{1}{1008.2015}\)

            \(y=\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)

Vì \(\frac{1}{1008.2015}>\frac{1}{1008.2017}\)

=> \(1+\frac{1}{1008.2015}>1+\frac{1}{1008.2017}\)

=> \(\frac{2015.2016+2}{2015.2016}>\frac{2016.2017+2}{2016.2017}\)

=> \(x>y\)

Ta có:

x = \(\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{2}{2015.2016}=1+\frac{1}{2015.1008}\)

y = \(\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)

Do \(\frac{1}{2015.1008}>\frac{1}{1008.2017}\) => \(1+\frac{1}{2015.1008}>1+\frac{1}{1008.2017}\)

     => x > y