đe có gì sai , bn 

đề sai rồi

Bài 3:

a, Đặt \(A=\left|2x-\frac{1}{5}\right|+2017\)

Để A đạt GTNN thì \(\left|2x-\frac{1}{5}\right|\)đạt GTNN

Mà \(\left|2x-\frac{1}{5}\right|\ge0\)

Do đó \(\left|2x-\frac{1}{5}\right|=0\)thì A đạt GTNN tức là A = 0 + 2017 = 2017 khi

\(2x-\frac{1}{5}=0=>2x=0+\frac{1}{5}=\frac{1}{5}=>x=\frac{1}{5}.\frac{1}{2}=\frac{1}{10}\)

b, Đặt \(B=\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|\)

Ta thấy \(\frac{1}{2}>\frac{1}{3}>\frac{1}{4}=>x+\frac{1}{2}>x+\frac{1}{3}>x+\frac{1}{4}\)

Do đó để B đạt GTNN thì \(x+\frac{1}{2}\)đạt GTNN

mà \(x+\frac{1}{2}\ge0\)

Từ 2 điều trên => \(x+\frac{1}{2}=0=>x=-\frac{1}{2}\)

Khi đó \(x+\frac{1}{3}=-\frac{1}{2}+\frac{1}{3}=-\frac{1}{6}\)

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

Vậy GTNN của \(B=\left|0\right|+\left|-\frac{1}{6}\right|+\left|-\frac{1}{4}\right|=0+\frac{1}{6}+\frac{1}{4}=\frac{10}{24}\)khi x = -1/2

Phần b này thì mình không chắc lắm bạn tự xem lại nhé

Bài 1: 

\(M=\frac{2017}{11-x}\)đạt GTLN <=> 11 - x đạt GTNN và 11 - x > 0 (nếu không thì M đạt giá trị âm (vô lí))

=> 11 - x = 1

=> x = 10

Vậy x = 10 thì M đạt GTLN tức là bằng \(\frac{2017}{1}=2017\)

Xét hiệu

(x+y+z)³-x³-y³-z³=[(x+y)+z]³-x³-y³-z³

=(x+y)³+3(x+y)²z+3(x+y)z²+z³-x³-y³-z³

=(x+y)³+3z(x+y)(x+y+z)+z³-x³-y³-z³

=x³+3x²y+3xy²+y³+3z(x+y)(x+y+z)+z³-x³-y³-z³

=(x³-x³)+(y³-y³)+(z³-z³)+3xy(x+y)+3z(x+y)(x+y+z)

=3(x+y)(xy+yz+xz+z²)

=3(x+y)[y(x+z)+z(x+z)]

=3(x+y)(x+z)(y+z

=>(x+y+z)³-x³-y³-z³=3(x+y)(y+z)(z+x)

=>(x+y+z)³=3(x+y)(y+z)(z+x)+x³+y³+z³

^{=> DPCM}

ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3\)\(=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)\)

\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+\left(x+y+z\right)x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz+x^2+xy+xz+x^2-y^2+yz-z^2\right)\)

\(=\left(y+z\right)\left(3x^2+3xy+3xz+3yz\right)\)

\(=\left(y+z\right)\left[3x\left(x+y\right)+3z\left(x+y\right)\right]\)

\(=\left(y+z\right)\left(x+y\right)\left(3x+3z\right)\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)(đpcm)

a.    5 : x = x : 125      

x=125:5

x=25  

b. 2x + 3  = 4 + 3x

2x-3x=4-3

-x=1

x=-1

\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)=\(\frac{\frac{8}{27}.\frac{9}{16}.-1}{\frac{4}{25}.\frac{-125}{1728}}\)=\(\frac{\frac{-1}{6}}{-\frac{5}{432}}\)=\(\frac{-1}{6}:\frac{-5}{432}=\frac{-1}{6}.-\frac{432}{5}=\frac{72}{5}\)

Bài này dễ mà bn

Bằng 72/5 nhé

\(x+\frac{2}{5.9}+\frac{2}{9.13}+\frac{2}{13.17}+...+\frac{2}{41.45}=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{2}{4}\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\frac{8}{45}\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\frac{4}{45}=\frac{-37}{45}\)

\(\Leftrightarrow x=-\frac{41}{45}\)