\(2.|5x-3|-x=\)\(7\)

\(\Leftrightarrow\) \(2.|5x-3|=7+x\)

\(\Leftrightarrow\) \(|5x-3|=\frac{7+x}{2}\)

\(\Leftrightarrow\)\(5x-3=\pm\left(\frac{7+x}{2}\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}5x-3=\frac{7+x}{2}\\5x-3=\frac{-7-x}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{2.\left(5x-3\right)}{2}=\frac{7+x}{2}\\\frac{2.\left(5x-3\right)}{2}=\frac{-7-x}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{10x-6}{2}=\frac{7+x}{2}\\\frac{10x-6}{2}=\frac{-7-x}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}10x-6=7+x\\10x-6=-7-x\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}10x-x=7+6\\10x+x=-7+6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}9x=13\\11x=-1\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{13}{9}\\x=\frac{-1}{11}\end{cases}}\)

Vậy x = \(\frac{13}{9}\) , x = \(\frac{-1}{11}\)

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow\frac{x+z}{2+4}=\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) mà x + z = 18

\(\Rightarrow\frac{18}{6}=\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow3=\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow\hept{\begin{cases}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{cases}}\)

#)Giải :

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3\)

\(\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{4}=3\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=12\end{cases}}}\)

Vậy x = 6; y = 9; z = 12

#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

đến 1/5 là xuống phần mẫu số nhé

a)\(=\frac{3^7}{2^2.3^4}=\frac{3^3}{2^2}=\frac{27}{4}\)

b)\(=\frac{2^{20}+2^{12}}{2^{10}+2^{18}}=\frac{2^{12}\left(2^8+1\right)}{2^{10}\left(1+2^8\right)}=\frac{2^{12}}{2^{10}}=2^2=4\)

ý vậy đó hở 

 láo mầy olm láo  

Thay \(x\ge4\) vào E:

\(E=1+8=9\) 

Vậy \(E\ge9\) tại  \(x\ge4\)

TL:

thay x=2 vào F:

\(F=3.\left|2-2\right|-2.2=-4\) 

Vậy F=-4 tại x=2

va go 123 =5=6=3=7=4 \6\543[ 645eu] Ơ -4387

a) \(\left|2y-3\right|-\frac{1}{7}=\frac{3}{4}\)

=> \(\left|2y-3\right|=\frac{3}{4}+\frac{1}{7}\)

=> \(\left|2y-3\right|=\frac{25}{28}\)

=> \(\orbr{\begin{cases}2y-3=\frac{25}{28}\\2y-3=-\frac{25}{28}\end{cases}}\)

=> \(\orbr{\begin{cases}2y=\frac{109}{28}\\2y=\frac{59}{28}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{109}{56}\\x=\frac{59}{56}\end{cases}}\)

Tính GTLN

a) Ta có: -|2x - 5| \(\le\)0 \(\forall\)x

=> -|2x - 5| + 32 \(\le\)32 \(\forall\)x

Hay A \(\le\)32 \(\forall\)x

Dấu "=" xảy ra khi : 2x - 5 = 0 <=> 2x = 5 <=> x = 5/2

Vậy Max của A = 32 tại x = 5/2

\(C=\left|y^2+1\right|+2020\)

Ta có: \(y^2\ge0\Leftrightarrow y^2+1\ge1\Leftrightarrow\left|y^2+1\right|\ge1\)

\(\Leftrightarrow C=\left|y^2+1\right|+2020\ge2021\)

Vậy \(C_{min}=2021\)

(Dấu "="\(\Leftrightarrow y^2+1=1\Leftrightarrow y^2=0\Leftrightarrow y=0\))

a) \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\frac{1}{3}:x=\frac{-1}{15}\)

\(x=-5\)

vậy ...

\(a,\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)

\(\frac{1}{3}:x=-\frac{1}{15}\)

\(x=\frac{1}{3}:\left(-\frac{1}{15}\right)\)

\(x=-5\)

câu b e chưa nghĩ ra =.=