a) \(3.\left(10.x\right)=111\)

\(10.x=37\)

\(x=\dfrac{37}{10}\)

b) \(3.\left(10+x\right)=111\)

\(10+x=37\)

\(x=27\)

c) \(3+\left(10.x\right)=111\)

\(10.x=108\)

\(x=\dfrac{54}{5}\)

d) \(3+\left(10+x\right)=111\)

\(x=111-3-10\)

\(x=98\)

Ta có:\(f\left(x\right)-1=\left(x-1\right)^3\)

\(=>A+\frac{1}{2}=\left(\frac{1}{112}-1\right)^3+\left(\frac{2}{112}-1\right)^3+\left(\frac{3}{112}-1\right)^3+...\left(\frac{111}{112}-1\right)^3\)

\(A+\frac{1}{2}=-\frac{1^3+2^3+3^3+...+111^3}{112^3}=-\frac{\frac{111^2\left(111+1\right)^2}{4}}{112^3}=-\frac{111^2}{4\cdot112}=-\frac{12321}{448}\)

\(A=-\frac{12321}{448}-\frac{1}{2}=-\frac{12545}{448}\)

à nhầm :v

cho \(a\)và \(1-a\), ta có:

\(f\left(1-a\right)=\frac{\left(1-a\right)^3}{3\left(1-a\right)^2-3\left(1-a\right)+1}=\frac{\left(1-a-1\right)^3}{3-6a+a^2-3+3a+1}+1=1-\frac{a^3}{3a^3-3a+1}=1-f\left(a\right)\)

hay \(f\left(a\right)+f\left(1-a\right)=1\)

\(=>A=f\left(\frac{1}{112}\right)+f\left(\frac{111}{112}\right)+f\left(\frac{2}{112}\right)+f\left(\frac{110}{112}\right)+...+f\left(\frac{55}{112}\right)+f\left(\frac{57}{112}\right)+f\left(\frac{56}{112}\right)-\frac{1}{2}\)

\(=>A=55+f\left(\frac{1}{2}\right)-\frac{1}{2}=55\) vì  \(f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=2f\left(\frac{1}{2}\right)=1\)nên \(f\left(\frac{1}{2}\right)=\frac{1}{2}\)

Vậy \(A=55\)

\(3^x+111=\left(y-3\right)\left(y-5\right)\)

\(3^x+111=y\left(y-5\right)-3\left(y-5\right)\)

\(3^x+111=y^2-5y-3y+15\)

\(3^x+111=y^2-8y+15\)

\(3^x+111-15=y^2-8y\)

\(3^x+96=y^2-8y\)

\(3\left(3^{x-1}+32\right)=y\left(y-8\right)\)

=> \(\hept{\begin{cases}y=3\\3^{x-1}+32=y-8\end{cases}}\)hoặc  \(\hept{\begin{cases}y-8=3\\3^{x-1}+32=y\end{cases}}\)

=> \(\hept{\begin{cases}y=3\\3^{x-1}+32=3-8=-5\end{cases}}\)hoặc \(\hept{\begin{cases}y=3+8=11\\3^{x-1}+32=11\end{cases}}\)

=> \(\hept{\begin{cases}y=3\\3^{x-1}=-5-32=-37\end{cases}}\)hoặc \(\hept{\begin{cases}y=11\\3^{x-1}=11-32=-21\end{cases}}\)

.............................................................................................................................................................

=> \(x,y\in\varnothing\)

.............................................................................................................................................................

hình như mình làm lộn rồi .............................

cái chỗ => ấy mình lộn 

SORRY

\(B=-5\left(3x+2\right)^4-\left(x+2y\right)^2+111\)

Ta có :

\(\left(3x+2\right)^4\ge0\Rightarrow-5\left(3x+2\right)^4\le0\left(1\right)\)

\(\left(x+2y\right)^2\ge0\Rightarrow-\left(x+2y\right)^2\le0\left(2\right)\)

Từ (1)(2) \(\Rightarrow-5\left(3x+2\right)^4-\left(x+2y\right)^2\le0\)

\(\Rightarrow-5\left(3x+2\right)^4-\left(x+2y\right)^2+111\le111\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}5\left(3x+2\right)^4=0\\\left(x+2y\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\-\dfrac{2}{3}+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy B đạt GTLN bằng 11 khi \(x=-\dfrac{2}{3};y=\dfrac{1}{3}\)

\(A=-5\left(3x+2\right)^4-\left(x+2y\right)^2+111\le111\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}-5\left(3x+2\right)^4=0\\-\left(x+2y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

a,(a-30)+(x-29)+...+110=0

(=)\(\frac{(110+x-30)}{x}=0\)

(=)\(80+x=0\)

(=)=-80

MIN_F=-111

MIN_G=18/25

để ý các đẳng thức có dấu gttđ luôn > 0 thôi

cach giai???