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\(3x+3x+1+3x+2=117\)
\(\Rightarrow3x+3x+3x=117-1-2\)
\(\Rightarrow3x+3x+3x=114\)
\(\Rightarrow x.\left(3+3+3\right)=114\)
\(\Rightarrow x.9=114\)
\(\Rightarrow x=\dfrac{38}{3}\)
Vậy \(x=\dfrac{38}{3}\)
=> 3x+3x+3x+1+2=117
=>9x+3=117
=>9x=117-3=114
=> x=\(\dfrac{114}{9}\)
\(3x+3x+1+3x+2=117\)
\(\Rightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)
\(\Rightarrow9x+3=117\)
\(\Rightarrow9x=117-3\)
\(\Rightarrow9x=114\)
\(\Rightarrow x=114:9\)
\(\Rightarrow x=\frac{38}{3}\)
Vậy \(x=\frac{38}{3}\)
P/s : Đúng nha
~ Ủng hộ nhé
Đặt bt trên là A nha
Đổi |x-1|=|1-x|
Suy ra A=|1-x|+x-2|+|x-3|
Áp dụng BĐTGTTĐ ta có
A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2
Dấu = xảy ra khi \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra
Vậy x =2
b,
\(\left|3x+\frac{1}{2}\right|\ge0\)
\(\left|3x+\frac{1}{6}\right|\ge0\)
..........
\(\left|3x+380\right|\ge0\)
Suy ra đề bài \(\ge\)0
suy ra 58x \(\ge\)0
Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)
Tự tính nhé hok tốt
a) x (3x - 2) - 3x (x + 5) = - 34
<=> 3x2-2x-3x2-15x=-34
<=>-17x=-34
<=>x=\(\frac{-34}{-17}\)
<=>x=2
b) (2x + 3) . (3x - 2) - 6x . (x - \(\frac{1}{2}\) ) = 26
<=>6x2-4x+9x-6-6x2 +3x=26
<=>8x=26+6=32
<=>x=32:8=4
nếu đúng nhớ k cho mk và kết bn nha!mk trả lời nhanh nhất đó. thank. chúc bn học tốt!
a) \(\Leftrightarrow x=2\)
b) \(\Leftrightarrow x=32\div8=4\)
Hk tốt
\(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{3}=\frac{z}{2}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x+y+z}{9}=\frac{117}{9}=13\)
\(\Rightarrow x=4.13=52;y=3.13=39;z=2.13=26\)
a, Vì \(\left|3x-2y\right|\ge0;\left|3y-4z\right|\ge0\Rightarrow\left|3x-2y\right|+\left|3y-4z\right|\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-2y=0\\3y-4z=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2y\\3y=4z\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{4}=\frac{z}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{9}\end{cases}\Leftrightarrow}\frac{x}{8}=\frac{y}{12}=\frac{z}{9}}\)
\(\Leftrightarrow\frac{x}{8}=\frac{2y}{24}=\frac{3z}{27}=\frac{x-2y+3z}{8-24+27}=\frac{5}{11}\)
từ đây tìm x,y,z
b,Ta có: \(\frac{2x+3}{2}=\frac{3x-6}{5}\Rightarrow5\left(2x+3\right)=2\left(3x-6\right)\Rightarrow10x+15=6x-12\Rightarrow4x=-27\Rightarrow x=\frac{-27}{4}\)
Thay x=-27/4 vào \(\frac{3x-6}{5}=\frac{3x+3y+1}{3x}\), ta được:
\(\frac{3\cdot\left(\frac{-27}{4}\right)-6}{5}=\frac{3.\left(\frac{-27}{4}\right)+3y+1}{3.\left(\frac{-27}{4}\right)}\)
\(\Rightarrow\frac{-21}{4}=\frac{\frac{-77}{4}+3y}{\frac{-81}{4}}\Rightarrow\frac{-77}{4}+3y=\frac{1701}{16}\Rightarrow3y=\frac{2009}{16}\Rightarrow y=\frac{2009}{48}\)
Vậy x=-27/4,y=2009/48
3x+3x+1+3x+2=177
9x+3=117
9x=117-3
9x=114
x=114/9
\(3x+3x+1+3x+2=117\)
\(\Leftrightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)
\(\Leftrightarrow9x+3=117\)\(\Rightarrow9x=114\Rightarrow x=\frac{114}{9}\)
\(\text{Vậy x=}\frac{114}{9}\)