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\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)
\(\Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left[{}\begin{matrix}3x-1=1\\3x-1=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)
Vì \(\left|x+2\right|+\left|2x+3\right|+\left|3x+4\right|\ge0\)
=> \(7x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+2\right|+\left|2x+3\right|+\left|3x+4\right|=x+2+2x+3+3x+4\)
\(\Rightarrow6x+7=7x\)
=> x=7
a) Ta có : 2x = 3y => \(\frac{x}{3}=\frac{y}{2}\)
7z = 5y => \(\frac{y}{7}=\frac{z}{5}\)
=> \(\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)
+) \(\frac{x}{3}=\frac{y}{2}\)=> \(\frac{x}{21}=\frac{y}{14}\)
+) \(\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\)
=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
=> \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
=> x = 2.21 = 42 , y = 2.14 = 28 , z = 2.10 = 20
b) Ta có : x : y : z = 3 : 5 : (-2) => \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)
Đặt \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=k\Rightarrow\hept{\begin{cases}x=3k\\y=5k\\z=-2k\end{cases}}\)
=> 5x = 15k , y = 5k , 3z = -6k
=> 5x - y + 3z = 15k - 5k + (-6k)
=> -16 = 10k - 6k
=> -16 = 4k
=> k = -4
Với k = -4 thì x = 3.(-4) = -12 , y = 5.(-4) = -20 , z = (-2).(-4) = 8
Vậy : ....
Ta có : |2x - 1| + 1 = x
=> |2x - 1| = x - 1
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x-1\\2x-1=1-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=-1+1\\2x+x=1+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
vì \(|x+1|+|2x+3|+|3x+4|>=0\)
suy ra 7x >=0
suy ra x>= 0
suy ra x+1 +2x+3+3x+4 = 7x
suy ra x=8
\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)
\(\left(3x-1\right)^{10}.\left[\left(3x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x-1=0\\3x-1^{10}=1\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{2}{3}ho\text{ặc}x=0\end{cases}}\)
Vậy \(x=\frac{1}{3}ho\text{ặc}x=\frac{2}{3}ho\text{ặc}x=0\)
Tham khảo nhé~
TA CÓ:\(\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)
\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{10}\times\left(3x-1\right)^{10}=0\)
\(\Rightarrow\left(3x-1\right)^{10}\times[1-\left(3x-1\right)^{10}]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{2}{3}\end{cases}}}\)
mik nha. cảm ơn nhìu!! ^^