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\(\dfrac{3x}{2\cdot5}+\dfrac{3x}{5\cdot8}+\dfrac{3x}{8\cdot11}+\dfrac{3x}{11\cdot14}=\dfrac{1}{21}\\ x\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\right)=\dfrac{1}{21}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ x\cdot\dfrac{3}{7}=\dfrac{1}{21}\\ x=\dfrac{1}{21}:\dfrac{3}{7}\\ x=\dfrac{1}{9}\)
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\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
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\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
\(3x.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)
\(3x.\frac{1}{7}=\frac{1}{21}\)
\(\frac{3}{7}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{3}{7}\)
\(x=\frac{7}{81}\)
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a,(2x-1)3 =23+102 b,(3x+1)+(3x+3)+...+(3x+99)=2800
(2x-1)3 =125 3x+1+3x+3+...+3x+99=2800
(2x-1)3=53 ( 3x+3x+.....+3x )+(1+3+...+99)=2800
2x-1=5 gọi A=3x+3x+...+3x ; B=1+3+...+99
2x=5+1 số số hạng của B là : (99-1):2+1=50 ( bằng số số hạng của A)
2x=6 B = (99+1) x 50:2
=2500
x=6:2 ta có: 150x + 2500=2800
x=3 150x=2800-2500
vậy x=3 150x=300
x=300:150
x=2
vậy x=2
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\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(=>\frac{3x}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\right]=\frac{1}{21}\)
\(=>x\left[\frac{1}{2}-\frac{1}{14}\right]=\frac{1}{21}\)
\(=>x\cdot\frac{3}{7}=\frac{1}{21}\Leftrightarrow x=\frac{1}{9}\)
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\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
=> \(x\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)=\frac{1}{21}\)
=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
=> \(x.\frac{3}{7}=21\)
=> x = 49
Vậy x = 49
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=>(3x-1)^5*[(3x-1)^3-1]=0
=>(3x-1)^5*(3x-2)=0
=>x=2/3 hoặc x=1/3
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mk giúp bạn câu cuối nhé:
3|x+2|-5=16
3|x+2|=16+5
3|X+2|=21
|x+2|=21:3
|x+2|=7
=>x+2=7 hoặc x+2=-7
+) với x+2=7 +) với x+2= -7
x=5. x=-9
vậy x€{5,-9}
nếu có TGian mk sẽ giải cho bạn mấy câu trên
cam ơn bạn nhé bạn có giup mình not câu trên trong vong ngay ko
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(3x - 1)^5 = (3x - 1)^8
=> 3x-1=0 hoặc 3x-1=1
3x=0+1 3x=1+1
3x=1 3x=2
x=1:3 x=2:3
x=1/3 x=2/3
vậy x=1/3 hoặc x=2/3