tìm x .(x-3)^3+x(1-x)=3 giúp mình nha
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\(3^{x+1}+3^{x+2}+3^{x+3}-4.3^x=315\)
\(\Leftrightarrow3^x.3+3^x.3^2+3^x.3^3-4.3^x=315\)
\(\Leftrightarrow3^x.3+3^x.9+3^x.27-4.3^x=315\)
\(\Leftrightarrow3^x.\left(3+9+27-4\right)=315\)
\(\Leftrightarrow3^x.35=315\)\(\Leftrightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
Bài làm :
Ta có :
\(3^{x+1}+3^{x+2}+3^{x+3}-4.3^x=315\)
\(\Leftrightarrow3^x.3+3^x.3^2+3^x.3^3-4.3^x=315\)
\(\Leftrightarrow3^x.3+3^x.9+3^x.27-4.3^x=315\)
\(\Leftrightarrow3^x.\left(3+9+27-4\right)=315\)
\(\Leftrightarrow3^x.35=315\)
\(\Leftrightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)
\(\Leftrightarrow x=2\)
Vậy x=2
3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21
=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21
=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21
x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21
x . ( 1/2 - 1/14 ) = 1/21
x . 3/7 = 1/21
x = 1/21 : 3/7
=> x = 1/9
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
a) \(\left(x+1\right)\left(x+2\right)=272\)
\(\Rightarrow x^2+3x+2=272\)
\(\Rightarrow x^2+3x-270=0\)
\(\Rightarrow x^2+18x-15x-270=0\)
\(\Rightarrow x\left(x+18\right)-15\left(x+18\right)=0\)
\(\Rightarrow\left(x+18\right)\left(x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+18=0\\x-15=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-18\\x=15\end{matrix}\right.\)
d) \(\left(x+4\right)\left(x+5\right)=552\)
\(\Rightarrow x^2+9x+20=552\)
\(\Rightarrow x^2+9x-532=0\)
\(\Rightarrow x^2+28x-19x-532=0\)
\(\Rightarrow x\left(x+28\right)-19\left(x+28\right)=0\)
\(\Rightarrow\left(x+28\right)\left(x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+28=0\\x-19=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-28\\x=19\end{matrix}\right.\)
3 . (-x + 1) - 2.(x + 7) - 32 = -(-2x - 9) - 158
=> -3x + 3 - 2x - 14 - 9 = 2x + 9 - 1
=> -3x - 2x - 2x = 9 - 1 + 9 + 14 - 3
=> -7x = 28
=> x = 28 : (-7)
=> x = -4
Vậy ...
=>(-3x+3)-(2x+14)-9=2x+9-1
=>(-3x-2x)-(14-3+9)=2x+8
=>-5x-20=2x+8
=>2x+5x=-8-20
=>-7x=-28
=>x=-28/7
Ta có: 7(x-3)-5(3-x)=11x-5 <=>7(x-3)+5(x-3)-11x+5=0 <=>12(x-3)-11x+5=0. <=>12x-36-11x+5=0 <=>x=31
Đặt \(A=\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{x\left(x+2\right)}\)(sửa đề)
\(\Rightarrow A=\frac{1}{2}.3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(\Rightarrow A=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)\)
\(\Rightarrow A=\frac{1}{2}-\frac{3}{2x+4}\)
`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
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`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
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`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
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`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
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`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
\(x+\dfrac{3}{4}=\dfrac{28}{6}+\dfrac{1}{2}\\ x+\dfrac{3}{4}=\dfrac{31}{6}\\ x=\dfrac{31}{6}-\dfrac{3}{4}\\ x=\dfrac{53}{12}\)
\(x+\dfrac{3}{4}=\dfrac{28+3}{6}=\dfrac{31}{6}\Leftrightarrow x=\dfrac{31}{6}-\dfrac{3}{4}=\dfrac{62-9}{12}=\dfrac{53}{12}\)
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