Tìm x € N
a) x20=x
b) 3x+2-5.3x=36
Cíu cíu
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\(a,5^x+5^{x+2}=650\\ \Rightarrow a,5^x+5^x.25=650\\ \Rightarrow26.5^x=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,3^{x.1}+5.3^{x.1}=162\\ \Rightarrow3^x+5.3^x=162\\ \Rightarrow6.3^x=162\\ \Rightarrow3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
Trả lời:
\(3x+1+5\times3x+2=144\)
\(\Leftrightarrow3x+1+15x+2=144\)
\(\Leftrightarrow18x=141\)
\(\Leftrightarrow x=\frac{47}{8}\)
Vậy \(x=\frac{47}{8}\)
3x - 17 = x + 3
3x - x = 17 + 3
3x - x = 20
2x = 20
x = 20 : 2
x = 10
tick cho tớ nha!
\(b,3\left(x-2\right)+2\left(3x-5\right)=10\\ \Leftrightarrow3x-6+6x-10=10\\ \Leftrightarrow3x+6x=10+10+6\\ \Leftrightarrow9x=26\\ \Leftrightarrow x=\dfrac{26}{9}\\ c,2x-\left(3x+1\right)=5x-2\\ \Leftrightarrow2x-3x-1=5x-2\\ \Leftrightarrow2x-3x-5x=-2+1\\ \Leftrightarrow-6x=-1\\ \Leftrightarrow x=\dfrac{1}{6}\\ d,3x+2=-5+6 \\ \Leftrightarrow3x=-5+6-2\\ \Leftrightarrow3x=-2\\ \Leftrightarrow x=-\dfrac{1}{3}\)
`-3x=2y `
`=> x/2 = -y/3 `
AD t/c của dãy tỉ số bằng nhau ta có
`x/2 =-y/3 = (x-y)/(2+3) = 6/5`
`=>{(x=2*6/5 = 12/5),(y=-3*6/5 =-18/5):}`
a) `6/x =-3/2`
`=>x =6 :(-3/2) = 6*(-2/3)=-4`
`b)`\(-3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)
Áp dụng t/c của DTSBN , ta đc :
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2+3}=\dfrac{6}{5}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{6}{5}\\\dfrac{y}{-3}=\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\y=-\dfrac{18}{5}\end{matrix}\right. \)
`a)`
`6/x=-3/2`
`x=6:(-3/2)`
`x=6*(-2/3)`
`x=-4`
(x - 2)² = (1 - 3x)²
x² - 4x + 4 = 1 - 6x + 9x²
9x² - x² - 6x + 4x + 1 - 4 = 0
8x² - 2x - 3 = 0
8x² + 4x - 6x - 3 = 0
(8x² + 4x) - (6x + 3) = 0
4x(2x + 1) - 3(2x + 1) = 0
(2x + 1)(4x - 3) = 0
2x + 1 = 0 hoặc 4x - 3 = 0
*) 2x + 1 = 0
2x = -1
x = -1/2
*) 4x - 3 = 0
4x = 3
x = 3/4
Vậy x = -1/2; x = 3/4
ĐKXĐ: ...
\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)
\(\Leftrightarrow\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)-\sqrt{x+1}\left(\sqrt{3x-1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{3x-1}-\sqrt{x+1}\right)\left(\sqrt{3x-1}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=\sqrt{x+1}\\\sqrt{3x-1}=x\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) x20 = x
=> x20 - x = 0
=> x(x19 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^{19}=1^{19}\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x = 0 hoặc x = 1
b) 3x + 2 - 5.3x = 36
=> 3x . 32 - 5.3x = 36
=> 3x.9 - 5.3x = 36
=> 3x.(9 - 5) = 36
=> 3x.4 = 36
=> 3x = 9
=> 3x = 32
=> x = 2
Vậy x = 2
\(a,\text{ }x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x^{19}=0+1=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }1\right\}\)