x.x^2.x^3. .... . x^10 = 355
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2X : X = X
2 = X
2X + X = 6X
câu này ko biết chắc sai đề
2X * X = 10X
2*X*X=10X
gạch X ở hai vế
ta có
2*X=10
X = 10: 2
X= 5
=x^2.x^8.x^4.x^6.x^10/x^1.x^9.x^3.x^7.x^3=3^5
=x^10.x^10.x^10 / x^10.x^10.x^5=3^5
=x^10/x^5=3^5
=x^5=3^5
x=3
\(x^2.x^4.x^6.x^8.x^{10}:\left(x.x^3.x^5.x^7.x^9\right)=243\)
\(x^{2+4+6+8+10}:x^{1+3+5+7+9}=243\)
\(x^{30}:x^{25}=243\)
\(x^{30-25}=243\)
\(x^5=243\)
\(x^5=3^5\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
1.x2+1+3=64
x6=64
Ta thấy:26=64
=>x=6
2.x2+4+6+8+10/ x1+3+5+7+9=243
x30/ x25=243
x30-25=243
x5=243
Ta thấy:35=243
=>x=5
a) Ta có: \(3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)
b) Ta có: \(5x-2=x+4\)
\(\Leftrightarrow5x-x=4+2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=1\frac{1993}{1991}\)
\(\Leftrightarrow\left(1\cdot\frac{1}{2}\right)+\left(\frac{1}{3}\cdot\frac{1}{2}\right)+\left(\frac{1}{6}\cdot\frac{1}{2}\right)+....+\left(\frac{1}{\frac{x\left(x+1\right)}{2}}\cdot\frac{1}{2}\right)=1\frac{1993}{1991}\div2\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=-\frac{1}{1991}\)
\(\Leftrightarrow x=-1992\)
làm ơn giúp mình ,bạn nào làm nhanh đúng mình chọn cho nhanh nha mọi người
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39
\(x.x^2.x^3...x^{10}=3^{55}\Rightarrow x^{1+2+...10}=3^{55}\)
\(\Rightarrow x^{55}=3^{55}\Rightarrow x=3\)
(335 sửa lại thành 355)