x^1 x^2 .... x^2016
giup mik voi
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\(=>x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1=\dfrac{13}{3}\)
\(=>x=\dfrac{13}{3}:\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{13}{3}:\dfrac{7}{6}=\dfrac{26}{7}\)
5/6x - 1 = 10/3
5/6x = 10/3 + 1 = 13/3
X = 13/3 : 5/6 = 26/5
a)x+x+2.x=76:2
=>2x+2x=38
=>2(x+x)=38
=>2x=19
=>x=19/2
b)2.x=x+14
=>2x-x=14
=>x=14
c)2.x-3/5=3/4
=>2x=3/4+3/5
=>2x=27/20
=>x=27/40
d)3/5-2.x=1/4
=>2x=3/5-1/4
=>2x=7/20
=>x=7/40
/2.x-3/=x-1 \(\Leftrightarrow\orbr{\begin{cases}2x-3=x-1\\2x-3=-(x-1)\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x-3-x=-1\\2x-3=-x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x(2-1)=2\\2x+x=4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\3x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}}\)
\(∘backwin\)
\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)
\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)
\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)
\(100 x + 5050 = 5750\)
\( 100 x = 5750 − 5050\)
\(100 x = 700\)
\(x = 700 : 100\)
\(x = 7\)
\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)
\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(B<1-\)\(\dfrac{1}{2021}\)
\(B<\)\(\dfrac{2020}{2021}\)
\(\dfrac{2020}{2021}< 1\)
\(B<1\)
a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7
a. 3^x=1-x^2
x=0 la nghiem
x>=1; VT>=3 VP<=0 vo nghiem
b. (de bai thieu n khac 0 vi neu n=0 dung voi moi x)
3x-14=1=> x=5
c.(5^2x5^x+1)=5^4
5^x+1=5^2=> x=1
\(x+\frac{4}{9}=\frac{1}{2}\)
\(x=\frac{1}{2}-\frac{4}{9}\)
\(x=\frac{9}{18}-\frac{8}{18}\)
\(x=\frac{1}{18}\)
x1.x2...x2016
= x1+2+...+2016
= x2 033 136