tim x thuoc Q biet
(8x-1)2n+1=52n+1
(3x)2:33=\(\frac{1}{243}\)
Những câu hỏi liên quan
Tim x biet : \(\left(8x-1\right)^{2n+1}=5^{2n+1}\)
nếu n E N thì 2n+1 là số lẻ lúc đó ta có:
(8x-1)2n+1 = 52n+1
<=> 8x - 1 = 5
<=> 8n = 6
<=> n = 6/8 = 3/4
Đúng 0
Bình luận (0)
Tim x thuoc Z, biet: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)(nhân mỗi vế với 1/2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\Rightarrow x=2010\)
Đúng 0
Bình luận (0)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\)
\(\Rightarrow x=2010\)
Đúng 0
Bình luận (0)
Bai 1 :cho A(x)=2x^2-8x
tim nghiem A(x)
b,tim B(x) biet B(x)=A(x)-3x+2x^2
c,tim ngiem B(x)
Bai2
Cho ham so y=f(x)=x+1
cac diem sau diem nao thuoc do thi ham so :A(0;1).B(1/2;-1). C(-1/2;0)
a,\(2x^2-8x=0\)
\(2x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)
\(=2x^2-8x-3x-2x^2\)
=\(-11x\)
c,\(-11x=0\)
\(\Rightarrow x=0\)
Đúng 0
Bình luận (0)
\(A\left(x\right)=2x^2-8x\)
\(\Rightarrow2x^2-8x=0\)
\(\Rightarrow x\left(2x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)
\(B\left(x\right)=-3x+2x^2\)
\(B\left(x\right)=2x^2-3x\)
\(2x^2-3x=0\)
\(\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (1)
bai1.tim x biet:
a,(x+2).(x+3)-(x-2).(x+5)=0
b,(2x+3).(x-4)+(x-5).(x-2)=(3x-5).(x-4)
c,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)=33
,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
tim x thuoc Z biet : 2x+\(\frac{1}{7}=\frac{1}{y}\)
1) Tim a, b thuoc Q biet: a-b=2(a+b)=a:b
2) Tim x thuoc Q sao cho: (x-1)(x+3)<0
Tim x biet :
a, \(4\frac{1}{3}\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}\left(\frac{1}{2}-\frac{1}{3}-\frac{3}{4}\right)\) x thuoc Z
b , \(|x-3|+1=x\)
tim x biet : \(\left(3^x\right)^2:3^2=\frac{1}{243}\)
\(\left(3^x\right)^2=\frac{1}{243}.3^2\)
\(\left(3^x\right)^2=\frac{1}{27}\)
=>x=1,5
Đúng 0
Bình luận (0)
Tim x,y thuoc Z biet: \(2x+\frac{1}{7}=\frac{1}{y}\)