Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=1002.12+1002.22+1002.32+...+1002.102
=1002(12+22+32+...+102)=1002.385=3850000
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
\(A=\frac{2}{3}+\frac{3}{4}\left(\frac{-4}{9}\right)\)
\(A=\frac{2}{3}+\frac{-1}{3}\)
\(A=\frac{1}{3}\)
\(B=\frac{2}{5}+\frac{3}{5}\div\left(-2\right)\)
\(B=\frac{2}{5}+\frac{-3}{10}\)
\(B=\frac{1}{10}\)
\(C=2\frac{3}{11}\cdot1\frac{1}{12}\cdot\left(-2,2\right)\)
\(C=\frac{325}{132}\cdot\left(-2,2\right)\)
\(C=\frac{-65}{12}\)
\(D=\left(\frac{3}{4}-0,2\right)\left(0,4-\frac{4}{5}\right)\)
\(D=\frac{11}{20}\cdot\frac{-2}{5}\)
\(D=\frac{-11}{50}\)
Dạng 1:
a: =>x(x-3)=0
=>x=3 hoặc x=0
b: =>x(3x-4)=0
=>x=4/3 hoặc x=0
c: =>2x-1=0
=>x=1/2
d: =>2x(2x+3)=0
=>x=0 hoặc x=-3/2
e: =>x(2x+5)=0
=>x=-5/2 hoặc x=0
bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
Câu 1:
a: \(A=1+3+3^2+\cdots+3^{11}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+\cdots+\left(3^{10}+3^{11}\right)\)
\(=\left(1+3\right)+3^2\left(1+3\right)+\cdots+3^{10}\left(1+3\right)\)
\(=4\left(1+3^2+\cdots+3^{10}\right)\) ⋮4
b: \(B=1+3+3^2+\cdots+3^{2024}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\cdots+\left(3^{2022}+3^{2023}+3^{2024}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+\cdots+3^{2022}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+\cdots+3^{2022}\right)\) ⋮13
Câu 2:
\(S=100^2+200^2+\cdots+1000^2\)
\(=100^2\left(1^2+2^2+\cdots+10^2\right)\)
\(=10000\cdot385=3850000\)