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\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)
\(x\times4+\frac{40}{81}=\frac{51}{81}\)
\(x\times4=\frac{51}{81}-\frac{40}{81}\)
\(x\times4=\frac{11}{81}\)
\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{11}{324}\)
[ 61 + ( 53 - x ) ] . 17 = 1785
61 + ( 53 - x ) = 1785 : 17
61 + ( 53 - x ) = 105
( 53 - x ) = 105 - 61
53 - x = 44
=> x = 53 - 44
=> x = 9
mk muốn xem bài của mk đúng hay sai thôi !
chứ làm thì mk làm xong rồi !
Bài 1:
Bài 2: Tính\(\frac{1-3-5-7-...-49}{89}=\frac{1-\left(3+5+7+9+...+49\right)}{89}=\frac{1-\left(\left(49+3\right)\times\left(\frac{49-3}{2}+1\right)\div2\right)}{89}=\frac{1-\left(52\times24\div2\right)}{89}=\frac{1-624}{89}=\frac{-623}{89}=-7\)
k cho mình nha Đô Mỹ Diệu Linh
1)
\(\frac{\left(-3\right)^x}{81}=-27\)
\(\left(-3\right)^x=-27.81=-2187\)
\(\left(-3\right)^x=\left(-3\right)^7\)
\(=>x=7\)
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)
=> \(A=\frac{121}{243}\)
=> \(2x+\frac{12}{25}=\frac{121}{243}\)
=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
=> x = ......
TA CÓ THỂ THẤY, VẾ TRÁI CÓ: 12 CẶP
=> \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
<=> \(x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\) (****)
Ta xét: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
=> \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\)
=> \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\)
=> \(2A=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Ta tiếp tục xét: \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
=> \(3B=1+\frac{1}{3}+...+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
=> \(2B=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(B=\frac{121}{243}\)
THAY CÁC GIÁ TRỊ A; B VÀO PT (****) TA ĐƯỢC:
=> \(x+\frac{12}{25}=\frac{121}{243}\)
<=> \(x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
\(\frac{1}{81}.27^x=3^x\Rightarrow\frac{1^2}{9^2}=\frac{3^x}{27^x}\)\(\Rightarrow\left(\frac{1}{9}\right)^2 =\left(\frac{1}{9}\right)^x\)\(\Rightarrow X=2\)
=>\(\frac{1}{3^4}\cdot\left(3^3\right)^x=3^x\)
=>\(\frac{3^{3x}}{3^4}=3^x\)
=>\(3x-4=x\)
=>\(3x=x+4\)
=>\(2x=4=>x=2\)
\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)
\(C=\frac{2}{3}\)
Bạn Cô nàng Thiên Bình làm đúng hết òi =.=
a=7.[1/8+1/27-1/49]
------------------------
11.[1/8+1/27-1/49]
=7/11
cau b,c tuong tu nha h mk
\(\left(x+3\right)^3=\frac{81}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=3\)
\(\Leftrightarrow x+3=\sqrt[3]{3}\)
\(\Leftrightarrow x=-3+\sqrt[3]{3}\)