Tìm x:
\(\frac{1}{2}.x+\frac{3}{5}.\left(x-2\right)=3\)
Giu1p em với khó quá àk
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( 3x - 1/2 ) + ( 1/2y + 3/5 ) = 0
=> ( 3 x - 1/2 ) = 0
3x = 0+1/2
3x = 1/2
x = 1/2 : 3
x = 1/6
=> ( 1/2 y + 3/5 ) = 0
1/2y = 0 - 3/5
1/2 y = -3/5
y = -3/5 : 1/2
y = -6/5
Đặt vế trái phương trình là A
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\)
\(3A=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{\left(x+3\right)-x}{x\left(x+3\right)}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}=\frac{x+2}{x+3}\Rightarrow A=\frac{x+2}{3\left(x+3\right)}\)
\(\Rightarrow\frac{x+2}{3\left(x+3\right)}=\frac{667}{2002}\Rightarrow2002\left(x+2\right)=3.667.\left(x+3\right)\)
\(\Leftrightarrow2002x+4004=2001x+6003\Leftrightarrow x=1999\)
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
x-[17/2-6/35]=-1/3
x-583/70=-1/3
x=-1/3+583/70
x=1679/210
vậy x=1769/210
[2/3-(x-7/4)]=9/2+5/4
[2/3-(x-7/4)]=23/4
(x-7/4)=23/4+2/3
(x-7/4)=77/12
x=77/12+7/4
x=49/6
vậy x=49/6
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(\frac{1}{2}.x+\frac{3}{5}.x-\frac{3}{5}.2=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right).x-\frac{6}{10}=3\)
\(\left(\frac{5}{10}+\frac{6}{10}\right).x=3+\frac{6}{10}\)
\(\frac{11}{10}.x=\frac{36}{10}\)
\(x=\frac{36}{10}:\frac{11}{10}\)
\(x=\frac{36}{10}.\frac{10}{11}\)
\(x=\frac{36}{11}\)