tìm x biết
2x phần 68 = 16 phần 32
5x phần 100 = 105 phần 25
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29 tháng 12 2020
FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
4 tháng 3 2022
CÁC EM TỰ LÀ ĐI MẤY BÀI NÀY DỄ LĂM CHỊ HỌC LỚP 8 NÊN CHỊ BIẾT NHƯNG ĐỪNG BẢO CHỊ LÀ CHO HAY GỢI Ý CÁCH LÀM NHA.
4 tháng 6 2022
c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
TG
12 tháng 4 2020
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
tìm x biết
\(\dfrac{2x}{68}\)=\(\dfrac{16}{32}\)
\(\Leftrightarrow\dfrac{x}{34}=\dfrac{1}{2}\)
\(\Leftrightarrow2x=1.34\)
\(\Leftrightarrow x=17\)
b)\(\dfrac{5x}{100}=\dfrac{105}{25}\)
\(\Leftrightarrow\dfrac{x}{20}=\dfrac{21}{5}\)
\(\Leftrightarrow5x=420\)
\(\Leftrightarrow x=84\)
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