Tìm x
a, (x-3).(x-4)=0
b, (\(\frac{1}{2}\)x - 4).(x - \(\frac{1}{4}\)) = 0
c, (\(\frac{1}{3}\)- x ).(\(\frac{1}{2}\)+1 : x)=0
d, (x+3).(x-4) +2 (x+3)=0
Cần gấp lắm luôn ạ! Giúp tớ nhé! cảm ơn! ><
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Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
B = \(\frac{-1}{9}\). \(\frac{-17}{29}\). \(\frac{58}{51}\)= \(\frac{2}{27}\)
D = \(\frac{-184}{9}\)
Nếu cần cách làm cụ thể thì ib tớ nhé, xl vì tớ không bt bấm cách làm cụ thể !!!:))
#Cụ_Maiz
\(B=\frac{-1}{9}.\frac{-17}{29}.\frac{58}{51}\)
\(=\frac{1.27.58}{9.29.51}\)
\(=\frac{1.27.29.2}{9.29.17.3}\)
\(=\frac{2}{27}\)
\(D=\left(1\frac{1}{27}.\frac{12}{23}.\frac{9}{14}\right):\left(\frac{-3}{23}\right)\)
\(=\frac{28}{27}.\frac{12}{23}.\frac{9}{14}.\left(\frac{-23}{3}\right)\)
\(=\frac{7.4.2.2.3.9.\left(-23\right)}{9.3.23.7.2.3}\)
\(=\frac{-8}{3}\)
Học tốt nhé
\(\left(\frac{3}{4}x-\frac{9}{16}\right)\left(1,5+\frac{3}{5}:x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{9}{16}=0\\\frac{3}{2}+\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x=\frac{9}{16}\\\frac{3}{5x}=-\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{2}{5}\end{cases}}\)
Có 2 trường hợp :
Trưởng hợp 1 :( \(\frac{3}{4}\)x - \(\frac{9}{16}\)) = 0
Ta có : \(\frac{3}{4}\)x = 0 + \(\frac{9}{16}\)
\(\frac{3}{4}\)x = \(\frac{9}{16}\)
x = \(\frac{9}{16}\): \(\frac{3}{4}\)
=> x = \(\frac{3}{4}\)
Trường hợp 2 : ( 1, 5 + \(\frac{3}{5}\): x ) = 0
( \(\frac{3}{2}\)+ \(\frac{3}{5}\): x ) = 0
\(\frac{3}{5}\) : x = 0 - \(\frac{3}{2}\)
\(\frac{3}{5}\): x = \(\frac{-3}{2}\)
x = \(\frac{3}{5}\): \(\frac{-3}{2}\)
= x = \(\frac{-2}{5}\)
Vậy : x = \(\frac{3}{4}\)Hoặc x = \(\frac{-2}{5}\)
Study well ^_^:))
# Cụ Maiz
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)
\(\Leftrightarrow x=45\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
Vậy x = 45.
a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
Bài làm :
Ta có :
\(\left|x+\frac{4}{15}\right|-\left|3,75\right|=-\left|2,15\right|\)
\(\Leftrightarrow\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\frac{4}{15}\right|=1,6\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{4}{15}=1,6\\x+\frac{4}{15}=-1,6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)
Vậy x=4/3 hoặc x=-28/15
\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)
\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=-1+1+\frac{2}{97}\)
\(=\frac{2}{97}\)
\(=\left(-\frac{1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{2}{7}+\frac{4}{35}\right)+\frac{1}{127}\)
\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\)
\(=\left(-\frac{18}{18}\right)+\frac{35}{35}+\frac{1}{127}\)
\(=-1+1+\frac{1}{127}\)
\(=\frac{1}{127}\)
Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)