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\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)
\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)
hay \(x=-\dfrac{41}{12}\)
Vậy: \(x=-\dfrac{41}{12}\)
\(a.\frac{2}{3}+\frac{1}{5}\cdot\frac{10}{7}\)
\(=\frac{2}{3}+\frac{1\cdot2}{1\cdot7}\)
\(=\frac{2}{3}+\frac{2}{7}=\frac{2\cdot7}{21}+\frac{2\cdot3}{21}=\frac{14}{21}+\frac{6}{21}=\frac{20}{21}\)
\(b.\frac{2}{7}\cdot\frac{4}{7}+\frac{2}{7}\cdot\frac{3}{7}\)
\(=\frac{2}{7}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)\)
\(=\frac{2}{7}\cdot\frac{7}{7}=\frac{2}{7}\cdot1=\frac{2}{7}\)
\(c.\left[-\frac{1}{4}+\frac{3}{10}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\left[-\frac{5}{20}+\frac{6}{20}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\frac{1}{20}:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\frac{1}{20}\cdot\left(-\frac{5}{3}\right)-\frac{7}{6}\)
\(=\frac{1\cdot\left(-1\right)}{4\cdot3}-\frac{7}{6}\)
\(=\left(-\frac{1}{12}\right)-\frac{7}{6}=\left(-\frac{1}{12}\right)-\frac{14}{12}=-\frac{15}{12}\)
bài 2 :a) \(2x-2\frac{2}{7}=2\frac{5}{7}\)
\(2x=2\frac{5}{7}-2\frac{2}{7}\)
\(2x=\left(2-2\right)+\left(\frac{5}{7}-\frac{2}{7}\right)\)
\(2x=0+\frac{3}{7}\)
\(2x=\frac{3}{7}\)
\(x=\frac{3}{7}:2=\frac{3}{7}\cdot\frac{1}{2}=\frac{3}{14}\)
\(b.\frac{17}{13x}=\frac{4}{39}\)
\(\Rightarrow13x\cdot4=17\cdot39\)
\(\Rightarrow13x\cdot4=663\)
\(\Rightarrow13x=663:4\)
\(\Rightarrow13x=165,75\)
\(\Rightarrow x=165,75:13\)
\(\Rightarrow x=12,75\)
\(****nha!!!!!!!!!!!!\)
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/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
a) \(\frac{x-5}{7}=\frac{5}{3}\)
\(\Rightarrow\left(x-5\right)\cdot3=7\cdot5\)
\(\Rightarrow3x-15=35\)
\(\Rightarrow3x=15+35\)
\(\Rightarrow3x=50\)
\(\Rightarrow x=\frac{50}{3}\)
b) \(\frac{x-3}{3}=\frac{12}{x-3}\)
\(\Rightarrow\left(x-3\right)\cdot\left(x-3\right)=3\cdot12\)
\(\Rightarrow\left(x-3\right)^2=36\)
\(\Rightarrow\left(x-3\right)^2=6^2\)hoặc \(\left(x-3\right)^2=\left(-6\right)^2\)
\(\Rightarrow x-3=6\) \(x-3=-6\)
\(\Rightarrow x=6+3\) \(x=-6+3\)
\(\Rightarrow x=9\) hoặc \(x=-3\)
c) \(\frac{5-2x}{4}=\frac{7}{3}\)
\(\Rightarrow\left(5-2x\right)\cdot3=4\cdot7\)
\(\Rightarrow15-6x=28\)
\(\Rightarrow6x=15-28\)
\(\Rightarrow6x=-13\)
\(\Rightarrow x=-\frac{13}{6}\)
d) \(\frac{x+1}{4}=\frac{7}{3}\)
\(\Rightarrow\left(x+1\right)\cdot3=2\cdot7\)
\(\Rightarrow3x+3=28\)
\(\Rightarrow3x=28-3\)
\(\Rightarrow3x=25\)
\(\Rightarrow x=\frac{25}{3}\)
Chúc bạn học tốt !!!
a ) \(-\frac{3}{7}.\frac{3}{11}+-\frac{3}{7}.\frac{8}{11}+1\frac{3}{7}\)
\(=-\frac{3}{7}.\left(\frac{3}{11}+\frac{8}{11}\right)+\frac{10}{7}\)
\(=-\frac{3}{7}.\frac{11}{11}+\frac{10}{7}\)
\(=-\frac{3}{7}.1+\frac{10}{7}\)
\(=\frac{10}{7}\)
b ) \(75\%.10,5=\frac{3}{4}.10,5=7,875\)
c ) \(5-3.\left(\left|-4\right|-30:15\right)\)
\(=5-3.\left(4-2\right)\)
\(=5-3.2\)
\(=5-6\)
\(=-1\)
d ) \(-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{11}+1\frac{5}{7}\)
\(=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=-\frac{5}{7}.1+\frac{12}{7}\)
\(=\frac{7}{7}\)
\(=1\)
Chúc bạn học tốt !!!
\(A=\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{100}{609}\\ \)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{1}{3}-\frac{1}{2x+3}\)\(=\frac{2x}{3\left(2x+3\right)}\)
\(A=\frac{x}{3\left(2x+3\right)}=\frac{100}{609}=\frac{100}{3.203}=\frac{100}{3\left(2.100+3\right)}\)\(\Rightarrow x=100\)