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\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
a,\(\frac{-2}{3}+\frac{1}{5}=\frac{-2.5}{3.5}+\frac{3.1}{3.5}=-\frac{10}{15}+\frac{3}{15}=-\frac{7}{15}\)
b,\(3\frac{11}{13}-5\frac{11}{3}=3+\frac{11}{13}-(5+\frac{11}{13})=\left(3-5\right)+\left(\frac{11}{13}-\frac{11}{13}\right)=-2+0=-2\)
c,\(1\frac{2}{3}:(-\frac{5}{3})=\frac{5}{3}:\left(-\frac{5}{3}\right)=\frac{5.3}{3.\left(-5\right)}=\frac{15}{-15}=-1\)
d,\(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)=1+\left(-1\right)=0\)
Tìm x:
a,x+12=8
x=8-12
x=-4
b,\(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{5}{10}\)
\(\frac{2}{3}x=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{2}{3}\)
\(x=\frac{6}{10}\)
\(x=\frac{3}{5}\)
a) \(\frac{-2}{3}+\frac{1}{5}=\frac{-10}{15}+\frac{3}{15}=\frac{-10+3}{15}=\frac{-7}{15}\)
b) \(3\frac{11}{13}-5\frac{11}{13}=\frac{50}{13}-\frac{76}{13}=\frac{50-76}{13}=\frac{-26}{13}=-2\)
c) \(1\frac{2}{3}:\frac{-5}{3}=\frac{5}{3}:\frac{-5}{3}=\frac{5}{3}\times\frac{3}{-5}=\frac{15}{-15}=-1\)
d) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)
\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)
\(=\left(\frac{31-14}{17}\right)+\left(\frac{-5-8}{13}\right)\)
\(=\frac{17}{17}+\frac{-13}{13}\)
\(=1+\left(-1\right)\)
\(=0\)
TÌM x
a) \(x+12=8\)
\(\Leftrightarrow x=8-12\)
\(\Leftrightarrow x=-4\)
b) \(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{1-5}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{-4}{10}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}:\frac{2}{3}=\frac{-2}{5}\times\frac{3}{2}\)
\(\Leftrightarrow x=\frac{-6}{10}=\frac{-3}{5}\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
Bài 1
a)\(\left(-\dfrac{2}{3}\right).\dfrac{3}{11}-\left(\dfrac{4}{3}\right)^2.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\left(\dfrac{4}{3}\right)^2\right]\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\dfrac{4}{3}.\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left[\left(-2\right).\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left(-\dfrac{8}{3}\right)\)
\(=-\dfrac{24}{33}\)
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
a, Không có số liệu để tìm x
b, x ϵ {1;2;3}
c, Không có x
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
a)\(x+\dfrac{3}{4}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{3}{4}=\dfrac{2}{4}-\dfrac{3}{4}=-\dfrac{1}{4}\)
b)\(\dfrac{5}{3}.x=\dfrac{4}{7}\\ x=\dfrac{4}{7}:\dfrac{5}{3}=\dfrac{4}{7}.\dfrac{3}{5}=\dfrac{12}{35}\)
c)\(\dfrac{2}{3}:x=\dfrac{-1}{4}\\ x=\dfrac{2}{3}:\dfrac{-1}{4}=\dfrac{2}{3}.\left(-4\right)=-\dfrac{8}{3}\)
d)\(\dfrac{5}{9}-x=\dfrac{4}{3}\\ x=\dfrac{5}{9}-\dfrac{4}{3}=\dfrac{5}{9}-\dfrac{12}{9}=\dfrac{-7}{9}\)
e)\(x+\dfrac{4}{17}=\dfrac{2}{34}\\ x=\dfrac{1}{17}-\dfrac{4}{17}=\dfrac{-3}{17}\)
f)\(3\dfrac{4}{7}:x=11\\ x=\dfrac{25}{7}:11=\dfrac{25}{7}.\dfrac{1}{11}=\dfrac{25}{77}\)
a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)
= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)
= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)
= \(\dfrac{7}{13}\times\dfrac{7}{12}\)
= \(\dfrac{49}{156}\)
b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)
= 0
Tương tự 5. HS tự làm