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\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
\(\left|x+5\right|=5\)
<=> \(\hept{\begin{cases}x+5=5\\x+5=-5\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=-10\end{cases}}\)
\(\left|x+1\right|+7=10\)
<=> \(\left|x+1\right|=3\)
<=> \(\hept{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)
<=> \(\hept{\begin{cases}x=2\\x=-4\end{cases}}\)
\(\left|x-3\right|-6=5\)
<=> \(\left|x-3\right|=11\)
<=> \(\hept{\begin{cases}x-3=11\\x-3=-11\end{cases}}\)
<=> \(\hept{\begin{cases}x=14\\x=-8\end{cases}}\)
\(\left|x+2\right|-6\left(x-4\right)=20-6x\)
<=> \(\left|x+2\right|-6x+24=20-6x\)
<=> \(\left|x+2\right|=-4\)
<=> \(\hept{\begin{cases}x+2=-4\\x+2=4\end{cases}}\)
<=> \(\hept{\begin{cases}x=-2\\x=2\end{cases}}\)
a) \(|x+5|=5\)
\(\Rightarrow\orbr{\begin{cases}x+5=5\\x+5=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)
Vậy x = 0 hoặc x = -10
b) \(|x+1|+7=10\)
\(\Rightarrow|x+1|=10-7\)
\(\Rightarrow|x+1|=3\)
\(\Rightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
Vậy x = 2 hoặc x = -4
c) \(|x-3|-6=5\)
\(\Rightarrow|x-3|=5+6\)
\(\Rightarrow|x-3|=11\)
\(\Rightarrow\orbr{\begin{cases}x-3=11\\x-3=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=14\\x=-8\end{cases}}\)
Vậy x = 14 hoặc x = -8
d) \(|x+2|-6\left(x-4\right)=20-6x\)
\(\Rightarrow|x+2|-6x+24=20-6x\)
\(\Rightarrow|x+2|=20-6x-24+6x\)
\(\Rightarrow|x+2|=\left(20-24\right)+\left(-6x+6x\right)\)
\(\Rightarrow|x+2|=-4\)
Vì \(|x|\ge0\)mà \(|x+2|=-4\)
\(\Rightarrow\)Không có giá trị x thỏa mãn
_Chúc bạn học tốt_
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa
\(x:(3\frac{1}{2}-5\frac{1}{6})=4\frac{1}{5}-6\frac{2}{3}\) b,\((2\frac{1}{3}+3\frac{1}{2})\cdot x=-4\frac{1}{4}+3\frac{1}{7}\)
\(x:\frac{-5}{3}\)=\(\frac{-37}{15}\) \(\frac{35}{6}\cdot x=\)\(\frac{-31}{28}\)
\(x=\frac{-37}{15}\cdot\frac{-5}{3}\) \(x=\frac{-31}{28}:\)\(\frac{35}{6}\)
\(x=\frac{37}{9}\) \(x=\frac{-93}{490}\)
Vậy x=\(\frac{37}{9}\) Vậy x=\(\frac{-93}{490}\)
7,5x:\((9-6\frac{13}{21})=2\frac{13}{25}\)
7,5x:\(\frac{50}{21}\)=\(\frac{63}{25}\)
7,5x=\(\frac{63}{25}:\frac{50}{21}=\frac{1323}{1250}\)
x=\(\frac{1323}{1250}:7,5\)
x=\(\frac{441}{3125}\)
Vậy x=\(\frac{441}{3125}\)
a, \(x:3\frac{1}{15}=1\frac{1}{2}\)
\(\Rightarrow x=1\frac{1}{2}\cdot3\frac{1}{15}\)
\(\Rightarrow x=\frac{3}{2}\cdot\frac{46}{15}=\frac{3\cdot46}{2\cdot15}=\frac{1\cdot23}{1\cdot5}=\frac{23}{5}=4\frac{3}{5}\)
\(b)x\cdot\frac{15}{28}=\frac{3}{20}\)
\(\Rightarrow x=\frac{3}{20}:\frac{15}{28}=\frac{3}{20}\cdot\frac{28}{15}=\frac{1}{5}\cdot\frac{7}{5}=\frac{7}{25}\)
Tự làm nốt câu cuối :>
a) x = 25 28
b) x = 5 12