Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
Chữ số tận cùng của:
a) A = 0
b) B = 8
c) C = 4
Mk ko chắc đâu! Mk nghĩ thế, ko bik đúng hay sai.
Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)
a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
a: \(=\dfrac{1}{3}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}+\dfrac{-5}{2}+\dfrac{3}{2}\)
\(=\dfrac{1}{3}+\dfrac{1}{8}-1-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{8+3-24}{24}-\dfrac{165}{100}\)
\(=\dfrac{-263}{120}\)
b: \(=\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{13}{8}-\dfrac{2}{11}+\dfrac{3}{2}\)
\(=\dfrac{5}{22}-\dfrac{4}{22}-\dfrac{13}{8}+\dfrac{12}{8}+\dfrac{3}{13}\)
\(=\dfrac{1}{22}-\dfrac{1}{8}+\dfrac{3}{13}=\dfrac{173}{1144}\)
Bạn gửi gì vậy không có câu hỏi thì đừng gửi nha