HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
\(Cm:\)\(Tacó:1\cdot2\cdot3\cdot4+4\cdot5\cdot6+3\cdot4\cdot8:\left(4\right)\)
\(=240\) \(:4\)
\(=60\)
\(Vậy...\)
Mỗi bao 120kg = ... kg
sai sai ấy
lx
\(a.\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)-4x\left(x+1\right)=20\)
\(-4x+20=20\)
\(-4x=20-20\)
\(-4x=0\)
\(x=0:\left(-4\right)\)
\(x=0.\)
\(b.\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(x^3+8-x^3-2x=15\)
\(-2x+8=15\)
\(-2x=15-8\)
\(-2x=7\)
\(x=7:\left(-2\right)\)
\(x=-\dfrac{7}{2}.\)
\(c.5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x-4\right)\left(x+4\right)=10\)
\(30x-235=10\)
\(30x=10+235\)
\(30x=245\)
\(x=245:30\)
\(x=\dfrac{49}{6}.\)
\(d.\left(3x-2\right)\left(9x^2+6x+4\right)+27x\left(\dfrac{1}{3}-x\right)\left(\dfrac{1}{3}+x\right)=1\)
\(3x-8=1\)
\(3x=1+8\)
\(3x=9\)
\(x=9:3\)
\(x=3\)
\(e.\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=5\)
\(-2x+8=5\)
\(-2x=5-8\)
\(-2x=-3\)
\(x=-3:\left(-2\right)\)
\(x=\dfrac{3}{2}.\)
\(f.\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(9x+7=17\)
\(9x=17-7\)
\(9x=10\)
\(x=10:9\)
\(x=\dfrac{10}{9}\)
\(Bài1\)
\(a.\dfrac{-3}{7}\cdot\dfrac{6}{13}+\dfrac{-4}{7}\cdot\dfrac{6}{13}+\dfrac{-7}{13}=-\dfrac{18}{91}-\dfrac{24}{91}-\dfrac{7}{13}\)
\(=\dfrac{-42}{91}-\dfrac{7}{13}=-\dfrac{6}{13}-\dfrac{7}{13}=\dfrac{-13}{13}=-1.\)
\(b.\dfrac{7}{8}+\dfrac{13}{16}:26-\dfrac{6}{24}\cdot\left(-2\right)^3=\dfrac{7}{8}+\dfrac{13}{16}:26-\dfrac{1}{4}\cdot\left(-8\right)\)
\(=\dfrac{7}{8}+\dfrac{1}{32}-\left(-2\right)=\dfrac{93}{32}=2\dfrac{29}{32}.\)
\(c.\dfrac{12\cdot3\cdot\left(-4\right)\cdot5\cdot36}{100\cdot54}=\dfrac{-12\cdot3\cdot4\cdot5\cdot36}{100\cdot3\cdot18}=\dfrac{-12\cdot4\cdot5\cdot36}{4\cdot25\cdot18}\)
\(=\dfrac{-12\cdot5\cdot36}{25\cdot18}=\dfrac{-12\cdot5\cdot18\cdot2}{5\cdot5\cdot18}=\dfrac{-12\cdot2}{5}=-\dfrac{24}{5}.\)
\(d.\dfrac{27}{60}-\dfrac{15}{120}+9\dfrac{3}{10}-\dfrac{33}{24}=\dfrac{27}{60}-\dfrac{15}{120}+\dfrac{93}{10}-\dfrac{33}{24}\)
\(=\dfrac{9}{20}-\dfrac{1}{8}+\dfrac{93}{10}-\dfrac{11}{8}=\dfrac{9}{20}+\dfrac{-12}{8}+\dfrac{93}{10}=\dfrac{9}{20}-\dfrac{3}{2}+\dfrac{93}{10}\)
\(=\dfrac{9}{20}-\dfrac{30}{20}+\dfrac{196}{20}=\dfrac{165}{20}=\dfrac{33}{4}=8\dfrac{1}{4}.\)
ảnh lỗi ko thấy bạn ơi
Biến có nghĩa là ví dụ nhé
-80x^6y^5
thì:
Bậc là 11. hệ số là -80; Biến là x^6y^5
vậy bạn xem lại bạn nói sai hay bạn ấy sai
bài này có mà
hình như bạn nói sai rồi đấy.