1/3+4/5. 5/4x4/6
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29 tháng 3 2022
a) \(\dfrac{13}{6}\)b) \(\dfrac{7}{18}\)c) \(5\) d) \(\dfrac{2}{3}\)
1.
a)x\(=\)\(\dfrac{7}{6}\)\(-\dfrac{2}{3}\)
x\(=\dfrac{1}{2}\)
b)x\(=\dfrac{9}{2}:\dfrac{3}{4}\)
x\(=6\)
20 tháng 8 2021
a, 5/9-1/6+1/3 =\(\dfrac{13}{18}\) b, 1/8:9/12+1/3=\(\dfrac{1}{2}\)
c, 5/4x4/3x3/4 = \(\dfrac{5}{4}\) d, 8/5x(3/7+1/4) =\(\dfrac{38}{35}\)
20 tháng 8 2021
a)10/18-3/18+6/18
=13/18
b)1/8.12/9+1/3
=1/6+1/3
=1/6+2/6
=1/2
c)5.4.3/4.3.4=5/4
d)8/5.(12/28+7/28)
=8/5.19/28=38/35
16 tháng 1 2023
1/2x^3y(2x^4y^3-4xy-6)
=1/2x^3y*2x^4y^3-1/2x^3y*4xy-1/2x^3y*6
=x^7y^4-2x^4y^2-3x^3y
16 tháng 1 2023
1/2x^3y(2x^4y^3-4xy-6)
=1/2x^3y*2x^4y^3-1/2x^3y*4xy-1/2x^3y*6
=x^7y^4-2x^4y^2-3x^3y
16 tháng 1 2023
1: \(=-3x^3-21x^2+x\)
2: \(=-15x^4y^7+10x^5y^6+5x^3y^5\)
3: \(=x^7y^4-2x^4y^2-3x^3y\)
5: \(=15x-6x^2\)
6: \(=4x^3-8x^2+10x\)
7: \(=-8x^5y^3+16x^7y^2-12x^3y^4\)
8: \(=x^7y^4-2x^4y^2-3x^3y\)
TT
4
25 tháng 8 2023
a)\(\left(\dfrac{5}{6}+\dfrac{3}{8}\right)\times\dfrac{2}{7}\)
\(=\dfrac{29}{24}\times\dfrac{2}{7}\)
\(=\) \(\dfrac{29}{84}\)
b) \(\dfrac{6}{7}\times\dfrac{2}{3}\div\dfrac{5}{7}\)
= \(\dfrac{4}{7}\div\) \(\dfrac{5}{7}\)
= \(\dfrac{4}{5}\)
c) \(\dfrac{8}{9}+\dfrac{3}{4}\times\dfrac{4}{9}\)
= \(\dfrac{8}{9}+\dfrac{1}{3}\)
= \(\dfrac{5}{9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
Cô tích xanh cho lisa blackpink vì câu a, b đúng. Còn câu c thì em làm như sau:
c, \(\dfrac{8}{9}\) + \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{9}\)
= \(\dfrac{8}{9}\) + \(\dfrac{1}{3}\)
= \(\dfrac{8}{9}\) + \(\dfrac{3}{9}\)
= \(\dfrac{11}{9}\)
28 tháng 6 2021
6) Ta có: \(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y-4\right)\left(x+y+3\right)\)
7) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
8) Ta có: \(4x^4-32x^2+1\)
\(=4x^4+12x^3+2x^2-12x^3-36x^2-6x+2x^2+6x+1\)
\(=2x^2\left(2x^2+6x+1\right)-6x\left(2x^2+6x+1\right)+\left(2x^2+6x+1\right)\)
\(=\left(2x^2+6x+1\right)\left(2x^2-6x+1\right)\)
9) Ta có: \(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left[x^4+2x^2+1-x^2\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)
\(=2\left(x-1\right)^2\cdot\left(x^2+x+1\right)\)
NH
0
6 tháng 6 2017
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
1/3+4/5. =1.13333333333
5/4x4/6=0.83333333333
toán lớp 4 ak ????
(^^) HT~
Trả lời :
\(\frac{1}{3}+\frac{4}{5}=\frac{17}{15}\)
\(\frac{5}{4}\times\frac{4}{6}=\frac{5}{6}\)