thu gọn các đa thức sau:
a,2a^3.(-1/2ab).a^2b
b,-2/1/3a^3c^2.1/7ac^2.6abc
c,2ab.4/3a^2b^4.7abc
d,2y.3y^2.d^2y^2
e,(-2/1/3.cd).(1/1/4c^2d).(-5/6cd)^2
g,(1/2a.1/4a^2.1/8^3)^2.2b.4b^2-8b^3
thu gọn các đơn thức sau
a)ab.4/3a^2b^4.7abc
b)a^3b^3.a^2b^2c
c)2/3a^3b.(-1/2ab).a^2b
d)-2 1/3a^3c^21/7ac^2 6abc
e)(-1,5ab^2)1/4bca^2b
a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)
b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)
c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)
d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)
e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)
Thu gọn đa thức sau:
a) A= \(5xy - y^2 - 2xy +4xy + 3x -2y\)
b) B= \(\dfrac{1}{2}ab^2 - \dfrac{7}{8}ab^2 + \dfrac{3}{4}a^2 b - \dfrac{3}{8}a^2b - \dfrac{1}{2}ab^2\)
c) C= \(2a^2b - 8b^2 + 5a^2b + 5c^2 - 3b^2 + 4c^2\)
Giúp mình với ạ. Cảm ơn các bạn nhiều!!
a: \(A=\left(5xy-2xy+4xy\right)+3x-2y-y^2\)
\(=7xy+3x-2y-y^2\)
b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2-\dfrac{1}{2}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)
\(=\dfrac{-7}{8}ab^2+\dfrac{3}{8}a^2b\)
c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)
\(=7a^2b-11b^2+9c^2\)
\(A=5xy-y^2-2xy+4xy+3x-2y\)
\(A=-y^2+7xy+3x-2y\)
\(B=\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2+\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b-\dfrac{1}{2}ab^2\)
\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)
\(C=2a^2b-8b^2+5a^2b+5c^2-3b^2+4c^2\)
\(C=7a^2b-11b^2+9c^2\)
\(A=7xy-y^2+3x-2y\)
\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)
\(C=7a^2b-11b^2+9c^2\)
Bài 1: Tính (rút gọn)
3)5x/42y^2 . 7y/x;
5) -25x^4y^3/14a^2 : (10x^3y^2/-21ab);
7) -25a^3b^5/3cd^2 : (15ab^2);
9) 5ab - 6b/ 9a^2 - 6ab . 2b - 3a/ b;
11) 4a^2 - 9b^2/a^2b^2 : 2ax + 3bx/ 2ab;
13) 2x^2 + 2xy/ 3y - 3x . y- x/y+x;
15) 2x - 2y/ 8 - b^3 . 4 + 2b + b^2/ x- y;
17) 3a + 3b/ b^3 - 1 : a + b/ b^2 + b+ 1;
19) 2a - 2/ 3 - 2b + 3a - 2ab: 1/ 4a + 4;
* Lưu ý: "/" nghĩa là phần
Giúp mik vs cần gấp sáng mai phải nộp bài cho cô r
Question Expandand simplify: 1. 8(x+5)-3(2x+7)
2. a(2b+c)+b(3c-2a)
3. 2y(y+5x)+x(3x+4y)
answer , 1. 8(x+5)-3(2x+7)=8x+40-6x+21=2x+61
2. a(2b+c)+b(3c-2a)=2ab+ac+3bc-2ab=ac+3bc=3abc^(2)
3. 2y(y+5x)+x(3x+4y)=2y^(2)+10xy+9x^(2)+4xy=9x^(2)+2y^(2)+14xy
a Explain what he has done wrong.
b work out the correct answer
Thực hiện các phép tính sau:
a) \(\dfrac{3x^2-3y^2}{5xy}.\dfrac{15x^2y}{2y-2x}\)
b) \(\dfrac{4a^2-3a+5}{a^3-1}-\dfrac{1-2a}{a^2+a+1}-\dfrac{6}{a-1}\)
c) \(\dfrac{2a^3-2b^3}{3a+3b}.\dfrac{6a+6b}{a^2-2ab+b^2}\)
d) \(x^2+1-\dfrac{x^4+1}{x^2+1}\)
e) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
a) = \(\dfrac{3\left(x^2-y^2\right)}{5xy}\) . \(\dfrac{15x^2y}{-2\left(x-y\right)}\)
= \(\dfrac{3\left(x-y\right)\left(x+y\right)}{1}\) . \(\dfrac{3x}{-2\left(x-y\right)}\)
= \(\dfrac{3\left(x+y\right)}{1}\) . \(\dfrac{3x}{-2}\) = \(\dfrac{-9x\left(x+y\right)}{2}\)
Bài 1: Rút gọn các phân thức
5) 15ab + 5b2 phần 9a^2 - b^2;
7) 3x^2 - 3y^2 phần 9x + 9y;
9) m^2 - 4m + 4 phần 2m - 4;
Bài 6 Tính (rút gọn )
1) 5by phần 12ax . 2x phần 5y;
3) 5x phần 42y^2 . 7y phần x;
5) -25x^4y^3 phần 14a^2 phần (10x^3y^2 phần -21ab);
7) -25a^3b^5 phần 3cd^2 : (15ab^2);
9) 5ab - 6b phần 9a^2 - 6ab . 2b - 3a phần b;
11) 4a^2 - 9b^2 phần a^2b^2 : 2ax + 3bx phần 2ab;
13) 2x^2 + 2xy phần 3y - 3x . y - x phần y + x;
15) 2x - 2y phần 8 - b^3 . 4b+ 2b + b^2 phần x - y;
17) 3a + 3b phần b^3 - 1 : a + b phần b^2 + B + 1;
19) 2a - 2 phần 3 - 2b + 3a - 2ab : 1 phần 4a + 4;
Giúp mik giải vs sắp phải KT òi TvT
Bài 1:
\(\frac{15ab+5b^2}{9a^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a\right)^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}=\frac{5b}{3a-b}\)
\(\frac{3x^2-3y^2}{9x+9y}=\frac{3\left(x^2-y^2\right)}{9\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{3\left(x+y\right)}=\frac{x-y}{3}\)
\(\frac{m^2-4m+4}{2x-4}=\frac{\left(x-2\right)^2}{2\left(x-2\right)}=\frac{x-2}{2}\)
Thu gọn các đa thức
a)\(2a^2x^3-ax^3-a^4-a^2x^3-ax^3+2a^4\) b)\(3xx^4+4xx^3-5x^2x^3-5x^2x^2\)
c)\(3a.4b^2-0,8b.4b^2-2ab.3b+b.3b^2-1\) d)\(5x.2y^2-5x.3xy-x^2y+6xy^2\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
1) \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2) \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3) \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4) \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
1. Rút gọn các biểu thức sau:
M = (2a+b)2-(b-2a)2
N = (3a+2)2+2a(1-2b)+(2b-1)2
A = (m-n)2+4mn
2. Tính:
a) (x+5)2 b) (5/2-t)2
c) (2u+3v)2 d) (-1/8 a+2/3 bc)2
e) (x/y-1/z)2 f) (mn/4-x/6)(mn/4+x/6)
Bài 2:
a) \(\left(x+5\right)^2=x^2+10x+25\)
b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
Bài 1:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
Bài 1:
a: Ta có: \(M=\left(2a+b\right)^2-\left(b-2a\right)^2\)
\(=4a^2+4ab+b^2-b^2+4ab-4a^2\)
\(=8ab\)
b: Ta có: \(N=\left(3a+2\right)^2+2a\left(1-2b\right)+\left(2b-1\right)^2\)
\(=\left(3a+2+1-2b\right)^2\)
\(=\left(3a-2b+3\right)^2\)
\(=9a^2+4b^2+9-12ab+18a-12b\)
c: Ta có: \(A=\left(m-n\right)^2+4nm\)
\(=m^2-2mn+n^2+4mn\)
\(=m^2+2mn+n^2\)
\(=\left(m+n\right)^2\)
2:
a: \(\left(x+5\right)^2=x^2+10x+25\)
b: \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)