Bài 1:

+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)

\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)

+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)

\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)

\(y=\dfrac{9}{50}\)

Bài 2:

+) \(=\dfrac{2}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{7}{7}=\dfrac{2}{5}\)

+) \(\dfrac{2}{9}:\dfrac{2}{3}:\dfrac{3}{9}\)

\(\dfrac{2}{9}\times\dfrac{3}{2}\times\dfrac{9}{3}=1\)

viết rứa ai mà biết

a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6

Câu đầu em xem lại đề bài sao có hai dấu bằng.

Câu 2: 

\(\dfrac{3}{2}\) \(\times\)y - \(\dfrac{3}{4}\) \(\times\)y + y = \(\dfrac{4}{5}\)

y \(\times\) ( \(\dfrac{3}{2}\) - \(\dfrac{3}{4}\) + 1) = \(\dfrac{4}{5}\)

y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)

y \(\times\) \(\dfrac{7}{4}\)            = \(\dfrac{4}{5}\)

y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)

y = \(\dfrac{16}{35}\)

\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)

Bậc:3

Thay x=-1, y=1 vào B ta có:

\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)

b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)

\(a,\dfrac{1}{3x-3y}=\dfrac{x-y}{3\left(x-y\right)^2};\dfrac{1}{x^2-2xy+y^2}=\dfrac{3}{3\left(x-y\right)^2}\\ b,\dfrac{3}{x^2-3x}=\dfrac{6}{2x\left(x-3\right)};\dfrac{5}{2x-6}=\dfrac{5x}{2x\left(x-3\right)}\\ c,\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{3-x}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{x^2-9}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}\)

\(d,\dfrac{1}{x^2+xy}=\dfrac{xy-y^2}{xy\left(x+y\right)\left(x-y\right)};\dfrac{1}{xy-y^2}=\dfrac{x^2+xy}{xy\left(x-y\right)\left(x+y\right)};\dfrac{2}{y^2-x^2}=\dfrac{-2xy}{xy\left(x-y\right)\left(x+y\right)}\)

Bài 4:

\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)

\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)

\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)

\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)

\(\Rightarrow y=\dfrac{5}{8}\)

________________

\(456+y:87=23987\)

\(\Rightarrow y:87=23987-456\)

\(\Rightarrow y:87=23531\)

\(\Rightarrow y=23531\cdot87\)

\(\Rightarrow y=2047197\)

a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)

\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)

\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)

b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)

\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)

\(=\dfrac{5}{6}+\dfrac{17}{15}\)

\(=\dfrac{59}{30}\)

Bài 2:

a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)

\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)

\(y:\dfrac{2}{5}=\dfrac{25}{16}\)

\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)

\(y=\dfrac{5}{8}\)

b)\(456+y:87=23987\)

\(y:87=23987-456\)

\(y:87=23531\)

\(y=23531\times87\)

\(y=2047197\)

a) 4/5 x 5/8 : 4/5

= 5/8

b) 5/6 + ( 1/2 : 3/2 + 4/5)

= 5/6 + (1/3 + 4/5)

= 5/6 + 17/15

= 59/30

B4:

3/4 + y : 2/5 = 37/16

y : 2/5 = 25/16

y = 5/8.

456 + y : 87 = 23987

y : 87 = 23531

y = 2047197.

a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

a) \(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{4}{3}+\dfrac{3}{4}\)

\(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{25}{12}\)

\(\dfrac{1}{2}:y=\dfrac{25}{12}:\dfrac{3}{5}\)

\(\dfrac{1}{2}:y=\dfrac{125}{36}\)

\(y=\dfrac{1}{2}:\dfrac{125}{36}\)

\(y=\dfrac{18}{125}\)

b) \(\dfrac{4}{3}-\dfrac{1}{2}\times y=1\)

\(\dfrac{1}{2}\times y=\dfrac{4}{3}-1\)

\(\dfrac{1}{2}\times y=\dfrac{1}{3}\)

\(y=\dfrac{1}{3}:\dfrac{1}{2}\)

\(y=\dfrac{2}{3}\)

c) \(\dfrac{1}{4}+y:\dfrac{1}{3}=\dfrac{5}{6}\)

\(y:\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{1}{4}\)

\(y:\dfrac{1}{3}=\dfrac{7}{12}\)

\(y=\dfrac{7}{12}\cdot\dfrac{1}{3}\)

\(y=\dfrac{7}{36}\)