11/9 + 18/5 -2/9 -3/5

= 11/9  - 2/ 9  + 18/5 - 3/5

= 9/9 + 15/5

= 1+ 3

= 4

=11/9-2/9+18/5-3/5=9/9+15/5

                               =1+3=4

\(\dfrac{x+2}{x-1}=\dfrac{x-1}{x-3}\) (1)

ĐKXĐ: \(x\ne1;x\ne3\)

(1) \(\Leftrightarrow\left(x+2\right)\left(x-3\right)=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-3x+2x-6=x^2-2x+1\)

\(\Leftrightarrow-3x+2x+2x=1+6\)

\(\Leftrightarrow x=7\) (nhận)

Vậy S = {7}

\(ĐK:x\ne\dfrac{1}{2};x\ne1;x\ne\dfrac{3}{2};x\ne2;x\ne\dfrac{5}{2}\\ PT\Leftrightarrow\dfrac{1}{\left(2x-1\right)\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(3x-2\right)}+\dfrac{1}{\left(3x-2\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(5x-2\right)}=\dfrac{4}{21}\\ \Leftrightarrow2\left[\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{1}{2}\right)\left(x-1\right)}+\dfrac{\dfrac{1}{2}}{\left(x-1\right)\left(x-\dfrac{3}{2}\right)}+\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{3}{2}\right)\left(x-2\right)}+\dfrac{\dfrac{1}{2}}{\left(x-2\right)\left(x-\dfrac{5}{2}\right)}\right]=\dfrac{4}{21}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{1}{2}}+\dfrac{1}{x-\dfrac{3}{2}}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-\dfrac{3}{2}}+\dfrac{1}{x-\dfrac{5}{2}}-\dfrac{1}{x-2}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{x-\dfrac{5}{2}-x+1}{\left(x-1\right)\left(x-\dfrac{5}{2}\right)}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{-\dfrac{3}{2}}{x^2-\dfrac{7}{2}x+\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow x^2-\dfrac{7}{2}x+\dfrac{5}{2}=-\dfrac{63}{4}\\ \Leftrightarrow4x^2-14x+10=-63\\ \Leftrightarrow4x^2-14x+73=0\\ \Leftrightarrow x\in\varnothing\)

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{3}< >-\dfrac{1}{m}\)

=>\(m^2\ne-3\)(luôn đúng)

Ta có: \(\left\{{}\begin{matrix}mx-y=2\\3x+my=3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-2\\3x+m\left(mx-2\right)=3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-2\\3x+m^2x-2m=3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-2\\x\left(m^2+3\right)=5m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m}{m^2+3}\\y=m\cdot\dfrac{5m}{m^2+3}-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5m}{m^2+3}\\y=\dfrac{5m^2-2m^2-6}{m^2+3}=\dfrac{3m^2-6}{m^2+3}\end{matrix}\right.\)

\(\left(x+y\right)\cdot\left(m^2+3\right)+8=0\)

=>\(\dfrac{5m+3m^2-6}{m^2+3}\cdot\left(m^2+3\right)+8=0\)

=>\(3m^2+5m-6+8=0\)

=>\(3m^2+5m+2=0\)

=>(m+1)(3m+2)=0

=>\(\left[{}\begin{matrix}m=-1\\m=-\dfrac{2}{3}\end{matrix}\right.\)

Ta có: \(\dfrac{\left(x+3\right)\left(x-3\right)}{3}+2=x\left(1-x\right)\)

\(\Leftrightarrow\dfrac{x^2-9}{3}+\dfrac{6}{3}=\dfrac{3x\left(1-x\right)}{3}\)

\(\Leftrightarrow x^2-9+6=3x-3x^2\)

\(\Leftrightarrow x^2-3-3x+3x^2=0\)

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Delta=9-4\cdot4\cdot\left(-3\right)=9+48=57\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{57}}{8}\\x_2=\dfrac{3+\sqrt{57}}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3-\sqrt{57}}{8};\dfrac{3+\sqrt{57}}{8}\right\}\)

Bài 3:

a: \(P=x\left(x^2-y\right)+y\left(x-y^2\right)\)

\(=x^3-xy+xy-y^3\)

\(=x^3-y^3\)

Thay \(x=-\frac12;y=-\frac12\) vào P, ta được:

\(P=\left(-\frac12\right)^3-\left(-\frac12\right)^3=\left(-\frac18\right)-\left(-\frac18\right)=-\frac18+\frac18=0\)

b: \(Q=x^2\left(y^3-xy^2\right)+x^2y^2\left(x-y+1\right)\)

\(=x^2y^3-x^3y^2+x^3y^2-x^2y^3+x^2y^2=x^2y^2\)

Thay x=-10; y=-10 vào Q, ta được:

\(Q=\left(-10\right)^2\cdot\left(-10\right)^2=100\cdot100=10000\)

c: \(A=x^3+2xy-2x^3+2y^3+2x^3-y^3\)

\(=\left(x^3-2x^3+2x^3\right)+2xy+\left(2y^3-y^3\right)\)

\(=x^3+2xy+y^3\)

Thay x=2; y=-3 vào A, ta được:

\(A=2^3+2\cdot2\cdot\left(-3\right)+\left(-3\right)^3\)

=8-12-27

=-4-27

=-31

d:

x=1; y=-1

=>\(xy=1\cdot\left(-1\right)=-1\)

\(B=xy+x^2y^2-x^4y^4+x^6y^6-x^8y^8\)

\(=\left(xy\right)+\left(xy\right)^2-\left(xy\right)^4+\left(xy\right)^6-\left(xy\right)^8\)

\(=\left(-1\right)+\left(-1\right)^2-\left(-1\right)^4+\left(-1\right)^6-\left(-1\right)^8\)

=-1+1-1+1-1

=-1

e: x=-1; y=1

=>xy=-1

\(C=xy+x^2y^2+x^3y^3+\cdots+x^{10}y^{10}\)

\(=xy+\left(xy\right)^2+\left(xy\right)^3+\cdots+\left(xy\right)^{10}\)

\(=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+\cdots+\left(-1\right)^{10}\)

=-1+1+(-1)+1+...+(-1)+1

=0

f: \(M=2x^2\left(x^2-5\right)+x\left(-2x^3+4x\right)+x^2\left(x+6\right)\)

\(=2x^4-10x^2-2x^4+4x^2+x^3+6x^2\)

\(=x^3\)

Khi x=-4 thì \(M=\left(-4\right)^3=-64\)

g: \(N=x^3\left(y+1\right)-xy\left(x^2-2x+1\right)-x\left(x^2+2xy-3y\right)\)

\(=x^3y+x^3-x^3y+2x^2y-xy-x^3-2x^2y+3xy\)

=2xy

Thay x=8; y=-5 vào N, ta được:

\(N=2\cdot8\cdot\left(-5\right)=-80\)

Bài 1:

d: \(x^2+2xy-3\cdot\left(-xy\right)\)

\(=x^2+2xy+3xy=x^2+5xy\)

e: \(\frac12x^2y\left(2x^3-\frac25xy^2-1\right)\)

\(=\frac12x^2y\cdot2x^3-\frac12x^2y\cdot\frac25xy^2-\frac12x^2y\)

\(=x^5y-\frac15x^3y^3-\frac12x^2y\)

f: \(\left(-xy^2\right)^2\left(x^2-2x+1\right)\)

\(=x^2y^4\left(x^2-2x+1\right)\)

\(=x^2y^4\cdot x^2-x^2y^4\cdot2x+x^2y^4\)

\(=x^4y^4-2x^3y^4+x^2y^4\)

g: (2xy+3)(x-2y)

\(=2xy\cdot x-2xy\cdot2y+3\cdot x-3\cdot2y\)

\(=2x^2y-4xy^2+3x-6y\)

h: \(\left(xy+2y\right)\left(x^2y-2xy+4\right)\)

\(=x^3y^2-2x^2y^2+4xy+2x^2y^2-4xy^2+8y\)

\(=x^3y^2+4xy-4xy^2+8y\)

i: \(4\left(x^2-\frac12y\right)\left(x^2+\frac12y\right)\)

\(=4\left(x^4-\frac14y^2\right)\)

\(=4\cdot x^4-4\cdot\frac14y^2=4x^4-y^2\)

k: \(2x^2\left(1-3x+2x^2\right)\)

\(=2x^2\cdot1-2x^2\cdot3x+2x^2\cdot2x^2\)

\(=2x^2-6x^3+4x^4\)

l: \(\left(2x^2-3x+4\right)\left(-\frac12x\right)\)

\(=-\frac12x\cdot2x^2+3x\cdot\frac12x-4\cdot\frac12x=-x^3+\frac32x^2-2x\)

m: \(\frac12xy\left(-x^3+2xy-4y^2\right)\)

\(=-\frac12xy\cdot x^3+\frac12xy\cdot2xy-\frac12xy\cdot4y^2\)

\(=-\frac12x^4y+x^2y^2-2xy^3\)

n: \(\frac12x^2y\left(2x^3-\frac25xy^2-1\right)\)

\(=\frac12x^2y\cdot2x^3-\frac12x^2y\cdot\frac25xy^2-\frac12x^2y\)

\(=x^5y-\frac15x^3y^3-\frac12x^2y\)

280 - x.9 = 450

x.9 = 280 - 450

x.9 = -170

x= -170/9

Câu 42

Phương trình chuyển động của vật

\(x=x_0+vt+\dfrac{1}{2}at^2=20t-\dfrac{1}{2}\cdot2t^2=5t-t^2\left(m,s\right)\)

Câu 45

< mình ko thấy hình nha bạn>

mik ko biết câu hỏi nên trl đại nha

Câu 26:

Đổi 36km/h = 10m/s; 54km/h = 15m/s

Gia tốc của tàu:

Ta có: \(v=v_0+at\Leftrightarrow a=\dfrac{v-v_0}{t}=\dfrac{15-10}{2}=2,5\left(m/s\right)\)

Quãng đường xe đi đc trong khoảng thời gian đó:

   \(s=v_0t+\dfrac{1}{2}at^2=10.2+\dfrac{1}{2}.2,5.2^2=25\left(m\right)\)