a) AB//CD \(\Rightarrow\widehat{B}+\widehat{C}=180^o\) (trong cùng phía)

\(\Rightarrow140^o+x=180^o\Rightarrow x=180^o-140^o=40^o\)

b) MN//PQ \(\Rightarrow\left\{{}\begin{matrix}\widehat{Q}+\widehat{M}=180^o\\\widehat{N}+\widehat{P}=180^o\end{matrix}\right.\) (trong cùng phía 

\(\widehat{Q}+\widehat{M}=180^o\Rightarrow60^o+x=180^o\Rightarrow x=120^o\)

\(\widehat{N}+\widehat{P}=180^o\Rightarrow\left(180^o-\widehat{N_n}\right)+\widehat{P}=180^o\\ \Rightarrow\left(180^o-70^o\right)+y=180^o\Rightarrow110^o+y=180^o\\ \Rightarrow y=70^o\) 

c) HG//KI \(\widehat{H}+\widehat{I}=180^o\) (trong cùng phía)

\(\Rightarrow x+4x=180^o\Rightarrow5x=180^o\Rightarrow x=\dfrac{180^o}{5}=36^o\) 

d) Xét tứ giác STUV có: 

\(\widehat{V}+\widehat{U}+\widehat{S}+\widehat{T}=360^o\\ \Rightarrow2x+x+90^o+90^o=360^o\\ \Rightarrow3x+180^o=360^o\Rightarrow3x=180^o\Rightarrow x=60^o\)

a) Vì AB//CD 

\(\Rightarrow\widehat{ABC}+\widehat{BCD}=180^o\)

\(\Rightarrow140^o+x=180^o\)

\(\Rightarrow x=40^o\)

b) Vì MN//PQ

\(\Rightarrow\widehat{PQM}+\widehat{QMN}=180^o\)

\(\Rightarrow60^o+x=180^o\)

\(\Rightarrow x=120^o\)

Vì MN//PQ

\(\Rightarrow\widehat{QPN}\) bằng góc ngoài tại đỉnh N của tứ giác MNPQ (SLT)

\(\Rightarrow y=70^o\)

c) Vì HG//KI

\(\Rightarrow\widehat{GHI}+\widehat{HIK}=180^o\)

\(\Rightarrow4x+x=180^o\)

\(\Rightarrow5x=180^o\)

\(\Rightarrow x=36^o\)

d) Vì \(\left\{{}\begin{matrix}VS\perp ST\\UT\perp ST\end{matrix}\right.\rightarrow\) VS//UT

\(\Rightarrow\widehat{UVS}+\widehat{VUT}=180^o\)

\(\Rightarrow2x+x=180^o\)

\(\Rightarrow3x=180^o\)

\(\Rightarrow x=60^o\)

\(4\dfrac{7}{5741}\cdot\dfrac{1}{3759}-\dfrac{4}{3759}\cdot1\dfrac{2}{5741}+\dfrac{1}{3759}+\dfrac{1}{3759\cdot5741}\\ =\dfrac{22971}{5741}\cdot\dfrac{1}{3759}-\dfrac{1}{3759}\cdot\dfrac{22972}{5741}+\dfrac{1}{3759}\cdot\dfrac{5741}{5741}+\dfrac{1}{3759}\cdot\dfrac{1}{5741}\\ =\dfrac{1}{3759}\cdot\left(\dfrac{22971}{5741}-\dfrac{22972}{5741}+\dfrac{5741}{5741}+\dfrac{1}{5741}\right)\\ =\dfrac{1}{3759}\cdot\dfrac{5741}{5741}=\dfrac{1}{3759}\cdot1=\dfrac{1}{3759}\)

Với `x=-19 => 1-x=20`

Thay `1-x=20` vào B, ta được:

`B=x^{10}+(1-x).x^9+(1-x).x^8+...+(1-x).x^2+(1-x).x+20`

`=x^{10}+x^9-x^{10}+x^8-x^9+...+x^2-x^3+x-x^2+20`

`=x+20=-19+20=1`

Cho góc bẹt xOy. Vẽ tia Oz thỏa mãn zOy = zOx. Gọi Om và On lần lượt là các tia phân giác của zOx;zOy a) Tính zOx;zOy b) zOm;zOn có phụ nhau không?

a) ta có: góc zOx + zOy = 180 độ
mà zOx = zOy
=> zOy + zOy = 180
2zOy = 180
zOy = 90
vậy zOy = zOx = 90
b) ta có: xOm + mOz = 90
mà xOm = mOz
=> 2mOz = 90
mOz = 45
vậy mOz = xOm = 45
ta có: nOy + nOz = 90
mà nOy = nOz
=> 2nOy = 90
nOy = 45
vậy nOy = nOz = 45
nOz + mOz = 45 + 45 = 90
vậy zOm; zOn có phụ nhau

Với x \(\ge0\) ta có:

\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{3x-6\sqrt{x}+x+3\sqrt{x}+2-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4\sqrt{x}}{\sqrt{x}+2}\)

\(\dfrac{5}{6}=\dfrac{5\times5}{6\times5}=\dfrac{25}{30}\)

\(1=\dfrac{30}{30}\)

4 phân số thỏa mãn đề bài là: \(\dfrac{26}{30};\dfrac{27}{30};\dfrac{28}{30};\dfrac{29}{30}\)

sai r bn ơi

9)

\(a)x+\dfrac{1}{3}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{3}\\ x=\dfrac{9}{12}-\dfrac{4}{12}\\ x=\dfrac{5}{12}\\ b)x-\dfrac{2}{5}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{2}{5}\\ x=\dfrac{25}{35}+\dfrac{14}{35}\\ x=\dfrac{39}{35}\\ c)-x-\dfrac{2}{3}=-\dfrac{6}{7}\\ x=-\dfrac{2}{3}+\dfrac{6}{7}\\ x=-\dfrac{14}{21}+\dfrac{18}{21}\\ x=\dfrac{4}{21}\\ d)\dfrac{4}{7}-x=\dfrac{1}{3}\\ x=\dfrac{4}{7}-\dfrac{1}{3}\\ x=\dfrac{12}{21}-\dfrac{7}{21}\\ x=\dfrac{5}{21}\)

6)
a) \(\dfrac{-1}{21}+\dfrac{-1}{28}\)                        b) \(\dfrac{-8}{18}-\dfrac{15}{27}\)                c) \(\dfrac{-5}{12}+0,75\)           d) \(3,5-\left(-\dfrac{2}{7}\right)\)
= \(\dfrac{-28}{84}+\dfrac{-21}{84}\)                     = \(\dfrac{-4}{9}-\dfrac{5}{9}\)                  = \(\dfrac{-5}{12}+\dfrac{3}{4}\)                 = \(\dfrac{7}{2}+\dfrac{2}{7}\)
= \(\dfrac{-49}{84}\)                                  = \(\dfrac{-9}{9}\)                         = \(\dfrac{-5}{12}+\dfrac{9}{12}\)               = \(\dfrac{49}{14}+\dfrac{4}{14}\)
= \(\dfrac{-7}{12}\)                                    = -1                            = \(\dfrac{4}{12}\)                         = \(\dfrac{53}{14}\)
                                                                                = \(\dfrac{1}{3}\)

Đặt \(\left\{{}\begin{matrix}\ln x=u\Rightarrow\dfrac{1}{x}dx=du\\dv=x^2dx\Rightarrow v=\dfrac{1}{3}x^3\end{matrix}\right.\)

Ta có: \(\int\limits^e_1x^2.\ln x=\dfrac{1}{3}x^3.\ln x|^e_1-\int\limits^e_1\dfrac{1}{3}x^2=\dfrac{e^3}{3}-\dfrac{1}{9}x^3|^e_1=\dfrac{e^3}{3}-\dfrac{e^3}{9}+\dfrac{1}{9}=\dfrac{2e^3}{9}+\dfrac{1}{9}\)

\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x+1}\left(x\ne2;x\ne-3;x\ne-1;x\ne-\dfrac{1}{2}\right)\\ \Leftrightarrow\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{3x+9}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow3\left(2x+1\right)=-3\left(x-2\right)\left(x+1\right)\\ \Leftrightarrow2x+1=-\left(x^2+x-2x-2\right)\\ \Leftrightarrow2x+1=-x^2+x+2\\ \Leftrightarrow x^2+2x-x+1-2=0\\ \Leftrightarrow x^2+x-1=0\)

\(\Delta=1^2-4\cdot1\cdot\left(-1\right)=5>0\)

\(x_1=\dfrac{-1+\sqrt{5}}{2\cdot1}=\dfrac{\sqrt{5}-1}{2}\\ x_2=\dfrac{-1-\sqrt{5}}{2\cdot1}=\dfrac{-\sqrt{5}-1}{2}\) (tm) 

Vậy: ... 

ĐK: \(x\notin\left\{-3;2;-1;-\dfrac{1}{2}\right\}\)

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

\(\Leftrightarrow\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{-3}{2x+1}\)

\(\Leftrightarrow\dfrac{5\left(x+1\right)-2\left(x-2\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\)

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

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

\(\Rightarrow2x+1=-\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow2x+1=-x^2+x+2\)

\(\Leftrightarrow x^2+x-1=0\)

\(\Delta=1^2-4\cdot\left(-1\right)=1+4=5>0\)

\(\rightarrow\) PT có 2 nghiệm pb 

\(x_1=\dfrac{-1+\sqrt{5}}{2}\) (T/m)

\(x_2=\dfrac{-1-\sqrt{5}}{2}\) (T/m)

Vậy \(S=\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)