a)A=(2;1)

b)Chị hong bíc:<<<

c)trục y e chấm 1 điểm(B) ngay số -1

Cj lên nhanh dzay, e bận đi học nên k cày dc

to be 

saying 

crying 

to answer

to find

traveling

to go

to quit

celebration

to know 

making

finishing 

to get 

paiting

to lock

informing

to visit

to exercise

talking

to smoke

to be 

to stop

recycling 

studying

to move 

to help

not to be 

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

X x ( 4,2 + 0,8 ) = 2,64
X x 5                   = 2,64
X                         = 2,64 : 5
X                         = 0,528

X x ( 4,2 + 0,8 ) = 2,64
X x 5                   = 2,64
X                         = 2,64 : 5
X                         = 0,528 ok

Bạn sẽ tính (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) : 9 = 5  

Cảm ơn bn nha

Bài 4: 

Theo định lý sin ta có:
\(\dfrac{AC}{sinB}=\dfrac{BC}{sinA}\)

\(\Rightarrow BC=a=\dfrac{b\cdot sinA}{sinB}=\dfrac{2\cdot sin60^o}{sin45^o}=\sqrt{6}\)

\(\Rightarrow\widehat{C}=180^o-60^o-45^o=75^o\)

\(\dfrac{AC}{sinB}=\dfrac{AB}{sinC}\)

\(\Rightarrow AB=c=\dfrac{b\cdot sinC}{sinB}=\dfrac{2\cdot sin75^o}{sin45^o}=1+\sqrt{3}\) 

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot AC\cdot AB\cdot sinA=\dfrac{1}{2}\cdot2\cdot\left(1+\sqrt{3}\right)\cdot sin75^o=\dfrac{\sqrt{6}+2\sqrt{2}}{2}\) (đvdt) 

Bán kình hình tròn tam giác ABC khi đó là:

\(S_{ABC}=\dfrac{abc}{4R}\)

\(\Rightarrow R=\dfrac{abc}{4S_{ABC}}=\dfrac{2\cdot\left(1+\sqrt{3}\right)\cdot\sqrt{6}}{4\cdot\left(\dfrac{\sqrt{6}+2\sqrt{2}}{2}\right)}=3-\sqrt{3}\) 

Bài 3:

a) Xét tam giác ABC theo định lý côsin ta có:
\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{8^2+10^2-13^2}{2\cdot8\cdot10}=-0,03125\)

\(\Rightarrow\widehat{C}=cos^{-1}-0,03125\approx91^o>90^o\)

Nên tam giác ABC có góc C là góc tù 

c) Theo hệ thức Heron ta có diện tích tam giác ABC là: 

\(S_{ABC}=\sqrt{p\cdot\left(p-a\right)\cdot\left(p-b\right)\cdot\left(p-c\right)}\)

\(\Rightarrow S_{ABC}=\sqrt{\dfrac{8+10+13}{2}\cdot\left(\dfrac{8+10+13}{2}-8\right)\cdot\left(\dfrac{8+10+13}{2}-10\right)\cdot\left(\dfrac{8+10+13}{2}-13\right)}\)

\(\Rightarrow S_{ABC}\approx40\) (đvdt) 

b) Bán kính đường tròn ngoại tiếp tam giác ABC là:
\(S_{ABC}=\dfrac{abc}{4R}\)

\(\Rightarrow R=\dfrac{abc}{4S_{ABC}}=\dfrac{8\cdot10\cdot13}{4\cdot40}=6,5\)

`2)`

`@` Xét `3x+6 >= 0<=>x >= -2`

         `=>A=[-2;+oo)`

`@` Xét `|x-2| < 3`

`<=>-3 < x-2 < 3`

`<=>-1 < x < 5=>B=(-1;5)`

Có: `A nn B=(-1;5)`

      `A uu B=[-2;+oo)`

      `R \\ B=(-oo;-1]uu[5;+oo)`

_______

`3)`

`@` Xét `x+3 >= 2x+7<=>x <= -4=>A=(-oo;-4]`

`@` Xét `4x+5 > 0<=>x > -5/4=>B=(-5/4;+oo)`

`@` Xét `|x+4| < 2<=>-2 < x+4 < 2<=>-6 < x < -2 =>C=(-6;-2)`

Có: `A nn B nn C=\emptyset`

      `A \\ B nn C=(-6;-4]`

       `C \\ A nn B=\emptyset`.