Lời giải:

$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{20-19}{19.20}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}$

$=1-\frac{1}{20}=\frac{19}{20}$

19/20

Bài 1: 

c: Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là 

x1=1; \(x2=\dfrac{c}{a}=\dfrac{3\sqrt{2}+1}{1-\sqrt{2}}\)

a: \(\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\3x+12y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-6y=-2\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Bài 2: 

a: \(\Leftrightarrow x^2+2x-x^2=0\)

=>2x=0

hay x=0

b: \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

hay \(x\in\left\{-3;5\right\}\)

`@` `\text {Ans}`

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`2,`

Vì số đó chia cho `54` được dư là `36`

`=>` Cần bớt ít nhất `36` đơn vị để số đó chia hết cho `54.`

`3,`

`x \times 36 = 972

`x = 972 \div 36`

`x=27`

Vậy, `x=27`

`(x-47) \times 21 = 840`

`x-47 = 840 \div 21`

`x - 47 =40`

`x = 40 + 47`

`x = 87`

Vậy, `x=87`

`x \times 22 + x \times 50 = 504`

`x \times (22+50) = 504`

`x \times 72 = 504`

`x = 504 \div 72`

`x=7`

Vậy, `x=7`

`729 \div x = 81

`x = 729 \div 81`

`x =9`

Vậy, `x=9`

`@` `\text {Kaizuu lv uuu}`

*Phần kết luận thêm hay k tùy bạn nha! K cần cũng dc!*

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They were disappointed that he didn't keep the secret

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he is worried that he will be punished

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Bài 2:

a. \(R=U:I=10:0,5=20\Omega\)

b. \(R=p\dfrac{l}{S}\Rightarrow S=\dfrac{p.l}{R}=\dfrac{0,50.10^{-6}.10}{20}=2,5.10^{-7}\left(m^2\right)\)

Bài 3:

a. \(R=\left(\dfrac{R1.R2}{R1+R2}\right)+R3=\left(\dfrac{30.15}{30+15}\right)+10=20\Omega\)

Bài 2:

\(b,\Delta=4\left(m+1\right)^2-4m\left(m+2\right)\\ =4m^2+8m+4-4m^2-8m=4>0,\forall m\)

Vậy PT có 2 nghiệm phân biệt với mọi m

Khi đó \(\left[{}\begin{matrix}x=\dfrac{2\left(m+1\right)-2}{2}=m\\x=\dfrac{2\left(m+1\right)+1}{2}=m+2\end{matrix}\right.\)

Bài 3:

Gọi nghiệm chung 2 PT là \(x_1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1^2-\left(m+4\right)x_1+m+5=0\left(1\right)\\x_1^2-\left(m+2\right)x_1+m+1=0\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left(1\right)-\left(2\right)=-\left(m+4\right)x_1+m+5+\left(m+2\right)x_1-m-1=0\\ \Leftrightarrow x_1\left(-m-4+m+2\right)+4=0\\ \Leftrightarrow-2x_1=-4\Leftrightarrow x_1=-2\)

Thay \(x_1=-2\) vào \(\left(2\right)\Leftrightarrow4+2\left(m+2\right)+m+1=0\)

\(\Leftrightarrow3m+9=0\Leftrightarrow m=-3\)

Bài 2:

\(a,\forall m=0\\ PT\Leftrightarrow-2\left(m-1\right)x=-2\left(m-1\right)\Leftrightarrow x=1\\ \forall m\ne0\\ \Delta=4\left(m-1\right)^2-8m\left(m-1\right)\\ =4m^2-8m+4-8m^2+8m\\ =4-4m^2\)

PT vô nghiệm \(\Leftrightarrow4-4m^2< 0\Leftrightarrow m^2>1\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)

PT có nghiệm kép \(\Leftrightarrow4-4m^2=0\Leftrightarrow\left(1-m\right)\left(1+m\right)=0\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-1\end{matrix}\right.\)

Khi đó \(x=\dfrac{2\left(m-1\right)}{2m}=\dfrac{m-1}{m}\)

PT có 2 nghiệm phân biệt \(\Leftrightarrow4-4m^2>0\Leftrightarrow-1< m< 1;m\ne0\)

Khi đó \(x=\dfrac{2\left(m-1\right)\pm\sqrt{4-4m^2}}{2m}=\dfrac{2\left(m-1\right)\pm2\sqrt{1-m^2}}{2m}=\dfrac{m-1\pm\sqrt{1-m^2}}{2}\)