a: =(-3/2)*(-2/3)+(5/2-3/4):7/4

=1+7/4:7/4=1+1=2

b: \(=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

=1/3*102/103=34/103

1 fun (make fun of sb: trêu chọc, biến ai thành trò đùa)

2 down (put sb down = criticise sb)

3 might have (I might have known -> show rằng tôi ko thấy bất ngờ tí nào về sự việc đc nhắc đến)

4 as I may (try as sb may/might = despite one's best efforts)

5 accept (accept an invitation = chấp nhận lời mời)

6 do as it was (do as sb be told = obey, làm như đc bảo)

7 waste of (be a waste of money = chỉ tổ phí tiền)

8 of looking after (look after one's interests = quan tâm đến lợi ích của ai)

9 have fell (fall asleep = start to sleep)

10 earth are you (what on earth are you doing? = m đang làm cái quái gì vậy? :'D)

1 him

2 her down

3 have

4 out

5 accept

6 be

7 waste of

8 of protecting

9 fall

10 chưa nghĩ ra

1 là fun

Câu I:

1) Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{x+\sqrt{x}-4}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\dfrac{x+x+\sqrt{x}-2-x-\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

2) Để P=3 thì \(\dfrac{x+2}{\sqrt{x}}=3\)

\(\Leftrightarrow x+2=3\sqrt{x}\)

\(\Leftrightarrow x-3\sqrt{x}+2=0\)

\(\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\)

\(\Leftrightarrow\sqrt{x}\cdot\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: Để P=3 thì x=4

Có

\(\left|x-2\right|+\left|x-4\right|=\left|x-2\right|+\left|4-x\right|\ge\left|x-2+4-x\right|=2\)

\(\left|x-3\right|\ge0\)

=> \(\left|x-2\right|+\left|x-4\right|+\left|x-3\right|\ge2\)

Dấu "=" xảy ra 

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x-2>0\\4-x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=3\\x-2< 0\\4-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\x>2\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\x< 2\\x>4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k2\pi\)

\(tan2x.tanx=1\)

\(\Leftrightarrow\dfrac{sin2x.sinx}{cos2x.cosx}=1\)

\(\Leftrightarrow sin2x.sinx=cos2x.cosx\)

\(\Leftrightarrow\dfrac{1}{2}\left(cosx-cos3x\right)=\dfrac{1}{2}\left(cos3x+cosx\right)\)

\(\Leftrightarrow cos3x=0\)

\(\Leftrightarrow3x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)

a: Xét ΔMEB vuông tại M và ΔACB vuông tại A có 

\(\widehat{B}\) chung

Do đó: ΔMEB\(\sim\)ΔACB

Xét ΔDMC vuông tại M và ΔABC vuông tại A có

\(\widehat{C}\) chung

Do đó: ΔDMC\(\sim\)ΔABC

b: BC=30cm

trả lời:

1996 x 1997 + 1998 x 3 + 1994 / 1997 x 1999 - 1997 x 1997

= 1996 x 1997 + ( 1997 + 1 x 3 ) + 1994 / 1997 x 1999 - 1997 x 1997

= 1996 x 1998 + 1 x 3 + 1994 / 1997 x 1999 - 1997 x 1997

= 3988008 + 3 + 1994 / 1997 x 1999 - 1997 x 1997

= 3990005 / 1997 x 1999 - 1997 x 1997

= 3990005 / ( 1999 - 1997 ) x 1997

= 3990005 / 3994

Bài làm:

\(\frac{1997.1996+1998.3+1944}{1994.1999-1997.1994}\)

\(=\frac{1997.1996+\left(1997+1\right).3+1994}{1994.\left(1999-1997\right)}\)

\(=\frac{1997.1996+1997.3+3+1994}{1994.2}\)

\(=\frac{1997.\left(1996+3\right)+1997}{1994.2}\)

\(=\frac{1997.\left(1999+1\right)}{1994.2}\)

\(=\frac{1997.2000}{1994.2}\)

\(=\frac{998500}{997}\)

Hc tốt!