Bài 1:

Bài 2:

Bài 3:

Bài 4:

Câu 2:

=>x=1-y và m(1-y)-y=2m

=>x=1-y và m-my-y=2m

=>x=1-y và y(-m-1)=m

=>x=1-y và y=-m/m+1

=>x=1+m/m+1=(m+2)/m+1 và y=-m/m+1

Để x,y nguyên thì m+1+1 chia hết cho m+1 và -m-1+1 chia hết cho m+1

=>\(m+1\in\left\{1;-1\right\}\)

mà m<>0

nên m=-2

\(a,\) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\\ \dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

b, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^4=\dfrac{b^4}{d^4}\\ \dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\\ \RightarrowĐpcm\)

c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\dfrac{b^2}{d^2}\\ \RightarrowĐpcm\)

a: Xét ΔDHE vuông tại H và ΔDKF vuông tại K có 

DE=DF

\(\widehat{D}\) chung

Do đó: ΔDHE=ΔDKF

Suy ra: DH=DK

hay ΔDKH cân tại D

uses crt;

var s:real;

a,i,n:integer;

begin

clrscr;

readln(a,n);

s:=1;

for i:=1 to n do 

  if i mod 2=0 then s:=s*(a+i);

writeln(s:0:0);

readln;

end.

45. I know how to use this machine, I can help you.

46. excited about the journey.

47. Peter is the best student in my class who can solve this difficult problem.

48. very good at typing.

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

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)-\left(1-x\right)\sqrt{3x-2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2+\sqrt{3x-2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2+\sqrt{3x-2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\\sqrt{3x-2}=2-x\left(1\right)\end{matrix}\right.\)

Xét (1) ta có: \(\left\{{}\begin{matrix}2-x\ge0\\3x-2=4-4x+x^2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2\ge x\\x^2-7x+6=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x\le2\\\left[{}\begin{matrix}x=6\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy nghiệm của phương trình là x=1

ĐKXĐ : x \(\ge\dfrac{2}{3}\)

Ta có \(\left(x-1\right)\left(x-2\right)=\sqrt{3x-2}\left(1-x\right)\)

<=> \(\left(x-1\right)\left(x-2+\sqrt{3x-2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{3x-2}=2-x\end{matrix}\right.\)

Khi x - 1 = 0 <=> x = 1 (tm)

Khi \(\sqrt{3x-2}=2-x\)

<=> \(\left\{{}\begin{matrix}3x-2=x^2-4x+4\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-7x+6=0\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-6\right)=0\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow x=1\)

Vậy phương trình 1 nghiêm \(x=1\)

\(ĐK:x\ge\dfrac{2}{3}\)

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

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

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

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

\(\Leftrightarrow2-x=\sqrt{3x-2}\)

\(\Leftrightarrow\left(2-x\right)^2=\left(\sqrt{3x-2}\right)^2\)

\(\Leftrightarrow4-4x+x^2=3x-2\)

\(\Leftrightarrow x^2-7x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\) (vi-et )

Vậy S=\(\left(1;6\right)\)