o ghi lại đề những câu chung mẫu 

a) \(\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)

b) \(\frac{4x-1-7x+1}{3x^2y}=\frac{-3x}{3x^2y}=\frac{-1}{xy}\)

c)\(\frac{xy+x^2-1}{2x-y}\)

d) \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)

MTC: x.(x-1).(x+1)

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

e) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{\left(3x\right)^2-2^2}=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

MTC: (3x-2).(3x+2)

\(=\frac{3x+2-4.\left(3x-2\right)+10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

=\(\frac{3x+2-12x+8+10x+8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x+18}{\left(3x-2\right)\left(3x+2\right)}\)

f) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

=\(\frac{3x+2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)

MTC: (x-1)².(x+1)²

=\(\frac{\left(3x+2\right).\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6.\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{\left(3x-2\right).\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\)

=\(\frac{\left(3x+2\right)\left(x+1\right)^2-6\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)^2}\)

=\(\frac{\left(3x+2\right)\left(x^2+2x+1\right)-6\left(x^2-1\right)-\left(3x-2\right)\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

=\(\frac{3x^3+6x^2+3x+2x^2+4x+1-6x^2+6-\left(3x^3-6x+3x-2x^2+4x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

=\(\frac{3x^3+6x^2+3x+2x^2+4x+1-6x^2+6-3x^3+6x-3x+2x^2-4x+2}{\left(x-1\right)\left(x+1\right)}\)

=\(\frac{4x^2+6x+9}{\left(x-1\right)\left(x+1\right)}\)

f làm cách đặt ntc cách này hơn lằng nhằng còn lại 1 số câu làm tắt nên b chú ý về dấu để mink check b lại luôn qwq

check lại luôn hộ nhé bài f cứ thấy sss =))) dud

TL:

Where is a question?

Hok tốt!

@Kaito Kid

a)Vì D dx với H qua AB => AB là trung trực của DH => AD = AH

cmtt có: AE = AH

mặt khác: A thuộc DE

=> A là trung điểm của DE => D đx E qua A 

Vì AH = AE = AD = 1/2DE

(đường trung tuyến = 1/2 cạnh huyền)

=> DHE là tam giác vuông tại H

-xét ΔBDA và ΔBHA có:

BA: chung

BD = BH (AB là trung trực...)

AH = AH (đã cm)

 xét ΔBDA và ΔBHA có:

BA: chung

BD = BH (AB là trung trực...)

AH = AH (đã cm)

=> ΔBDA = ΔBHA (c.c.c)

=> ˆBDA=ˆBHA=90o⇒BD⊥EDBDA^=BHA^=90o⇒BD⊥ED (1)

Cmtt có: ˆAEC=ˆAHC=90o⇒CE⊥EDAEC^=AHC^=90o⇒CE⊥ED(2)

Từ (1) và (2) => BD // CE => BDEC là hình thang

 b)Ta có: BD = BH; CE = CH (đã cm)

=> BH + CH = BD + CE hay BC = BD + CE (đpcm)

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

\(\frac{3x}{x^3+2x^2+x}=\frac{3x}{x\left(x^2+2x+1\right)}=\frac{3}{\left(x+1\right)^2}\)

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

rzddddddfg

a) \(ĐK:x\ne\pm3\)

 \(P=\left(\frac{3+x}{3-x}-\frac{3-x}{3+x}+\frac{4x^2}{x^2-9}\right):\left(\frac{2x+1}{x+3}-1\right)\)

<=> \(P=\frac{9+6x+x^2-9+6x-x^2+4x^2}{\left(x-3\right)\left(x+3\right)}:\frac{2x+1-x-3}{x+3}\)

<=> \(P=\frac{4x^2+12x}{\left(x-3\right)\left(x+3\right)}:\frac{x-4}{x+3}\)

<=> \(P=\frac{4x\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)\left(x-4\right)}\)

<=> \(P=\frac{4x\left(x+3\right)}{\left(x-3\right)\left(x-4\right)}\)

b) 2x² - 5x + 2 = 0

<=> (2x-1)(x-2) = 0

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=2\end{cases}}\)

Thay x = 1/2 vào P => P = 4/5

         x = 2 vào P => P = 20