\(a,PT\left(1\right)=\dfrac{75y^4}{42x^2y^5};PT\left(2\right)=\dfrac{28x}{42x^2y^5}\\ b,PT\left(1\right)=\dfrac{11y^2}{102x^4y^3};PT\left(2\right)=\dfrac{9x^3}{10x^4y^3}\\ c,PT\left(1\right)=\dfrac{3x\left(3x+1\right)}{36x^2y^4};PT\left(2\right)=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ d,PT\left(1\right)=\dfrac{6y^2}{36x^3y^4};PT\left(2\right)=\dfrac{4x\left(x+1\right)}{36x^3y^4};PT\left(3\right)=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\)

\(e,PT\left(1\right)=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};PT\left(2\right)=\dfrac{75x^2y^3}{120x^4y^5};PT\left(3\right)=\dfrac{8x^3}{120x^4y^5}\\ f,PT\left(1\right)=\dfrac{3\left(x+1\right)\left(4x-4\right)}{6x\left(x+3\right)\left(x+1\right)};PT\left(2\right)=\dfrac{2\left(x+3\right)\left(x-3\right)}{6x\left(x+1\right)\left(x+3\right)}\)

\(g,PT\left(1\right)=\dfrac{4x^2}{2x\left(x+2\right)^3};PT\left(2\right)=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\\ h,PT\left(1\right)=\dfrac{5}{3x\left(x-2\right)\left(x+2\right)}=\dfrac{10\left(x+3\right)}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\\ PT\left(2\right)=\dfrac{3}{2\left(x+2\right)\left(x+3\right)}=\dfrac{9x\left(x-2\right)}{6x\left(x+2\right)\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{5}{3}-\dfrac{5a-6}{3a}=\dfrac{5a-5a+6}{3a}=\dfrac{6}{3a}=\dfrac{2}{a}\)

\(=\dfrac{2x-3a}{2a}+\dfrac{3}{2}=\dfrac{2x-3a+3a}{2a}=\dfrac{2x}{2a}=\dfrac{x}{a}\)

\(=\dfrac{2-a}{2a}+\dfrac{1}{2x}=\dfrac{4x-2xa+2a}{4xa}=\dfrac{2x-xa+a}{xa}\)

bn đg thi hả

đr bn ạ

a: \(BC=\sqrt{6^2+8^2}=10\left(cm\right)\)

b: Xét ΔABC vuông tại A và ΔHBA vuông tại H có

góc B chung

Do đó:ΔABC\(\sim\)ΔHBA

c: Xét ΔABC vuông tại A có AH là đường cao

nên \(AB^2=BH\cdot BC\)

=>BH=3,6cm

=>CH=6,4cm

a: Xét ΔADB vuông tại D và ΔAEC vuông tại E có

góc DAB chung

Do đó: ΔADB\(\sim\)ΔAEC

b: Xét ΔHEB vuông tại E và ΔHDC vuông tại D có

\(\widehat{EHB}=\widehat{DHC}\)

Do đó: ΔHEB\(\sim\)ΔHDC

Suy ra: HE/HD=HB/HC

hay \(HE\cdot HC=HD\cdot BH\)

b-c-b-a-d

Câu 3:B

Câu 4:C

Câu 5:B

Câu 6:A

Câu 7:D