a,Xét \(\Delta\)AHB và \(\Delta\)CAB, có:

^AHB=^CAB(=90⁰)

^ABC: chung

=> \(\Delta\)AHB \(_{\infty}\)\(\Delta\)CAB(g.g)

Vậy ......

đa tạ đa tạ bn. cx thế nhé Chắc bn là 1 ng iu hoa, 1 ng iu thiên nhiên và có 1 trí tưởng tượng phong phú 

a, 5(2-3n)+42+3n\(\ge\)0

<=> 10-15n+42+3n\(\ge\)0

<=> 52-12n\(\ge\)0

<=> -12n\(\ge\)-52

<=>n\(\le\)\(\dfrac{13}{3}\)

Vậy bft có tập nghiệm là S={n/ n\(\le\)\(\dfrac{13}{3}\)}

a) -10 ; -15; -20 

b) 1;-1;2;-2;5;-5;10;-10

phần b là tam giác EBD hay EBC vậy bạn?

a, (3x-1)(x²+2)=(3x-1)(7x-10)

<=>(3x-1)(x²+2)-(3x-1)(7x-10)=0

<=>(3x-1)(x²+2-7x+10)=0

<=>(3x-1)(x²-7x+12)=0

<=>(3x-1)(x²-3x-4x+12)=0

<=>(3x-1)(x-3)(x-4)=0

<=>\(\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy ft có tập nghiệm S=\(\left\{\dfrac{1}{3},3,4\right\}\)

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (ĐKXĐ:t\(\ne2;t\ne-3\))

<=>\(\dfrac{\left(t+3\right)^2+\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{t^2-2t+3t-6}\)

<=>\(\dfrac{t^2+6t+9+t^2-4t+4}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

=>2t²+2t+13=5t+15

<=>2t²+2t-5t+13-15=0

<=>2t²-3t-2=0

<=>2t²-4t+t-2=0

<=>(t-2)(2t+1)=0

<=>\(\left[{}\begin{matrix}t-2=0\\2t+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-1}{2}\left(tmđkxđ\right)\end{matrix}\right.\)

Vậy ft có nghiệm duy nhất x=\(\dfrac{-1}{2}\)

\(\left(x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\right)^2\le\left(x+y+z\right)\left(xy+yz+zx\right)\le\frac{\left(x+y+z\right)^3}{3}\ge\frac{12^3}{3}\)

a) Ta có:

A=(\(\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\)) . \(\dfrac{x^2-1}{x^2+2}\)

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

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

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

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

=\(\dfrac{2x^2+4}{x\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2}\)

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

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

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

Vậy để giá trị của biểu thức A xác định thì x\(\ne\)0

23+36=59

=> 59<a+37<61

=> a+37=60 

a=60-37

a=23