Question

Giải phương trình:

a. (x - 3)(x - 5)(x - 6)(x - 10) = \(^{24x^2}\)

b. (x + 1)(x + 2)(x + 4)(x + 5) = 40

Answer 1

a) ta có:

(x-3)(x-5)(x-6)(x-10)=24x²

^{<=> \(\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)-\left(x-6\right)\right]=24x^2\)}

<=> \(\left(x^2-13x+30\right)\left(x^2-11x+30\right)=24x^2\)

<=> \(\left(x^2-12x+20-x\right)\left(x^2-12x+30+x\right)=24x^2\)

<=> \(\left(x^2-12x+30\right)^2-x^2=24x^2\)

<=> \(\left(x^2-12x+30\right)^2-x^2-24x^2=0\)

<=> \(\left(x^2-12x+30\right)^2-25x^2=0\)

<=> \(\left(x^2-17x+30\right)\left(x^2-7x+30\right)=0\)

mà x²-7x+30=(x-\(\dfrac{7}{2}\))²+\(\dfrac{71}{4}\)> 0

=> x²-17x+30=0

<=> (x-15)(x-2)=0

=>\(\left[{}\begin{matrix}x-15=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)

Vậy S=\(\left\{15;2\right\}\)

b) ta có:

(x+1)(x+2)(x+4)(x+5)=40

<=> \(\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)

<=> (x²+6x+5)(x²+6x+8)=40

<=> (x²+6x+6,5-1,5)(x²+6x+6,5+1,5)=40

<=> (x²+6x+6,5)^{2 _} 2,25=40

<=> (x²+6x+6,5)^{2 _} 42,25=0

<=> (x²+6x+6,5-6,5)(x²+6x+6,5+6,5)=0

<=> (x²+6x)(x²+6x+13)=0

mà x²+6x+13=(x+3)²+4>0

=> x²+6x=0

<=> x(x+6)=0

=>\(\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy S=\(\left\{0;-6\right\}\)

Answer 2

b)

\(\Rightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)=40\)

\(\Rightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)

\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt: \(a=x^2+6x+5\)

\(\Rightarrow a.\left(a+3\right)=40\)

Mà:\(40=5.8\)

\(\Rightarrow a=5\)

Học tốt !!! :)

Answer 3

cau a bang 5 nha bn

chuc bn hoc tot

happy new year