\(A=\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72=\left(x^2-4\right)\left(x^2-10\right)-72=x^4-14x^2+40-72\)

\(A=x^4-14x^2-32=x^4+2x^2-16x^2-32=x^2\left(x^2+2\right)-16\left(x^2+2\right)\)

\(A=\left(x^2+2\right)\left(x^2-16\right)=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

Nguyễn TrươngTruong Viet TruongAkai Harumasoyeon_Tiểubàng giảiMysterious PersonMashiro Shiina

(x-2)(x+2)(x²-10)=72

<=>x⁴-14x²+40=72

<=>x⁴-14x²-32=0

<=>\(\left\{{}\begin{matrix}x^2=-2\left(VN\right)\\x^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

b​; x(x-1)(x+1)(x+2)-24

=(x²+x)(x²+x-2)-24

​Đ​ặt x²​+x=k khi đ​ó​ k(k-2)-24=k²-2k-24

​=(k²-2k+1)-25=(k-1)²-5²

​ =(k-1-5)(k-1+5)=(k-6)(k+4)

c; (x+2)(x-2)(x²-10)-72

​ =(x²-4)(x²-10)-72

​Đ​ặt x²​-7=k khi đ​ó​ (k-3)(k+3)-72=k²-9-72

=k²-81=(k-9)(k+9)=(x²-7-9)(x²-7+9)

=(x²-16)(x^​2+2)=(x-4)(x+4)(x²+2)

d; (x-7)(x-5)(x-4)(x-2)-72

=(x²-9x+14)(x²-9x+8)-72

Đ​ặt x²-9x​+11=k khi đ​ó​ (k+3)(k-3)-72=k²-9-72

=k²-81=(k-9)(k+9)=(x²-9x+11-9)(x²-9x+11+9)

=(x²-9x+2)(x²-9x+20)

=(x²-9x+2)(x²-4x-5x​+20)

=(x²-9x+2)(x-4)(x-5)

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72\)

\(=\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-10x^2-4x^2+40-72\)

\(=x^4-14x^2-32\)

\(=x^4-16x^2+2x^2-32\)

\(=\left(x^4-16x^2\right)+\left(2x^2-32\right)\)

\(=x^2\left(x^2-16\right)+2\left(x^2-16\right)\)

\(=\left(x^2+2\right)\left(x^2-16\right)\)

\(=\left(x^2+2\right)\left(x-4\right)\left(x+4\right)\)

Câu 1:

\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)-112\)

\(=\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+5\right)-112\)

\(=\left(x^2+3x-4\right)\left(x^2+3x-10\right)-112\)

\(=\left(x^2+3x-7\right)^2-3^2-112\)

\(=\left(x^2+3x-7\right)^2-11^2\)

\(=\left(x^2+3x+4\right)\left(x^2+3x-18\right)\)

\(=\left(x^2+3x+4\right)\left(x+6\right)\left(x-3\right)\)

Câu 2:

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72\)

\(=\left(x^2-4\right)\left(x^2-10\right)-2\)

\(=\left(x^2-7\right)^2-3^2-72\)

\(=\left(x^2-7\right)^2-81\)

\(=\left(x^2-16\right)\left(x^2+2\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

(x−1)(x−2)(x+4)(x+5)−112

=(x−1)(x+4)(x−2)(x+5)−112

=(x^2+3x−4)(x^2+3x−10)−112

=(x^2+3x−7)^2−32−112

=(x^2+3x−7)^2−112

=(x^2+3x+4)(x^2+3x−18)

=(x^2+3x+4)(x+6)(x−3)

Câu 2:

(x−2)(x+2)(x^2−10)−72

=(x2−4)(x^2−10)−2

=(x^2−7)^2−32−72

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left[\left(x^2-7\right)+3\right]\left[\left(x^2-7\right)-3\right]=72\)

\(\Leftrightarrow\left(x^2-7\right)^2-9=72\)

\(\Leftrightarrow\left(x^2-7\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-7=9\\x^2-7=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=16\\x^2=-2\left(loai\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

Đặt \(y=x^2-7\) Pt trở thành: \(\left(y+3\right)\left(y-3\right)=73\)

\(\Leftrightarrow y^2-9=72\)

\(\Leftrightarrow y^2=81\)

\(\Leftrightarrow y=\pm9\)

Từ đó tìm được: \(x_1=-4;x_2=4\)

\(\left(x^2-4\right)\left(x^2-10\right)=72\)

<=> \(x^4-14x^2+40-72=0\)

<=> \(x^4-14x^2-32=0\)

<=> \(\left(x^2-16\right)\left(x^2+2\right)=0\)

<=> \(\left[\begin{array}{nghiempt}x^2-16=0\\x^2+2=0\end{array}\right.\)=> x=\(\pm\)4

vậy tập nghiệm S={4;-4}

\(\left(x^2-4\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow x^4-10x^2-4x^2+40=72\)
\(\Leftrightarrow x^4-14x^2+40-72=0\)
\(\Leftrightarrow x^4-14x^2-32=0\)
\(\Leftrightarrow x^4+2x^2-16x^2-32=0\)
\(\Leftrightarrow x^2\left(x^2+2\right)-16\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-4^2\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x-4\right)\left(x+4\right)=0\left(1\right)\)
\(Có:x^2\ge0\)\(\text{ với mọi x}\)
\(\Rightarrow x^2+2\ge0+2=2\ne0\text{ với mọi x}\)
\(\left(1\right)\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x+4=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)
\(\text{Vậy }x=\pm4\)
 

(x²-4)(x²-10)=72

=>x⁴-14x²+40=72

=>x⁴-14x²-32=0

=>(x-4)(x³+4x²+2x+8)=0

=>(x-4)(x+4)(x²+2)=0

=> (x-4) = 0 hoặc (x+4)=0 hoặc (x²+2)=0

=> x = 4 hoặc x=-4

x=4 hoặc x=-4