(x-2)(x+2)(\(x^2\)-10)=72
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\(a,\left(2\times x-40\right):10=-55\)
\(2\times x-40=-55\times10\)
\(2\times x-40=-550\)
\(2\times x=-550+40\)
\(2\times x=-510\)
\(x=-510:2\)
\(\Rightarrow x=-255\)
\(b,\left(2\times x+15\right):5=-40\)
\(2\times x+15=-40\times5\)
\(2\times x+15=-200\)
\(2\times x=-200-15\)
\(2\times x=-215\)
\(x=-215:2\)
\(\Rightarrow x=-\dfrac{215}{2}\)
\(c,\left(2\times x-72\right):8=25\)
\(2\times x-72=25\times8\)
\(2\times x-72=200\)
\(2\times x=200+72\)
\(2\times x=272\)
\(x=272:2\)
\(\Rightarrow x=136\)
\(A=\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72=\left(x^2-4\right)\left(x^2-10\right)-72=x^4-14x^2+40-72\)
\(A=x^4-14x^2-32=x^4+2x^2-16x^2-32=x^2\left(x^2+2\right)-16\left(x^2+2\right)\)
\(A=\left(x^2+2\right)\left(x^2-16\right)=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
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
Đặt x2+x=k khi đó 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
Đặt x2-7=k khi đó (k-3)(k+3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-7-9)(x2-7+9)
=(x2-16)(x2+2)=(x-4)(x+4)(x2+2)
d; (x-7)(x-5)(x-4)(x-2)-72
=(x2-9x+14)(x2-9x+8)-72
Đặt x2-9x+11=k khi đó (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: \(x_1=-4;x_2=4\)
a) 172 + 10 x 2 = 170 +20 = 192
b) 10 x 2 + 300 = 20 + 300 = 320
c) 69 – 54 : 6 = 69 – 9 = 60
d) 900 + 9 x 10 = 900 +90 = 990
e) 20 x 6 + 70 = 120 + 70 = 190
f) 72 + 300 x 3 = 72 + 900 = 972