(8x2+3)(8x2-3)-(8x2-1)2=22
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1/8*2/6:3/4+5/3
=1/8*1/3:3/4+5/3
=1/24*4/3+5/3
=1/18+30/18
=31/18
\(3xy^3+6x^3y+xy=xy\left(3y^2+6x^2+1\right)\)
\(4x^3+8x^2+4x=4x\left(x^2+2x+1\right)=4x\left(x+1\right)^2\)
\(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5
B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5
B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5
B = 7-5=2
`B = x^15 - 7x^14 - x^14 + 7x^13 + x^13 - .... +7x + x - 7 + 2`
`<=> x^14(x-7) - x^13(x-7) + ... + x - 7 + 2`
`<=> (x^14-x^13 + ... + 1)(x-7) + 2`
Thay `x = 7 <=> (x^14 - x^13 + ... + 1) xx 0 + 2 = 2`.
ta có: 8=7+1=x+1
\(B=x^{15}-8x^{14}+8x^{13}-...-8x^2+8x-5\)
\(\Rightarrow B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(\Rightarrow B=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)
\(\Rightarrow B=x-5\)
\(\Rightarrow B=7-5\)
\(\Rightarrow B=2\)
ta có :
\(\left(8x^2+3\right)\left(8x^2-3\right)-\left(8x^2-1\right)^2=22\)
\(\Leftrightarrow64x^4-9-\left(64x^4-16x^2+1\right)=22\Leftrightarrow16x^2=32\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
\(\Leftrightarrow64x^4-9-64x^4+16x^2-1=22\Leftrightarrow16x^2=32\Rightarrow x^2=2\Rightarrow\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)