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Bài 3:
a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)
\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)
\(\Leftrightarrow x^3-27-x^3+16x=21\)
=>16x=48
hay x=3
b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\)
=>-2x=4-8=-4
hay x=2
\(CMR:\forall m,n\in Z\)
\(a)n^2\times(n^2-1)⋮12\)
\(b)n^2\times(n^4-1)⋮60\)
\(c)mn\times(m^4-n^4)⋮30\)
a)\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
b) k hiểu đề
a) \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1\right)^2+2\left(9x-1\right)\left(1-5x\right)+\left(1-5x\right)^2\)
\(=\left(9x-1+1-5x\right)^2=\left(4x\right)^2\)
b) \(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^2\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^4-16x^2-x^4+1=-16x^2+1\)
a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)
b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)
dễ thì giải cho người ta đi,bạn thông minh hơn thì thay vì ns người khác thì giúp người khác sẽ tốt hơn đó
a/ \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^2}=\dfrac{2\left(x^2-10x+25\right)\cdot\left(x^2-1\right)}{3\left(x+1\right)\cdot4\left(x-5\right)^2}=\dfrac{2\left(x-5\right)^2\left(x-1\right)\left(x+1\right)}{12\left(x+1\right)\left(x-5\right)^2}=\dfrac{x+1}{6}\)
b/ \(\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^2}=-\dfrac{3\left(1-2x\right)\cdot\left(5x+1\right)^2}{x\left(5x+1\right)\left(1-2x\right)\left(1+2x+4x^2\right)}=\dfrac{3\left(5x+1\right)}{x\left(4x^2+2x+1\right)}\)
c/ \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}=\dfrac{x-3x^2}{1-x^2}\cdot\dfrac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-3x\right)^3}=\dfrac{x\left(1-3x\right)\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(1-3x\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(1-3x\right)^3}\)
\(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\)
\(\ge\sqrt{a^2\cdot4}\cdot\sqrt{b^2\cdot4}\cdot\sqrt{c^2\cdot4}\cdot\sqrt{d^2\cdot4}\)
\(=4\left|a\right|\cdot4\left|b\right|\cdot4\left|c\right|\cdot4\left|d\right|\)
\(\ge256abcd\)
Dấu "=" xảy ra tại \(a=b=c=2\)
Khoan đã,fix lại cả bài nốt:v
\(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\)
\(\ge2\sqrt{a^2\cdot4}\cdot2\sqrt{b^2\cdot4}\cdot2\sqrt{c^2\cdot4}\cdot2\sqrt{d^2\cdot4}\)
\(=256\left|abcd\right|\ge256abcd\)
Dấu "=" xảy ra \(a=b=c=d=2\)