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Khôi phục các đa thức sau:
1,\(\left(2x-\dfrac{3}{2}y\right)^2=4x^2-6xy+\dfrac{9}{4}y^2\)
2,\(\left(x+2y\right)^3=x^3+6x^2y+12xy^2+8y^3\)
3,\(\left(3x+5y\right)^2=9x^2+30xy+25y^2\)
4,\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=x^3-8y^3\)
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
1,(x+2)(x+5)(x+3)(x+4)-24=(x2+7x+10)(x2+7x+12)-24
Đặt x2+7x+10= t ta có t(t+2)-24=t2+2t-24=(t-4)(t+6)
hay (x2+7x+6)(x2+7x+16)
2,x(x+10)(x+4)(x+6)+128=(x2+10x)(x2+10x+24)+128
Đặt x2+10x=t ta có t(t+24)+128=t2+24t+128=(t+8)(t+16)
hay (x2+10x+8)(x2+10x+16)
3,(x+2)(x-7)(x+3)(x-8)-144=(x2-5x-14)(x2-5x-24)-144
Đặt x2-5x-14=t ta có t(t-10)-144=t2-10t-144=(t-18)(t+8)
Hay (x2-5x-32)(x2-5x-6)=(x2-5x-32)(x+1)(x-6)
\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
a. Ta có:
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c+a-b\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
và \(ab^2-ac^2-b^3+bc^2=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
Vậy, \(A=\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{c-a}{-c-b}=\frac{a-c}{c+b}\)
8x3+36x2+54x+27=(2x+3)3
hok tốt!
_Lan Lan_
8x3+36x2+54x+27=(2x+3)3