Tính nhẩm: a2 - 86a + 13. Thay a = 87
Tính giá trị bt:
a 10a - a2 - 25
b (x + y)2 - 2(x + y)+1
c -x2 - 2xy - y2
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a) 3x2 – 7x + 2
\(=3x^2-6x-x+2\)
\(=\left(3x^2-6x\right)-\left(x-2\right)\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) a(x2 + 1) – x(a2 + 1)
\(=ax^2+a-\left(a^2x+x\right)\)
\(=a\left(x^2+1\right)-x\left(a^2+1\right)\)
.......?
a) Ta có: \(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)
\(=x^2a+a-a^2x-x\)
\(=\left(x^2a-a^2x\right)+\left(a-x\right)\)
\(=xa\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(xa-1\right)\)
c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)
\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)
\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)
\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)
\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
a: \(=\left(x-y\right)\left(x+y\right)\)
\(=74\cdot100=7400\)
c: \(=\left(x+2\right)^3\)
\(=10^3=1000\)
a) \(=\left(x-y\right)\left(x+y\right)\)
Thay \(x=87;y=13\) ta đc: \(\left(87-13\right)\left(87+13\right)=74\cdot100=7400\)
b)\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
Thay \(x=10;y=-1\) ta đc:
\(10^3-\left(-1\right)^3=1000-1=999\)
c)\(=\left(x+2\right)^3\)
Thay \(x=8\) ta đc: \(\left(8+2\right)^3=10^3=1000\)
d)\(=x^2-8x+16+1=\left(x-4\right)^2+1\)
Thay \(x=104\) ta đc: \(\left(104-4\right)^2+1=100^2+1=10001\)
\(1.a,\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=\left(ac\right)^2+2abcd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(b,\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ad-bc\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow-\left(ad-bc\right)^2\le0\left(luôn-đúng\right)\)
\(dấu"='\) \(xảy\) \(ra\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(c2:x+y=2\Rightarrow\left(x+y\right)^2=4\)
\(\Rightarrow\left(x+y\right)^2+\left(x-y\right)^2\ge4\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy+y^2\ge4\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge4\Leftrightarrow x^2+y^2\ge2\)
\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=y=1\)
Câu 1:
a)Ta có (ac+bd)2+(ad-bc)2=(ac)2+2abcd+(bd)2+(ad)2-2abcd+(bc)2
=(ac)2+(bd)2+(ad)2+(bc)2
=a2(c2+d2)+b2(c2+d2)
=(a2+b2)(c2+d2) (đpcm)
b)Ta có (ac+bd)2 = (ac)2+2abcd+(bd)2
Lại có (a2+b2)(c2+d2) = (ac)2+(bd)2+(ad)2+(bc)2
Ta có (ac+bd)2 ≤ (a2+b2)(c2+d2)
<=>(a2+b2)(c2+d2) - (ac+bd)2 ≥ 0
<=>(ac)2+(bd)2+(ad)2+(bc)2-[(ac)2+2abcd+(bd)2]
<=>(ad)2 - 2abcd +(bc)2 ≥ 0
<=>(ad-bc)2 ≥ 0 (Luôn đúng) => đpcm
Câu 2:
Áp dụng BĐT Bunhiacôpxki, ta có (x+ y)2 ≤ (x2 + y2)(12 + 12) => 4 ≤ 2.S => 2 ≤ S
Dấu ''='' xảy ra <=> x=y=1
Vậy Min S=2 <=> x=y=1
a) Rút gọn M = -6ab(-2b + a). Tính được M = 60.
b) Rút gọn M = 6xy – 7. Tính được N = -10.
\(a,VT=\left(a^2-1\right)^2+4a^2\\ =a^4-2a^2+1+4a^2\\ =a^4+2a^2+1\\ =\left(a^2+1\right)^2 =VP\\ b,VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)\\ =x^2-2xy+y^2+x^2+y^2+2xy+2x^2-2y^2\\ =4x^2=VP\)