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a) A = a3 + b3 = (a + b)(a2 - ab + b2) = (a + b)3 - 3ab(a + b)
= 23 - 3.(-1).2 = 8 + 6 = 14
b) B = a4 + b4 = a4 - 2a2b2 + b4 + 2a2b2 = (a2 - b2)2 + 2a2b2
= (a - b)2(a + b)2 + 2(ab)2 = (a2 - 2ab + b2)(a + b)2 + 2(ab)2
= (a + b)4 + 2(ab)2 - 4ab(a + b)2 = 24 + 2.(-1)2 - 4.(-1).22 = 16 + 2 + 16 = 34
c) Ta có: a2 + b2 = (a2 + 2ab + b2) - 2ab = (a + b)2 - 2ab = 22 - 2.(-1) = 4 + 2 = 6
=> (a2 + b2)(a3 + b3) = 6.14 = 84
=> a5 + a2b3 + a3b2 + b5 = a5 + b5 + a2b2(a + b) = 84
=>C = 84 - (ab)2(a + b) = 84 - (-1)2.2 = 82
d) D = a6 + b6 = a6 + 3a4b2 + 3a2b4 + a6 - 3a2b2(a2 + b2) = (a2 + b2)3 - 3(ab)2(a2 + b2) = 63 - 3(-1)2. 6 = 198
a) Ta có : a + b = 2
=> (a + b)3 = 8
=> a3 + b3 + 3a2b + 3ab2 = 8
=> a3 + b3 + 3ab(a + b) = 8
=> a3 + b3 - 6 = 8
=> a3 + b3 = 14
b) Ta có a + b = 2
=> (a + b)4 = 16
=> a4 + b4 + 4a3b + 4ab3 = 16
=> a4 + b4 + 4ab(a2 + b2) = 16 (1)
Lại có a + b = 2
=> (a + b)2 = 4
=> a2 + b2 + 2ab = 4
=> a2 + b2 = 6
Khi đó (1) <=> a4 + b4 - 24 = 16
=> a4 + b4 = 40
c) a + b = 2
=> (a + b)5 = 32
=> a5 + b5 + 5a4b + 5ab4 = 32
=> a5 + b5 + 5ab(a3 + b3) = 32
Vận dụng kết quả câu b
=> a5 + b5 - 70 = 32
a5 + b5 = 102
d) a + b = 2
=> (a + b)6 = 64
=> a6 + b6 + 6a5b + 6ab5 = 64
=> a6 + b6 + 6ab(a4 + b4) = 64
Vận dụng kết quả câu c
=> a6 + b6 - 240 = 64
=> a6 + b6 = 304
a/ \(\left(-4xy\right)\left(2xy^2-3x^3y\right)=-8x^2y^3+12x^4y^2\)
b/ \(\left(-5x\right)\left(3x^3+7x^2-x\right)=-15x^4-35x^3+5x^2\)
c/ \(\left(\frac{1}{2}a^3b^2-\frac{3}{4}ab^4\right)\left(\frac{4}{3}a^3b\right)=\frac{2}{3}a^6b^3-a^4b^5\)
d/ \(\left(-a^5x^5\right)\left(-a^6x+2a^3x^2-11ax^5\right)=a^{11}x^6-2a^8x^7+11a^6b^{10}\)
làm a) còn b);c) tương tự
A = (a + b)2 - 2ab = 100 - 8 = 92
=a, a(b2+c2)+b(a2+c2)+c(a2+b2)+2abc
= ab2+ac2+ba2+bc2+ca2+cb2+2abc
= c2(a+b)+ab(a+b)+c(a2+b2+2ab)
= c2(a+b)+ab(a+b)+c(a+b)2
= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)
= (a+b)(c2+ab+ca+cb)
= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
=(a+b)(a+c)(b+c)
b, a(b-c)3+b(c-a)3+c(a-b)3
= a(b-c)3-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)2(a-b)-3b(b-c)(a-b)2-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(b-c+a-b)-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(a-c)-b(a-b)3+c(a-b)3
= (b-c)3(a-b)-3b(b-c)(a-b)(a-c)-(a-b)3(b-c)
= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)
=(b-c)(a-b)(b2-2bc+c2-3ab+3bc-a2+2ab-b2)
= (b-c)(a-b)(c2-a2+bc-ab)
= (b-c)(a-b)\(\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
= (b-c)(a-b)(c-a)(c+a+b)
c, a2b2(a-b)+b2c2(b-c)+c2a2(c-a)
= a2b2(a-b)-b2c2\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c2a2(c-a)
= a2b2(a-b)-b2c2(a-b)-b2c2(c-a)+c2a2(c-a)
= b2(a-b)(a2-c2)+c2(c-a)(a2-b2)
= b2(a-b)(a-c)(a+c)-c2(a-c)(a-b)(a+b)
= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)
= (a-c)(a-b)(b2a+b2c-c2a-c2b)
= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)
= (a-c)(a-b)(b-c)(ab+ac+bc)
d, a4(b-c)+b4(c-a)+c4(a-b)
= a4(b-c)-b4[(b-c)+(a-b)]+c4(a-b)
= (b-c)(a4-b4)+(a-b)(c4-b4)
= (b-c)(a2-b2)(a2+b2)+(a-b)(c2-b2)(c2+b2)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b2+c2)
= (b-c)(a-b)(a3+ab2+ba2+b3-bc2-b3-cb2-c3)
= (b-c)(a-b)(a3+ab2+ba2-bc2-c3-cb2)
= (b-c)(a-b)(a3-c3)+b2(a-c)+b(a2-c2)
= (b-c)(a-b)(a-c)(a2+ac+c2)+b2(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a2+ac+c2+b2+ab+ac)
= (a-b)(b-c)(c-a)(a2+b2+c2+ab+bc+ca)
B1:a)(3x-5)2-(3x+1)2=8
[(3x-5)+(3x+1)].[(3x-5)-(3x+1)]=8
(3x-5+3x+1)(3x-5-3x-1)=8
9x2-15x-9x2-3x-15x+25+15x+5+9x2-15x-9x2-3x+3x-5-3x-1=8
-36x+24=8
-36x=8-24=16
x=16:(-36)=\(\dfrac{-4}{9}\)
Bài 5:
a: \(=\left(xy-u^2v^3\right)\left(xy+u^2v^3\right)\)
b: \(=\left(2xy^2-3xy^2+1\right)\left(2xy^2+3xy^2-1\right)\)
\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)
Bài 6:
a: \(\left(a+b+c-d\right)\left(a+b-c+d\right)\)
\(=\left(a+b\right)^2+\left(c-d\right)^2\)
\(=a^2+2ab+b^2+c^2-2cd+d^2\)
b: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=a^2-2ad+d^2-b^2+2bc-c^2\)
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
Ý 3 bạn bỏ dòng áp dụng....ta có nhé
\(a^2+b^2+c^2+d^2\ge a\left(b+c+d\right)\)
\(\Leftrightarrow\left(\frac{a^2}{4}-2.\frac{a}{2}b+b^2\right)+\left(\frac{a^2}{4}-2.\frac{a}{2}c+c^2\right)+\)\(\left(\frac{a^2}{4}-2.\frac{a}{d}d+d^2\right)+\frac{a^2}{4}\ge0\forall a;b;c;d\)
\(\Leftrightarrow\left(\frac{a}{2}-b\right)+\left(\frac{a}{2}-c\right)+\)\(\left(\frac{a}{2}-d\right)^2+\frac{a^2}{4}\ge0\forall a;b;c;d\)( luôn đúng )
Dấu " = " xảy ra <=> a=b=c=d=0
6) Sai đề
Sửa thành:\(x^2-4x+5>0\)
\(\Leftrightarrow\left(x-2\right)^2+1>0\)
7) Áp dụng BĐT AM-GM ta có:
\(a+b\ge2.\sqrt{ab}\)
Dấu " = " xảy ra <=> a=b
\(\Leftrightarrow\frac{ab}{a+b}\le\frac{ab}{2.\sqrt{ab}}=\frac{\sqrt{ab}}{2}\)
Chứng minh tương tự ta có:
\(\frac{cb}{c+b}\le\frac{cb}{2.\sqrt{cb}}=\frac{\sqrt{cb}}{2}\)
\(\frac{ca}{c+a}\le\frac{ca}{2.\sqrt{ca}}=\frac{\sqrt{ca}}{2}\)
Dấu " = " xảy ra <=> a=b=c
Cộng vế với vế của các BĐT trên ta có:
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{2}=\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}\)
Dấu " = " xảy ra <=> a=b=c
1)\(x^3+y^3\ge x^2y+xy^2\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)
\(\Leftrightarrow x^2-xy+y^2\ge xy\) ( vì x;y\(\ge0\))
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng )
\(\Rightarrow x^3+y^3\ge x^2y+xy^2\)
Dấu " = " xảy ra <=> x=y
2) \(x^4+y^4\ge x^3y+xy^3\)
\(\Leftrightarrow x^4-x^3y+y^4-xy^3\ge0\)
\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)( luôn đúng )
Dấu " = " xảy ra <=> x=y
3) Áp dụng BĐT AM-GM ta có:
\(\left(a-1\right)^2\ge0\forall a\Leftrightarrow a^2-2a+1\ge0\)\(\forall a\Leftrightarrow\frac{a^2}{2}+\frac{1}{2}\ge a\forall a\)
\(\left(b-1\right)^2\ge0\forall b\Leftrightarrow b^2-2b+1\ge0\)\(\forall b\Leftrightarrow\frac{b^2}{2}+\frac{1}{2}\ge b\forall b\)
\(\left(a-b\right)^2\ge0\forall a;b\Leftrightarrow a^2-2ab+b^2\ge0\)\(\forall a;b\Leftrightarrow\frac{a^2}{2}+\frac{b^2}{2}\ge ab\forall a;b\)
Cộng vế với vế của các bất đẳng thức trên ta được:
\(a^2+b^2+1\ge ab+a+b\)
Dấu " = " xảy ra <=> a=b=1
4) \(a^2+b^2+c^2+\frac{3}{4}\ge a+b+c\)
\(\Leftrightarrow\left[a^2-2.a.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\)\(+\left[b^2-2.b.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\)\(+\left[c^2-2.c.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\ge0\forall a;b;c\)
\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2\)\(+\left(b-\frac{1}{2}\right)^2\)\(+\left(c-\frac{1}{2}\right)^2\ge0\forall a;b;c\)( luôn đúng)
Dấu " = " xảy ra <=> a=b=c=1/2
a, \(a+b=10\Rightarrow\left(a+b\right)^2=10^2\Rightarrow a^2+2ab+b^2=100\)
\(\Rightarrow a^2+b^2=100-2ab\Rightarrow a^2+b^2=100-2.4\Rightarrow a^2+b^2=100-8\)
\(\Rightarrow a^2+b^2=92\). Vậy \(a^2+b^2=92\)
b, \(a+b=10\Rightarrow\left(a+b\right)^3=10^3\Rightarrow a^3+3a^2b+3ab^2+b^3=1000\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=1000\Rightarrow a^3+b^3+3.4.10=1000\)
\(\Rightarrow a^3+b^3+120=1000\Rightarrow a^3+b^3=880\). Vậy \(a^3+b^3=880\)
c, \(a+b=10\Rightarrow\left(a+b\right)^4=10000\)
\(\Rightarrow a^4+4a^3b+6a^2b^2+4ab^3+b^4=10000\)
\(\Rightarrow a^4+b^4+4ab\left(a^2+b^2\right)+6\left(ab\right)^2=10000\)
\(\Rightarrow a^4+b^4+4.4.92+6.4^2=10000\Rightarrow a^4+b^4+992+96=10000\)
\(\Rightarrow a^4+b^4=8912\). Vậy \(a^4+b^4=8912\)
d, \(a+b=10\Rightarrow\left(a+b\right)^5=100000\)
\(\Rightarrow a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=100000\)
\(\Rightarrow a^5+b^5+5ab\left(a^3+b^3\right)+10a^2b^2\left(a+b\right)=100000\)
\(\Rightarrow a^5+b^5+5.4.880+10.4^2.10=100000\)
\(\Rightarrow a^5+b^5+17600+1600=100000\Rightarrow a^5+b^5=80800\)
Vậy \(a^5+b^5=80800\)
tks bạn ^^