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8 tháng 9 2018

a ) \(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)

\(\Leftrightarrow a^2+b^2+c^2=0\)

Do \(a^2\ge0;b^2\ge0;c^2\ge0\)

\(\Rightarrow a^2+b^2+c^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )

Thay * vào biểu thức M , ta được :

\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)

\(=-1^{1999}+0+1^{2001}\)

\(=-1+0+1\)

\(=0\)

Vậy \(M=0\)

8 tháng 9 2018

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)

\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)

\(\Leftrightarrow bc+ac+ab-1=0\)

\(\Leftrightarrow bc+ac+ab=1\)

\(a^2+b^2+c^2=1\)

\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)

\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)

\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

\(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)

\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)

\(\Rightarrow P=1+1+1=3\)

Vậy \(P=3\)

21 tháng 7 2018

a) \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{64}-1\right)\)

\(=\dfrac{3^{64}-1}{2}\)

b) \(\left(a+b+c\right)2+\left(a-b-c\right)2+\left(b-c-a\right)2+\left(c-a-b\right)2\)

\(=2\left[\left(a+b+c\right)+\left(a-b-c\right)+\left(b-c-a\right)+\left(c-a-b\right)\right]\)

\(=2\left(a+b+c+a-b-c+b-c-a+c-a-b\right)\)

\(=2.0\)

\(=0\)

c)\(\left(a+b+c+d\right)2+\left(a+b-c-d\right)2+\left(a+c-b-d\right)2+\left(a+d-b-c\right)2\)

\(=2\left(a+b+c+d+a+b-c-d+a+c-b-d+a+d-b-c\right)\)

\(=2.4a\)

\(=8a\)

5 tháng 7 2018


Thân heo vừa béo lại vừa ù
Bảy nổi ba chìm với nước lu
Chết đuối quẫy chân không ai cứu
Đứa nào mà cứu, đứa ấy ngu


 

5 tháng 7 2018

a, a2+b2+c2 >= ab+bc+ca

<=>a2+b2+c2-ab-bc-ca >= 0

<=>2(a2+b2+c2-ab-bc-ca) >= 0

<=>2a2+2b2+2c2-2ab-2bc-2ca >= 0

<=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ca+a2) >= 0

<=>(a-b)2+(b-c)2+(c-a)2 >= 0 (luôn đúng)

Dấu "=" xảy ra chỉ khi và khi \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow a=b=c}\)

Vậy...

b, a2+b2+1 >= ab+a+b

<=>a2+b2+1-ab-a-b >= 0

<=>2(a2+b2+1-ab-a-b) >= 0

<=>2a2+2b2+2-2ab-2a-2b >= 0

<=>(a2-2ab+b2)+(a2-2a+1)+(b2-2b+1) >= 0

<=>(a-b)2+(a-1)2+(b-1)2 >= 0 (luôn đúng)

Dấu "=" xảy ra chỉ khi và khi \(\hept{\begin{cases}a-b=0\\a-1=0\\b-1=0\end{cases}\Leftrightarrow a=b=1}\)

Vậy...

c, a2+b2+c2+3 >= 2(a+b+c)

<=>a2+b2+c2+3-2a-2b-2c >= 0

<=>(a2-2a+1)+(b2-2b+1)+(c2-2c+1) >= 0

<=>(a-1)2+(b-1)2+(c-1)2 >= 0 (luôn đúng)

Dấu "=" xảy ra chỉ khi và khi \(\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\Leftrightarrow a=b=c=1}\)

Vậy...

d, a2+b2+c2 >= 2(ab+bc-ca)

<=>a2+b2+c2-2ab-2bc+2ca >= 0

<=>(a-b-c)2 >= 0 (luôn đúng)

Dấu "=" xảy ra khi a=b=c

Vậy...

e,ta có:  \(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\Leftrightarrow\frac{a^2+b^2}{2}-\left(\frac{a+b}{2}\right)^2\ge0\)

\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{4}-\frac{a^2+2ab+b^2}{4}\ge0\)

\(\Leftrightarrow\frac{2a^2+2b^2-a^2-2ab-b^2}{4}\ge0\)

\(\Leftrightarrow\frac{a^2-2ab+b^2}{4}\ge0\Leftrightarrow\left(\frac{a-b}{2}\right)^2\ge0\) (luôn đúng) (1)

Lại có: \(\left(\frac{a+b}{2}\right)^2\ge ab\Leftrightarrow\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}\ge0\)

\(\Leftrightarrow\frac{a^2+2ab+b^2-4ab}{4}\ge0\Leftrightarrow\left(\frac{a-b}{2}\right)^2\ge0\) (luôn đúng) (2)

Từ (1) và (2) => \(ab\le\left(\frac{a+b}{2}\right)^2\le\frac{a^2+b^2}{2}\)

Dấu "=" xảy ra khi a = b

21 tháng 3 2019

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

21 tháng 3 2019

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

* Phân tích đa thức thành nhân tử: 1/ 25x2 - 10xy + y2 2/ 8x3 + 36x2y + 54xy2 + 27y3 3/ (a2 + b2 - 5)2 - 4 (ab + 2)2 4/ (a + b + c)3 - a3 - b3 - c3 5/ 2x3 + 3x2 + 2x + 3 6/ x3z + x2yz - x2z2 - xyz2 7/ x3 + y (1 - 3x2) + x (3y2 - 1) - y3 8/ x3 + 3x2y + 3xy2 + y + y3 9/ x2 - 6x + 8 10/ x2 - 8x + 12 11/ a2 (b - c) + b2 (c - a) + c2 (a - b) 12/ x3 - 7x - 6 13/ x4 + 4 14/ a4 + 64 15/ x5 + x + 1 16/ x5 + x - 1 17/ (x2 + x)2 - 2 (x2 + x) - 15 18/ (x + 2) (x + 3) (x + 5) -...
Đọc tiếp

* Phân tích đa thức thành nhân tử:

1/ 25x2 - 10xy + y2

2/ 8x3 + 36x2y + 54xy2 + 27y3

3/ (a2 + b2 - 5)2 - 4 (ab + 2)2

4/ (a + b + c)3 - a3 - b3 - c3

5/ 2x3 + 3x2 + 2x + 3

6/ x3z + x2yz - x2z2 - xyz2

7/ x3 + y (1 - 3x2) + x (3y2 - 1) - y3

8/ x3 + 3x2y + 3xy2 + y + y3

9/ x2 - 6x + 8

10/ x2 - 8x + 12

11/ a2 (b - c) + b2 (c - a) + c2 (a - b)

12/ x3 - 7x - 6

13/ x4 + 4

14/ a4 + 64

15/ x5 + x + 1

16/ x5 + x - 1

17/ (x2 + x)2 - 2 (x2 + x) - 15

18/ (x + 2) (x + 3) (x + 5) - 24

19/ (x2 + 8x + 7) (x2 + 8x + 15) + 15

20/ (x2 + 3x + 1) (x2 + 3x + 2) - 6

21/ x2 + 4xy + 3y2

22/ 2x2 - 5xy + 2y2

23/ x2 (y - z) + y2 (z - x) + z2 (x - y)

24/ 2x2 - 7xy + 3y2 + 5xz - 5yz + 2z2

25/ x2 - 7x + 10

26/ 4x2 - 3x - 1

27/ x2 - x - 12

28/ bc (b + c) + ac (c - a) - ab (a + b)

29/ x2y + xy2 + x2z + xz2 + y2z + yz2 + 2xyz

30/ (a - b)3 + (b - c)3 + (c - a)3

31/ ab (a - b) + bc (b - c) + ca (c - a)

32/ bc (b + c) + ca (c + a) + ba (a + b) + 2abc

Giúp mình với, giải chi tiết nha, nhiều bài mà mình đang cần gấp lắm!

3
18 tháng 9 2018

1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)

2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)

4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

5, \(2x^3+3x^2+2x+3\)

\(=x^2\left(2x+3\right)+2x+3\)

\(=\left(x^2+1\right)\left(2x+3\right)\)

6, \(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^3z-x^2z^2+x^2yz-xy^2\)

\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)

\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)

\(=xz\left(x+y\right)\left(x-z\right)\)

8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)

9, \(x^2-6x+8\)

\(=x^2-4x-2x+8\)

\(=x\left(x-4\right)-2\left(x-4\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

10, \(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-2\right)\left(x-6\right)\)

Chỗ còn lại mai làm nốt nha.

19 tháng 9 2018

Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha

11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)

\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)

\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)

\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)

\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

12, \(x^3-7x-6\)

\(=x^3-3x^2+3x^2-9x+2x-6\)

\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+3x+2\right)\)

\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)

\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)

13, \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

14, \(a^4+64\)

\(=a^4+16a^2+64-16a^2\)

\(=\left(a^2+8\right)^2-16a^2\)

\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)

15, \(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

\(=x^2\left(x^3-1\right)+x^2+x+1\)

\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

16, \(x^5+x-1\)

\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)

\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)

17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)

19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)

Đặt \(x^2+8x+7=a\) ta có:

(*) \(\Leftrightarrow a\left(a+8\right)+15\)

\(\Leftrightarrow a^2+8a+15\)

\(\Leftrightarrow a^2+3a+5a+15\)

\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)

\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)

Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)

Đặt \(x^2+3x+1=a\) ta có:

(*) \(\Leftrightarrow a\left(a+1\right)-6\)

\(\Leftrightarrow a^2+a-6\)

\(\Leftrightarrow a^2+3a-2a-6\)

\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)

\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)

Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)