Chứng minh rằng:
a) \(\left(a+b\right)^2=a^2+2ab+b^2\)
b) \(\left(a-b\right)^2=a^2-2ab+b^2\)
c) \(\left(a-b\right)\left(a+b\right)=a^2-b^2\)
Khẳng định nào sau đây là đúng?
A. \(\left( {A - B} \right)\left( {A + B} \right) = {A^2} + 2AB + {B^2}\)
B. \(\left( {A - B} \right)\left( {A + B} \right) = {A^2} - 2AB + {B^2}\)
C. \(\left( {A - B} \right)\left( {A + B} \right) = {A^2} + {B^2}\)
D. \(\left( {A - B} \right)\left( {A + B} \right) = {A^2} - {B^2}\)
\({A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)\)
Chọn D.
PTĐTTNT:\(3abc+a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-b-a\right)-c\left(b-c\right)\left(a-c\right)\)
\(=3abc+a^3-a^2b-a^2c+b^3-b^2a-b^2c+c^3-c^2b-c^2a-\left(abc-bc^2-c^2a+c^3\right)\)
\(=2abc+a^3-a^2b-a^2c+b^3-b^2c-b^2a\)
\(=\left(a^3+a^2b-a^2c\right)-\left(2a^2b+2ab^2-2abc\right)+\left(ab^2+b^3-b^2c\right)\)
\(=a^2\left(a+b-c\right)-2ab\left(a+b-c\right)+b^2\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b-c\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b-c\right)\left(a-b\right)^2\) nha !
P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.
Trả lời
Ở phần kết quả bạn vẫn chưa thu gọn hết đâu nha
\(=\left(a+b+c\right).\left(a-b\right)^2\)
Mk góp ý thôi mong mọi người đừng có đáp gạch đáp đá nha
Study well
CHỨNG MINH RẰNG PHƯƠNG TRÌNH BẠC 2: \(\left(a+b\right)^2.x^2-\left(a-b\right).\left(a^2-b^2\right).x-2ab.\left(a^2+b^2\right)=0.\)LUÔN CÓ 2 NGHIỆM PHÂN BIỆT
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-2^2\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a^2-2ab+b^2\right)-9\right]\left[\left(a^2+2ab+b^2\right)-1\right]\)
\(=\left[\left(a-b\right)^2-3^2\right]\left[\left(a+b\right)^2-1^2\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a-b-1\right)\left(a-b+1\right)\)
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(5-x-4y\right)\left(3+3x+2y\right)\)
7 Chứng minh các đẳng thức sau
a) \(a^2+b^2=\left(a+b\right)^2-2ab\) ; b) \(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\)
c) \(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)
d) \(a^6-b^6=\left(a^2-b^2\right)\left[\left(a^2+b^2\right)^2-a^2b^2\right]\)
a) \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(VP=\left(a+b\right)^2-2ab=a^2+2ab+b^2-2ab\)\(=a^2+b^2=VT\)
\(\Rightarrowđpcm\)
b)\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\)
\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)
c) \(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)
\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)
\(VP=VT\Rightarrowđpcm\)
d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)
\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)
\(VP=VT\Rightarrowđpcm\)
1/Cho các số thực dương chứng minh:\(\frac{3\left(a^4+b^4+c^4\right)}{\left(a^2+b^2+c^2\right)^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge2\)
2/Cho a,b dương.Chứng minh:\(\left(\frac{a}{b}+\frac{b}{a}\right)+4\sqrt{2}\frac{a+b}{\sqrt{a^2+b^2}}\ge10\)
3/ Cho các số thực dương. Chứng minh: \(\left(a^2+2bc\right)\left(b^2+2ca\right)\left(c^2+2ab\right)\ge abc\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\)
Cho a, b>0. Chứng minh rằng:
a) \(\dfrac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
b) \(\dfrac{2ab}{a+b}+\sqrt{\dfrac{a^2+b^2}{2}}\ge\sqrt{ab}+\dfrac{a+b}{2}\)
c) \(\dfrac{1}{\left(1+a\right)^2}+\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{1}{1+ab}\)
chứng minh\(\frac{a\cdot\left(b+c\right)}{a^2+2bc}+\frac{b\cdot\left(a+c\right)}{b^2+2ac}+\frac{c\cdot\left(a+b\right)}{c^2+2ab}< =2\)2 với a,b,c là độ dài 3 cạnh tam giác
BĐT cần CM tương đương:
\(3-VT\ge1\)
\(\Leftrightarrow\frac{a^2+2bc-a\left(b+c\right)}{a^2+2bc}+...\ge1\) (1)
\(VT\left(1\right)=\frac{\left[a^2+2bc-a\left(b+c\right)\right]^2}{\left(a^2+2bc\right)\left[a^2+2bc-a\left(b+c\right)\right]}+...\)
\(\ge\frac{\left[a^2+2bc-a\left(b+c\right)+b^2+2ca-b\left(c+a\right)+c^2+2ab-c\left(a+b\right)\right]^2}{\left(a^2+2bc\right)\left[a^2+2bc-a\left(b+c\right)\right]+...}\)
\(=\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+2bc\right)\left[a^2+2bc-a\left(b+c\right)\right]+...}\) (2)
Ta cần chứng minh mẫu của (2) \(\le\left(a^2+b^2+c^2\right)^2\)
... Tự biến đổi ra thôi thi ta được 1 biểu thức không âm luôn đúng
=> BĐT trên đúng
=> đpcm
Dấu "=" xảy ra khi: a = b = c
Chứng minh các hằng đẳng thức sau :
a, \(\left(a^2-b^2\right)+\left(2ab\right)^2=\left(a^2+b^2\right)^2\)
b, \(\left(a^2+b^2\right).\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
c, \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2=\left(a^2+b^2+c^2\right).\left(x^2+1\right)\)
d, \(\dfrac{1}{2}.\left(a+b+c\right).\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=a^3+b^3+c^3-3abc\)
e, \(1000^2+1003^2+1005^2+1006^2=1001^2+1002^2+1004^2+1007^2\)
a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)