Chứng minh các đẳng thức sau
a) \(\left(2x+3\right)\left(4x^2+9\right)\left(2x-3\right)=16x^4-81\)
b) \(\left(a+b\right)^2+2\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2=4a^2\)
Phân tích các đa thức sau thành nhân tử:
\(A=4x^2+6x\). \(B=\left(2x+3\right)^2-x\left(2x+3\right)\). \(C=\left(9x^2-1\right)-\left(3x-1\right)^2\).
\(D=x^3-16x\). \(E=4x^2-25y^2\). \(G=\left(2x+3\right)^2-\left(2x-3\right)^2\).
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
1.Chứng tỏ các đa thức sau không phụ thuộc vào biến x
a)\(x\cdot\left(2x+1\right)-x^2\left(x\cdot2\right)+\left(x^3-x+3\right)\)
b)\(4\cdot\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
2.Chứng minh đẳng thức sau :
a)\(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)=-2bc\)
b)\(a\left(1-b\right)+a\left(a^2-1\right)=a\left(a^2-b\right)\)
câu 2:
a(b-c)-b(a+c)+c(a-b)=-2bc
ta có:
a( b-c ) - b ( a +c )+ c(a-b)
=ab-ac-(ba+bc)+(ca-cb)
=ab-ac-ba-bc+ca-cb
=ab-ba-ac+ca-bc-cb
=0-0-bc-cb
=bc+(-cb)
=-2cb hay -2bc
b)a(1-b)+a(a^2-1)=a(a^2-b)
Ta có:
a(1-b) + a(a^2-1)
=a-ab+(a^3-a)
=a-ab+a^3-a
=a-a-ab+a^3
=0-ab+a^3
=-ab+a^3
=a(-b +a^2) hay a(a^2-b)
Chứng minh các đẳng thức sau:
a) \(\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2=4\left(a^2+b^2+c^2\right)\)
b) \(\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3-\left(a+b-c\right)^3=24abc\)
a) Ta có: \(\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2bc-2ab-2ac+a^2+b^2+c^2-2ab-2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ca\)
\(=a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2\)
\(=4a^2+4b^2+4c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
b) Đặt x = b + c - a
y = c + a - b
z = a + b - c
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Rightarrow a+b+c=x+y+z\)
Ta có: \(\left(a+b+c\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left[xy+\left(x+y\right)z+z^2\right]\)
\(=3\left(x+y\right)\left[z^2+xy+xz+yz\right]\)
\(=3\left(x+y\right)\left[z\left(x+y\right)+y\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=3.2a.2b.2c\)
\(=24abc\) (đpcm)
a, \(VP=\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2\)
\(=\left(a^2+b^2+c^2+ab+bc+ac\right)+\left(a^2+b^2+c^2+bc-ab-ac\right)+\left(a^2+b^2+c^2+ac-ab-bc\right)+\left(a^2+b^2+c^2+ab-ac-bc\right)\)\(=4a^2+4b^2+4c^2+\left(ab-ab-ab+ab\right)+\left(bc+bc-bc-bc\right)+\left(ac-ac+ac-ac\right)\)
\(VP=4\left(a^2+b^2+c^2\right)\)
So VP với VT ta thấy: \(VP=VT=4\left(a^2+b^2+c^2\right)\)
=> đpcm.
Bài đó cm tương tự h buồn ngủ quá
Chứng minh rằng biểu thức sau không phụ thuộc vào biến x:
a/A= \(\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
b/B=\(\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)
a) \(A=\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b) \(B=\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
Giải các phương trình sau:
\(a.\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3=0\)
\(b.\left(x^2+4x+2\right)^2+4x^2+16x+11=0\)
a) Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2+\left(x^2-2x\right)-3\left(x^2-2x\right)-3=0\)
\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-2x+1\right)-3\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x^2-2x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Vậy: S={1;-1;3}
Bài 2: Chứng minh bất đẳng thức:
a) \(\left(a+b+c+d\right)-\left(a-b-c+d\right)+1=a-\left(a-2b-2c-d\right)+\left(d+1\right)\)
b)\(\left(4x-3y+2\right)-\left(3x-4y+2\right)=\left(2x+2y\right)-\left(x+y\right)\)
b) Ta có :
\(VT=\left(4x-3y+2\right)-\left(3x-4y+2\right)\)
\(=4x-3y+2-3x+4y-2\)
\(=\left(4x-3x\right)-\left(3y-4y\right)+\left(2-2\right)\)
\(=x+y\)
\(VP=\left(2x+2y\right)-\left(x+y\right)=2x+2y-x-y\)
\(=\left(2x-x\right)+\left(2y-y\right)\)
\(=x+y\)
\(\Rightarrow VT=VP\)
\(\Rightarrow\)đpcm
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
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: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
Chứng minh rằng biểu thức sau không phụ thuộc vào biến x:
a/A=\(\left(3x-2\right)\left(9x^2+6x+4\right)-3\left(9x^2-2\right)\)
b/B=\(\left(3x+5\right)^2+\left(6x+10\right)\left(2x-3x\right)+\left(2-3x\right)^2\)
a) Sửa đề: \(A=\left(3x-2\right)\left(9x^2+6x+4\right)-3x\left(9x^2-2\right)\)
\(=27x^3-8-27x^3+6=-2\)
b: Ta có: \(B=\left(3x+5\right)^2+\left(6x+10\right)\left(2-3x\right)+\left(2-3x\right)^2\)
\(=\left(3x+5+2-3x\right)^2\)
=49