Cho a2+b2+c2=ab+bc+ca.Chứng minh rằng a=b=c
Cho a2+b2+c2+d2+1=a+b+c+d.Tính giá trị của biểu thức
K=(a-1,5)2+(b-1,5)3+(c-1,5)4+(d-1,5)5
Cho a2+b2+c2=ab+bc+ca.Chứng minh rằng a=b=c
Cho a2+b2+c2+d2+1=a+b+c+d.Tính giá trị của biểu thức
K=(a-1,5)2+(b-1,5)3+(c-1,5)4+(d-1,5)5
Viet cac da thuc sau duoi dang tich :
a, 100x2−(x2+25)2
b, 1+(x−y+5)2−2(x−y+5)
c, (x2+4y2−5)2−16(x2y2+2xy+1)
d, (x2+8x−34)2−(3x2−8x−2)2
a) \(100x^2-\left(x^2+25\right)^2\)
\(=\left(10x\right)^2-\left(x^2+25\right)^2\)
\(=\left(10x+x^2+25\right)\left(10x-x^2-25\right)\)
\(=\left(x+5\right)^2\left(10x-x^2-25\right)\)
b) \(1+\left(x-y+5\right)^2-2\left(x-y+5\right)\)
\(=\left[\left(x-y+5\right)-1\right]^2\)
\(=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
\(=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2\)
\(=\left[\left(x^2+4y^2-5\right)-4\left(xy+1\right)\right]\left[\left(x^2+4y^2-5\right)+4\left(xy+1\right)\right]\)
\(=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)\)
\(=\left(x^2-4xy+4y^2-9\right)\left(x^2+4xy+4y^2-1\right)\)
\(=\left[\left(x-2y\right)^2-3^2\right]\left[\left(x+2y\right)^2-1^2\right]\)
\(=\left(x-2y-3\right)\left(x-2y+3\right)\left(x+2y-1\right)\left(x+2y+1\right)\)
d) \(\left(x^2+8x-34\right)^2-\left(3x^2-8x-2\right)^2\)
\(=\left[\left(x^2+8x-34\right)-\left(3x^2-8x-2\right)\right]\left[\left(x^2+8x-34\right)+\left(3x^2-8x-2\right)\right]\)
\(=\left(x^2+8x-34-3x^2+8x+2\right)\left(x^2+8x-34+3x^2-8x-2\right)\)
\(=\left(-2x^2+16x-32\right)\left(4x^2-36\right)\)
\(=-2\left(x^2-8x+16\right)\left[\left(2x\right)^2-6^2\right]\)
\(=-2\left(x-4\right)^2\left(2x-6\right)\left(2x+6\right)\)
\(=-2\left(x-4\right)^24\left(x-3\right)\left(x+3\right)\)
\(=8\left(x-4\right)^2\left(x-3\right)\left(x+3\right)\)
tính theo các hằng thức đã học
1, (x-4)^2 -36 =0
2, x^2 -25-(x+5)^2
3, (2x-3)^2 = (x+5)^2
4, (2x -1)^1 - (4x^2 - 1^2) = 0
5, (x+8)^2 = 191
6, x^2 +4 -(x-2)^2 =0
\(1,\left(x-4\right)^2-36=0\)
\(\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
\(2,x^2-25-\left(x+5\right)^2\)
\(=\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=\left(x+5\right)\left(x-5-x-5\right)\)
\(=-10\left(x+5\right)\)
\(3,\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(5,\left(x+8\right)^2=191\)
\(\Leftrightarrow\left(x+8\right)^2-191=0\)
\(\Leftrightarrow\left(x+8-\sqrt{191}\right)\left(x+8+\sqrt{191}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{191}-8\\x=-\sqrt{191}-8\end{matrix}\right.\)
\(6,x^2+4-\left(x-2\right)^2=0\)
\(\Leftrightarrow x^2+4-x^2+4x-4=0\)
\(\Leftrightarrow4x=0\Leftrightarrow x=0\)
Chứng minh:
a) \(\left(a-b\right)^3=-\left(b-a\right)^3\)
b) \(\left(-a-b\right)^2=\left(a+b\right)^2\)
c) \(\left(x+y\right)^3=x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
d) \(\left(x+y\right)^3-\left(x-y\right)^3=2y\left(y^2+3x^2\right)\)
a. -(b-a)3= -b3+a3 (phá ngoặc trước có dấu trừ nên đổi dấu)
= a3 - b3 = (a-b)3
b)
\(\left(-a-b\right)^2=\left(-a\right)^2-2.\left(-a\right)b+b^2\\ =a^2+2ab+b^2=\left(a+b\right)^2\)
c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2\\ =x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\\ =x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\\ =x^3+3x^2y+3xy^2+y^3\\ =\left(x+y\right)^3\)
Tìm giá trị nhỏ nhất của biểu thức sau:
x2-4x-7
\(A=x^2-4x-7\)
\(A=x^2-2.x.2+4-11\)
\(A=\left(x-2\right)^2-11\)
Vì \(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-2\right)^2-11\ge-11\)
\(\Rightarrow Amin=-11\Leftrightarrow x=2\)
Ta có: \(x^2-4x-7=x^2-2.x.2+2^2-4-7\)
\(=\left(x-2\right)^2-11\)
Do \(\left(x-2\right)^2\ge0\) với mọi x (dấu "=" xảy ra <=> x=2)
\(\Rightarrow\left(x-2\right)^2-11\ge-11\) hay \(x^2-4x-7\ge-11\) (dấu "=" xảy ra <=> x=2)
Vậy giá trị nhỏ nhất của biểu thức trên là -11 tại x=2
Chứng minh rằng:
\(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(2c-b\right)^2=2a^2+3b^2+6c^2\)
\(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(2c-b\right)^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2-2a\left(b+c\right)+\left(b+c\right)^2+4c^2-4bc+b^2=2a^2+2b^2+5c^2+2ab+2bc+2ac-2ab+b^2+2bc+b^2=2a^2+3b^2+6c^2+4bc\)
=> Đề sai
1. Tìm x, biết
a) (x-1)4 + (x-2)4 = 1
Đặt: \(a=x-1\Rightarrow x-2=a-1\)
Ta có:
\(a^4+\left(a-1\right)^4=1\)
\(\Leftrightarrow a^4+a^4-4a^3+6a^2-4a+1=1\)
\(\Leftrightarrow2a^4-4a^3+6a^2-4a=0\)
\(\Leftrightarrow a\left(2a^3-4a^2+6a-4\right)=0\)
\(\Leftrightarrow a\left(2a^3-2a^2-2a^2+2a+4a-4\right)=0\)
\(\Leftrightarrow a\left[2a^2\left(a-1\right)-2a\left(a-1\right)+4\left(a-1\right)\right]=0\)
\(\Leftrightarrow a\left(a-1\right)\left(2a^2-2a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=1\\2a^2-2a+4=2\left(a-\dfrac{1}{2}\right)^2+\dfrac{7}{2}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(x-1\right)^4+\left(x-2\right)^4=1\)
Đặt: \(a=x-1\Rightarrow x-2=a-1\)
Ta có:
\(a^4+\left(a-1\right)^4=1\)
\(\Leftrightarrow a^4-a^4+4a^3-6a^2+4a-1=1\)
\(\Leftrightarrow4a^3-6a^2+4a-2=0\)
\(\Leftrightarrow4a^3-4a^2-2a^2+2a+2a-2=0\)
\(\Leftrightarrow4a^2\left(a-1\right)-2a\left(a-1\right)+2\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(4a^2-2a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\4a^2-2a+2=4\left(a-\dfrac{1}{4}\right)^2+\dfrac{7}{4}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow a=1\Rightarrow x=2\)
Đặt x - 2 = t
=> t4 + (t - 1)4 = 1
<=> t4 + t4 - 4t3 + 6t2 - 4t + 1 - 1 = 0
<=> 2t4 - 4t3 + 6t2 - 4t = 0
<=> t(2t4 - 4t3 + 6t - 4) = 0
<=> t( 2t3 - 2t2 + 4t - 2t2 + 2t - 4 ) = 0
<=> t[t(2t2 - 2t + 4) - 1(2t2 - 2t + 4)] = 0
<=> t(t - 1)(2t2 - 2t + 4) = 0
=> t = 0
=> t - 1 = 0
=> 2t2 - 2t + 4 = 0
=> t = 0
=> t = 1
=> Không có nghiệm
=> x - 2 = 0
=> x = 2
=> x - 2 = 1
=> x = 3
Bài 8: a)Chứng minh rằng ( a + b + c)3- a3 – b3 – c3 = 3( a +b)(b +c)( c+ a)
b)a3 +b3 +c3 – 3abc = ( a + b + c)( a2 +b2 + c2)
a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)
b) Ta có:
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi
a) Ta có: (a + b + c)3 - a3 - b3 - c3 = [ (a + b + c)3 - a3 ] - ( b3 + c3)
= (a + b + c - a) ( a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2) - (b + c) ( b2 - bc + c3)
= (b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac) - (b + c) ( b2 - bc + c3)
= ( b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac - b2 + bc - c3)
= ( b + c) ( 3a2 + 3ab + 3bc + 3ac)
= 3 (b + c) [a (a + b) + c (a + b)]
= 3 (b + c) (a + b) (a + c) (đpcm)
Tìm giá trị nhỏ nhất của biểu thức:
\(A=x^2-4x+1\)
\(B=4x^2+4x+11\)
\(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy Min A là : \(-3\Leftrightarrow x=2\)
\(B=4x^2+4x+11=\left(2x\right)^2+2.2x+1+10=\left(2x+1\right)^2+10\ge10\forall x\)Dấu " = " xảy ra
\(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy Min B là : \(11\Leftrightarrow x=-\dfrac{1}{2}\)
\(A=x^2-4x+1\)
\(\Rightarrow A=x^2-4x+4-3\)
\(\Rightarrow A=\left(x-2\right)^2-3\)
Do \(\left(x-2\right)^2\ge0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow x-2=0\Rightarrow x=2\))
\(\Rightarrow\left(x-2\right)^2-3\ge-3\) hay \(A\ge-3\) (dấu "=" xảy ra \(\Leftrightarrow x=2\))
Vậy \(A_{min}=-3\) tại \(x=2\)
\(B=4x^2+4x+11\)
\(\Rightarrow B=\left(2x\right)^2+4x+1^2+10\)
\(\Rightarrow B=\left(2x+1\right)^2+10\)
Do \(\left(2x+1\right)^2\ge0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow2x+1=0\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\))
\(\Rightarrow\left(2x+1\right)^2+10\ge10\) hay \(B\ge10\) (dấu ''='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))
Vậy \(B_{min}=10\) tại \(x=\dfrac{1}{2}\)
Chúc Bạn Học Tốt!!!
Biết số tự nhiên a chia 5 dư 4.Chứng minh rằng \(a^2\) chia 5 dư 1
Ta có: a chia 5 dư 4 suy ra a có dạng 5k+4 (k \(\in\) Z)
\(\Rightarrow\) a2 = (5k+4)2 = 25k2+40k+16
= 25k2+40k+15+1
Vì 25k2; 40k và 15 chia hết cho 5 nên suy ra 25k2+40k+15+1 chia 5 dư 1.
Vậy a2 chia 5 dư 1.
a : 5 dư 4 \(\Rightarrow a=5k+4\left(k\ge0\right)\)
\(\Rightarrow a^2=\left(5k+4\right)^2\)
\(\Rightarrow a^2=25k^2+40k+16\)
\(\Rightarrow a^2=5\left(5k^2+8k+3\right)+1\)
\(\Rightarrow a^2:5\) dư 1
( đpcm )