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d )
\(B=5\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{64}-1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{128}-1\right)\)
Sửa lại dấu \(\Rightarrow\)dòng 3 :
\(B=\frac{5}{3}\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
Bài 1:
\(A=23^2+46\cdot37+37^2=23^2+2\cdot23\cdot37+37^2=\left(23+37\right)^2=60^2=3600\)
\(B=27^2-44\cdot27+22^2=27^2-2\cdot27\cdot22+22^2=\left(27-22\right)^2=5^2=25\)
Bài 2:
\(A=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\)
Vì: \(\left(x-2\right)^2\ge0\) với mọi x
=> \(\left(x-2\right)^2+1\ge1\)
Vậy GTNN của A là 1 khi x=2
\(A=23^2+2.23.37+37^2=\left(23+37\right)^2=60^2=3600\)
\(B=27^2-2.27.22+22^2=\left(27-22\right)^2=5^2=25\)
\(A=x^2-4x+5=\left(x-2\right)^2+1\ge1\)
=> A min=1 khi x=2
\(18x^2y^2\left(?\right)4x^2y\)
câu b)
\(\left(b\right)6x^3-9x^2=3x^2\left(x-3\right)\)
\(\left(c\right)4x^2-1=\left(2x-1\right)\left(2x+1\right)\)
Bài 1: Rút gọn
a) Ta có: \(A=\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
\(=2x+15\)
b) Ta có: \(B=\left(2x-1\right)^2+2\left(2x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(2x-1+x+1\right)^2\)
\(=\left(3x\right)^2=9x^2\)
Bài 2: Tính nhanh
a) Ta có: \(A=138^2+124\cdot138+62^2\)
\(=138^2+2\cdot138\cdot62+62^2\)
\(=\left(138+62\right)^2\)
\(=200^2=40000\)
b) Ta có: \(B=\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+3^2+1^2\right)\)
\(=100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+..+2+1\)
\(=5050\)
Bài 3: Chứng minh rằng các biểu thức sau luôn nhận giá trị dương với mọi giá trị của biến
a) Ta có: \(x^2-5x+10\)
\(=x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{75}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{75}{4}\)
Ta có: \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{5}{2}\right)^2+\frac{75}{4}\ge\frac{75}{4}\forall x\)
hay \(x^2-5x+10>0\forall x\)(đpcm)
b) Ta có: \(\left(x-1\right)\left(x-2\right)+5\)
\(=x^2-3x+2+5\)
\(=x^2-3x+7\)
\(=x^2-2\cdot x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{19}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
hay \(\left(x-1\right)\left(x-2\right)+5>0\forall x\)(đpcm)
\(40^2-39^2+38^2-37^2+..........+2^2-1^2\)
\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+..........+\left(2^2-1^2\right)\)
\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...........+\left(2-1\right)\left(2+1\right)\)
\(=40^2+39^2+38^2+37^2+.........+2^2+1^2\)
\(=\dfrac{40.41}{2}=820\)
Câu 1 :
a) \(x^3-5x^2-14x\)
\(=x^3-7x^2+2x^2-14x\)
\(=x^2\left(x-7\right)+2x\left(x-7\right)\)
\(=\left(x-7\right)\left(x^2+2x\right)\)
\(=x\left(x-7\right)\left(x+2\right)\)
b) \(a^4+a^2+1\)
\(=\left(a^2\right)^2+2a^2+1-a^2\)
\(=\left(a^2+1\right)-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
c) \(x^4+64\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot8+8^2-2\cdot x^2\cdot8\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
Câu 2 :
a) \(\left(a-b\right)^2=a^2-2ab+b^2\)
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot14=25\)
\(\Rightarrow\left(a-b\right)^2=25-2\cdot12=1\)
b) tương tự
Ta có : B = 202 - 192 + 182 - 172 + ..... + 22 - 12
=> B = (20 - 19)(20 + 19) + (18 - 17)(18 + 17) + ..... + (2 - 1)(2 + 1)
=> B = 39 + 35 + 31 + ..... + 3
Số số hạng của dãy trên là :
(39 - 3) : 4 + 1 = 10 (số)
Tổng B là :
(39 + 3) x 10 : 2 = 210
Vậy B = 210
Ta có : \(C=\left(15^4-1\right)\left(15^4+1\right)-3^8.5^8\)
\(\Rightarrow C=\left(15^4\right)^2-1-15^8\)
\(\Rightarrow C=15^8-1-15^8\)
=> C = -1
Vậy C = - 1
a,Ta có A=\(\dfrac{\left(54-23\right)\left(54+23\right)}{\left(36,5-25,5\right)\left(36,5+25,5\right)}=\dfrac{31.77}{11.62}=\dfrac{7}{2}\)
b,Ta có B=\(\dfrac{\left(82-34\right)\left(82+34\right)}{\left(30,5-1,5\right)\left(30,5+1,5\right)}=\dfrac{48.116}{29.32}=\dfrac{6.8.4.29}{29.8.4}=6\)
c,Ta có C=\(\dfrac{\left(86-54\right)\left(86^2+86.54+54^2\right)}{32}+86.54=86^2+86.54+54^2+86.54=\left(86+54\right)^2=19600\)
Lời giải:
$A=64^2-54^2=(64-54)(64+54)=10.118=1180$
$B=76^2+2.76.23+23^2=(76+23)^2=99^2=(100-1)^2=100^2-2.100+1$
$=10000-200+1=9801$
$C=88^2+62^2-38^2-12^2=(88^2-38^2)+(62^2-12^2)$
$=(88-38)(88+38)+(62-12)(62+12)$
$=50.126+50.74=50(126+74)=50.200=10000$