viết các biểu thức sau dưới dạng lũy thừa
a) \(127^2+146\times127+73^2\)
b) \(9^8\times2^8-\left(18^4-1\right)\left(18^4+1\right)\)
Tính nhanh giá trị các biểu thức sau
a. \(127^2+146\times127+73^2\)
b.\(9^8\times2^8-\left(18^4-1\right)\times\left(18^4+1\right)\)
c.\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
d. \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)
e. \(\dfrac{780^2-220^2}{125^2+150\times125+75^2}\)
a) \(127^2+146.127+73^2=127^2+2.73.127+73^2=\left(127+73\right)^2=40000\)b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^8-1\right)=1\)
c) \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=100+99+98+97+...+2+1\)
\(=\dfrac{100\left(100+1\right)}{2}=5050\)
d) \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\) \(=20^2-19^2+18^2-17^2+16^2-15^2+...+4^2-3^2+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(2-1\right)\left(2+1\right)\)\(=20+19+18+17+...+2+1\)
\(=\dfrac{20\left(20+1\right)}{2}=210\)
e) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)
\(=\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560.1000}{200}=2800\)
Bài 1:
a) Tính giá trị của đa thức
A= 2(x\(^3\) + y\(^3\)) - 3(x\(^2\) + y\(^2\) ) Với: x + y = 1
b) Biểu thức sau đây có phụ thuộc biến x không ?
B= (x + 1)\(^3\) - (x - 1)\(^3\) - 6 (x + 1) (x - 1)
Bài 2: Tính nhanh
a) \(127^2\) + \(146\times127\) + \(73^2\)
b) \(9^8\times2^8-\left(18^4-1\right)\left(18^4+1\right)\)
1.
a/ \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2.1.\left[1^2-3xy\right]-3\left[1^2-2xy\right]\)
\(=2-6xy-3+6xy\)
\(=-1\)
Vậy...
2.
a. \(127^2+146.127+73^2\)
\(=127^2+2.73.127+73^2\)
\(=\left(127+73\right)^2=200^2=40000\)
b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1\)
\(=1\)
1. Rút gọn các biểu thức sau:
a) \(\left(x+y\right)^2-\left(x-y\right)^2\)
b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)
c) \(9^8\times2^8-\left(18^4-1\right)\left(18^4+1\right)\)
a) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)
b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\\ =6ab^2\)
c) \(9^8\times2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-18^8+1=1\)
viết các biểu thức sau dưới dạng lũy thừa của 1 số hữu tỉ
a. \(8^{15}.4^{13}\)
b. \(\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{4}\right)^{28}\)
c. \(9^{12}.27^{10}\)
a) \(8^{15}.4^{13}=\left(2^3\right)^{15}.\left(2^2\right)^{13}=2^{45}.2^{26}=2^{71}\)
b) \(\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{4}\right)^{28}=\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{2^2}\right)^{28}=\frac{1}{2^{18}}.\frac{1}{2^{56}}=\frac{1}{2^{74}}=\left(\frac{1}{2}\right)^{74}\)
c) \(9^{12}.27^{10}=\left(3^2\right)^{12}.\left(3^3\right)^{10}=3^{24}.3^{30}=3^{54}\)
a) = (23)15. (22)13 = 245.226 = 271
b) = \(\left(\frac{1}{2}\right)^{18}.\left(\left(\frac{1}{2}\right)^2\right)^{28}=\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{2}\right)^{56}=\left(\frac{1}{2}\right)^{18+56}=\left(\frac{1}{2}\right)^{74}\)
c) = (32)12.(33)10 = 324.330 = 324+30 = 354
Viết các biểu thức sau dưới dạng lũy thừa
a) \(2^2.9.\dfrac{1}{54}.\left(\dfrac{4}{9}\right)^2\)
b) \(2^3.2^5.\left(\dfrac{3}{2}\right)^4\)
c) \(\dfrac{\left(\dfrac{1}{2}\right)^3.\dfrac{1}{2^2}.8}{\left(-2^3\right)^2.16}.\left(2^2\right)^3\)
a: \(=2^2\cdot9\cdot\dfrac{1}{6\cdot9}\cdot\dfrac{4^2}{9^2}=\dfrac{2^2}{6}\cdot\dfrac{2^4}{3^4}=\dfrac{2^6}{2\cdot3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\left(\dfrac{1}{2}\right)^3\cdot2^3\cdot\left(\dfrac{1}{2}\right)^2}{\left(-8\right)^2\cdot16}\cdot2^6=\dfrac{\dfrac{1}{2^2}}{64\cdot16}\cdot64=\dfrac{1}{4}:16=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\)
Viết các biểu thức sau dưới dạng lũy thừa
a) \(2^2.9.\dfrac{1}{54}.\left(\dfrac{4}{9}\right)^2\)
b) \(2^3.2^5.\left(\dfrac{3}{2}\right)^4\)
c) \(\dfrac{\left(\dfrac{1}{2}\right)^3.\dfrac{1}{2^2}.8}{\left(-2^3\right)^2.16}.\left(2^2\right)^3\)
a: \(=2^2\cdot9\cdot\dfrac{1}{3^3\cdot2}\cdot\dfrac{2^4}{3^4}=\dfrac{2^4\cdot2^2}{2}\cdot\dfrac{9}{3^3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\dfrac{1}{2^3}\cdot\dfrac{1}{2^2}\cdot8}{\left(-8\right)^2\cdot2^4}\cdot2^6=\dfrac{1}{2^2}\cdot2^6:2^{10}=\dfrac{2^4}{2^{10}}=\dfrac{1}{2^6}=\left(\dfrac{1}{8}\right)^2\)
\(I\)Tính nhanh
\(a.127^2+146.127+73^2\)
\(b.9^8.2^8\left(18^4-1\right)\left(18^4+1\right)\)
\(c.100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(d.\frac{780^2-220^2}{125^2+150.125+75^2}\)
\(II.\)Rút gọn các biểu thức
\(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
Bài 11.Tính giá trị biểu thức
1) \(A=127+146.127+73^2\)
2) \(B=98.28-\left(18^4-1\right).\left(18^4+1\right)\)
3) \(C=-1^2+2^2-3^2+4^2+.....-99^2+100^2\)
4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
5) \(E=1^2-2^2+3^2-4^2+.....-2004^2+2005^2\)
Bài 11:
1) Sửa lại đề là: \(A=127^2+146.127+73^2\)
\(\Rightarrow A=127^2+2.127.73+73^2\)
\(\Rightarrow A=\left(127+73\right)^2\)
\(\Rightarrow A=200^2\)
\(\Rightarrow A=40000\)
Vậy \(A=40000.\)
2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)
\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(\Rightarrow B=18^8-\left(18^8-1\right)\)
\(\Rightarrow B=18^8-18^8+1\)
\(\Rightarrow B=0+1\)
\(\Rightarrow B=1\)
Vậy \(B=1.\)
4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(\Rightarrow D=\frac{3^{32}-1}{2}\)
Viết các số \({\left( {\frac{1}{4}} \right)^8};{\left( {\frac{1}{8}} \right)^3}\) dưới dạng lũy thừa cơ số \(\frac{1}{2}\)
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)