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Tính nhanh:
a)
153^2 + 94 * 153 + 47^2
= 153 * (153 + 94) + 47^2
= 153 * 247 + 47^2
= 153 * (200 + 47) + 47^2
= 153 * 200 + 153 * 47 + 47^2
= 153 * 200 + 47 * (153 + 47)
= 153 * 200 + 47 * 200
= 200 * (153 + 47)
= 200 * 200
= 40000
b)126^2 - 152.126 + 5776
= 126 . 126 - 152.126 +126. 2888/63
= 126 . ( 126 - 152 + 2888/63)
= 126 . 1250/63
= 2500
Câu c bn tự làm nha
a) \(153^2+94.153+47^2=153^2+2.47.153+47^2\)
\(=\left(153+47\right)^2=200^2=40000\)
b) \(126^2-152.126+5776=126^2-2.76.126+76^2\)
\(=\left(126-76\right)^2=50^2=2500\)
c) \(3^8.5^8-\left(15^4-1\right)\left(15^4+1\right)=15^8-\left[\left(15^4\right)^2-1\right]\)
\(=15^8-\left(15^8-1\right)=15^8-15^8+1=1\)
A: \(135^2+94\cdot153+47^2=135^2\cdot2\cdot47+153\cdot47\)
\(=47\left(36450+153\right)=36603\cdot47=1720341\)
B: \(126^2-152\cdot126+5776=126^2-2\cdot126\cdot76+76^2=\left(126-76\right)^2=50^2=2500\)
C: \(3^8\cdot5^8-\left(15^4-1\right)\left(15^4+1\right)=15^8-\left(15^8-1\right)=15^8-15^8+1=1\)
1. Câu hỏi của Koy Pham - Toán lớp 8 | Học trực tuyến
2.
a. \(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-27\right)-\left(54+x^3\right)\)
\(=x^3-27-54-x^3\)
\(=-81\)
b. \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+xy+y^2\right)\)
\(=\left[\left(3x\right)^3+y^3\right]-\left[\left(3x\right)^3-y^3\right]\)
\(=2y^3\)
a, \(53^2+47^2+94.53=53^2+2.53.47+47^2=\left(53+47\right)^2=100^2=10000\)
b, \(50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50+49\right).\left(50-49\right)+\left(48+47\right).\left(48-47\right)+...+\left(2+1\right).\left(2-1\right)\)
\(=50+49+48+47+...+2+1=\dfrac{\left(50+1\right).50}{2}\)
\(=51.25=1275\)
bạn giải thích chỗ \(\frac{\left(50+1\right).50}{2}\) từ đâu ra được không ạ
\(a,\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}=\frac{16.110}{110.320}=\frac{1}{20}\)
\(b,\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}=\frac{74.800}{74.1000}=0,8\)
\(c,2^{32}-\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-2^{32}+1=1\)
\(d,100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)
\(=\left(100^2-98^2\right)+\left(103^2-101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)
\(=\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)-\left(107-105\right)\left(107+105\right)-\left(96-94\right)\left(96+94\right)\)
\(=2.198+2.204-2.212-2.190\)
\(=2\left(198+204-212-190\right)=2.0=0\)
a) \(A=5^4.3^4-\left(15^2-1\right)\left(15^2+1\right)=\left(5.3\right)^4-\left(\left(15^2\right)^2-1^2\right)\)
\(=15^4-\left(15^4-1\right)=15^4-15^4+1=1\)
b) \(C=50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=1.99+1.95+...+1.3=99+95+...+3\)
\(=\left(99+3\right)+\left(95+7\right)+...+\left(55+47\right)+51\)
\(=102+102+...+102+51\)
số lượng con số \(102\) là \(\dfrac{25-1}{2}=12\)
\(\Rightarrow C=102.12+51=1224+51=1275\)
Bài 1:
a) \(100^2-99^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+...+2+1\)
=> tự làm tiếp :))
b) tương tự
Bài 2 :
a) \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\)
\(A=2^{16}-1< 2^6=B\)
b) Phân tích \(2004\cdot2006=\left(2005-1\right)\left(2005+1\right)=\left(2005^2-1\right)\)rồi áp dụng hđt thứ 3 tự làm tiếp như câu a)
Bài 3:
a) Cứ khai triển hết ra
b) \(a^2+b^2+c^2=ab+bc+ac\)
\(a^2+b^2+c^2-ab-bc-ac=0\)
Nhân 2 vào cả 2 vế được :
\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+c^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mà mũ 2 luôn lớn hơn hoặc bằng 0
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c\left(đpcm\right)}\)
P.s: toàn bài nâng cao làm hơi ẩu tí ^^
532 + 106 * 47 + 472
= 532 + 2 * 53 * 47 + 472
= ( 53 + 47 )2 = 1002 = 10000