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Ta có : 642 + 128.36 + 362
= 642 + 2.64.36 + 362
= (64 + 36)2
= 1002
= 10000
Đúng 0 Báo cáo sai phạm
Ta có : 642 + 128.36 + 362
= 642 + 2.64.36 + 362
= (64 + 36)2
= 1002
= 10000
a/ A = 1002 - 992 + 982 -...+22 - 12
= (1002 - 992) + (982 - 972) +...+ (22 - 12)
= 199 + 195 + 191 + ... + 1
= (\(\frac{199-1}{4}+1\))(\(\frac{199+1}{2}\)) = 5050
b/ Y chang câu a luôn nha
c/ \(C=\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}\)
\(=\frac{560.1000}{200^2}=14\)
a) Đặt: x = a- b; y = b - c ; z = c- a
Ta có: x + y + z = 0
=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)
=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) Đặt: \(a=x^2-2x\)
Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)
d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)
Đặt: \(x^2-8=t\)
Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)
\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)
\(=\left(2x^2+9x-16\right)^2\)
a) Ta có F = \(\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
=> 8F = \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^8-1\right)\left(3^8+1\right)-3^{16}=3^{16}-1-3^{16}=-1\)
=> F = -1/8
b) Ta có G = \(\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)-\frac{2^{24}}{7}\)
=> 7G = 7(23 + 1)(26 + 1)(212 + 1) - 224
=> 7G = (23 - 1)(23 + 1)(26 + 1)(212 + 1) - 224
=> 7G = (26 - 1)(26 + 1)(212 + 1) - 224
=> 7G = (212 - 1)(212 + 1) - 224
=> 7G = 224 - 1 - 224
=> 7G = -1
=> G = -1/7
\(F=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
<=> \(\left(3^2-1\right)F=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\left(3^2-1\right)\frac{3^{16}}{8}\)
<=> \(8F=\left(3^4-1\right)\left(3^4+1\right)\left(3^8-1\right)-3^{16}\)
<=> \(8F=\left(3^8+1\right)\left(3^8-1\right)-3^{16}\)
<=> \(8F=\left(3^{16}-1\right)-3^{16}=-1\)
<=> F = -1/8
Câu G làm tương tự
a) Áp dụng hằng đẳng thức ta đc:
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100+99\right)\left(100-99\right)+\left(98-97\right)\left(98+87\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=199+195+191+...+3\)
\(=\left[\left(199-3\right):4+1\right]\cdot\left(199+3\right):2=50\cdot101=5050\)
a) Áp dụng hằng đẳng thức ta đc:
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100+99\right)\left(100-99\right)+\left(98-97\right)\left(98+87\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=199+195+191+...+3\)
\(=\left[\left(199-3\right):4+1\right]\cdot\left(199+3\right):2=50\cdot101=5050\)
b) mk nghĩ bước đầu tiên là phải bỏ ngoặc:
\(=20^2+18^2+16^2+...4^2+2^2-19^2-17^2-....-3^2-1^2\)
\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+...+\left(4^2-3^2\right)-1^2\)
\(=\left(20+19\right)\left(20-19\right)+\left(18+17\right)\left(18-17\right)+...+\left(4-3\right)\left(4+3\right)-1\)
\(=\left(39+35+31+...+7\right)-1\)
\(=\left(\left[\left(39-7\right):4+1\right]\cdot\left(39+7\right):2\right)-1=207-1=206\)
a) 1002-992+....+22-12
=(100+99)(100-99)+(98+97)(98-97)+...+(2+1)(2-1)
=100+99+98+...+2+1
b) bieu thuc tren =
202-192+182-172+...+22-12
tinh tuong tu cau a
Bài 1:
F=(x-1)3-x2(x-3)
=x3-3x2+3x-1-x3-3x2
=(x3-x3)-(3x2-3x2)+3x-1
=3x-1
Bài 2:
a)(x+3)2=(x-2)(x+4)
<=>x2+6x+9=x2+2x-8
<=>4x=-17
<=>x=-17/4
b)(x+4)2=2x2+16
<=>x2+8x+16=2x2+16
<=>8x=x2
<=>8x-x2=0
<=>x(8-x)=0
<=>x=0 hoặc x=8
Bài 1:
F=(x-1)3-x2(x-3)=x3-3x2+3x-1-x3+3x2=3x-1
Bài 2:
a, <=>(x+3)2-(x-2)(x-4)=0
<=>x^2+6x+9-x^2-4x+2x+8=0
<=>4x+17=0
<=>x=-4,25
b,<=>(x+4)2-2x2-16=0
<=>x2+8x+16-2x2-16=0
<=>8x-x2=0
<=>x(8-x)=0
<=>\(\orbr{\begin{cases}x=0\\x=8\end{cases}}\)
Bài 3:(đợi một xíu)
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) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)
\(=\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560.1000}{200}=2800\)
mik ko chép lại đề bài nha
a) = (123)2- 12- (36. 46)
= (126-1)- (3.4)6
= 126-1-126
= -1