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Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :
\(y^2+2y+4^x-2^{x+1}+2=0\)
\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)
\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)
\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)
CHÚC BẠN HỌC TỐT...........
a)
$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$
$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$
$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$
$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$
$=2005^2-1002.2005=2005(2005-1002)=2011015$
b)
$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{32}-1)(2^{32}+1)-2^{64}$
$=2^{64}-1-2^{64}=-1$
c) Do $x=16$ nên $x-16=0$
$R(x)=x^4-17x^3+17x^2-17x+20$
$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$
$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$
$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$
d) Do $x=12$ nên $x-12=0$. Khi đó:
$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$
$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$
$=(x-12)(x^9-x^8+x^7-....+x)-x+10$
$=0-x+10=-x+10=-12+10=-2$
a) \(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=.............................................................\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1=B-1\)
Suy ra A < B
b) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1=B-1\)
Suy ra A < B
Phần a bạn nhân thêm ở A là (2-1) là ra hằng đẳng thức, cứ thế mà triển. (Kết quả: A<B)
Phần b: phân tích A, ta có:
2015.2017= (2016-1).(2016+1)= 2016^2 -1 <2016^2
Suy ra: A<B
\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=5^{32}-1< 5^{32}\)
Vậy \(B< A\)
\(B=10^2+8^2+...+2^2-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(B=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(B=\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2-1\right)\left(2+1\right)\)
\(B=19+15+...+3\)
Đến đây dễ rồi. Câu a) đang suy nghĩ
\(A=1+\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+4\cdot\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(4A=4+5^{64}-1\)
\(4A=5^{64}+3\)
\(A=\frac{5^{64}+3}{4}\)
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)
Vậy \(B< A\)
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt
Lời giải:
Ta sử dụng các hằng đẳng thức đáng nhớ, cụ thể là công thức:
\((a-b)(a+b)=a^2-b^2\)
a)
\(2003.2005=(2004-1)(2004+1)=2004^2-1^2=2004^2-1< 2004^2\)
Vậy \(2003.2005< 2004^2\)
b)
\(8(7^8+1)(7^4+1)(7^2+1)=(7+1)(7^2+1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}.(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}(7^2-1)(7^2+1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}(7^4-1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}(7^8-1)(7^8+1)=\frac{1}{6}(7^{16}-1)< 7^{16}-1\)
Tks