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Bài 2:
a: \(\left(a-b-2\right)^2-\left(2a-2b\right)\left(a-b-2\right)+a^2-2ab+b^2\)
\(=\left(a-b\right)^2-4\left(a-b\right)+4+\left(a-b\right)^2-2\left(a-b\right)\left(a-b-2\right)\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left[\left(a-b\right)^2-2\left(a-b\right)\right]\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left(a-b\right)^2+4\left(a-b\right)\)
\(=4\)
b: \(\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(=2^{512}-1+1=2^{512}\)
c: \(24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}\)
=-1
Bài 1:
\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^2\left(x^2-1\right)\)
\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
\(=x^6+27-27-27x^2-9x^4-x^6\)
\(=-9x^2\left(3-x^2\right)\)
Bài 5:
\(A=x^2-2x+1\)
\(=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
Vậy Min A = -2
Để A = -2 thì \(x-1=0\Rightarrow x=1\)
b, \(B=4x^2+4x+5\)
\(=\left(4x^2+4x+1\right)+4\)
\(=\left(2x+1\right)^2+4\)
Với mọi giá trị của x ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)
Vậy Min B = 4
Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
c, \(C=2x-x^2-4\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3
để C = -3 thì \(x-1=0\Rightarrow x=1\)
ý a pạn đưa về dạng ax+b=0 khi chuyển 16 sang và rút gọn 2 biểu thức còn lại đưa về dạng (a+b)2+(a-b)2-16=0. thế thôi. hai biểu thức (x+3)4+(x-2) 4 tự phân tích nhé
a, 4x.(x - 2017 ) - x + 2017 = 0
\(\Leftrightarrow\) 4x ( x - 2017 ) - ( x - 2017 ) = 0
\(\Leftrightarrow\) ( x - 2017 ) ( 4x - 1 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 2017 hoặc x = \(\dfrac{1}{4}\) .
b) \(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-x-1\right)=0\)
\(x+1=0\)
x = -1
c) \(x\left(x-5\right)-\left(4x-20\right)=0\)
\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
Bài 2:
a: \(x^2-16-\left(x+4\right)=0\)
=>(x+4)(x-4)-(x+4)=0
=>(x+4)(x-5)=0
=>x=5 hoặc x=-4
b: \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+1=0\)
=>-6x+2=0
=>-6x=-2
hay x=1/3
c: \(4x^2+9=-12x^2\)
\(\Leftrightarrow4x^2+12x^2=-9\)
\(\Leftrightarrow16x^2=-9\)(vô lý)
Do đó: \(x\in\varnothing\)
d: \(4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
=>x=1 hoặc x=1/4
e: \(4x^2-4x+3=0\)
\(\Leftrightarrow4x^2-4x+1+2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=-2\)(vô lý)
Do đó: \(x\in\varnothing\)
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)....\left(2^{64}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)....\left(2^{64}+1\right)+1\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)......\left(2^{64}+1\right)+1\)
\(A=\left(2^8-1\right)\left(2^8+1\right)......\left(2^{64}+1\right)+1\)
\(A=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(A=2^{128}-1+1=2^{128}\)
Ribi Nkok Ngok hãy cảm nhận sự khác biệt cách tiếp cận sử lý bài toán (B là huyền thoại)\(B=1^2-2^2+3^2-4^2+....-2010^2+2011^2\)
\(\text{B=1+(3-2)(3+2) +(5-4)(5+4)+....+(2011-2010)(2011+2010)}\)\(\text{B=1+(3+2) +(5+4)+....+(2011+2010)}\)\(B=1+2+3+...+2011\)
\(B=\dfrac{2011.2012}{2}=2011.1006=2023066\)
Giúp vs @@Phạm Hoàng GiangTrần Quốc LộcTrần Thị Hươnghattori heijiTRẦN MINH HOÀNGAn Nguyễn BáRibi Nkok NgokKien Nguyen
Trần Đăng NhấtHung nguyen
Sửa đề bài 1 : Rút gọn
a,\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right).........\left(2^{32}+1\right)-2^{64}\)