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\(x^5+x^4+1\)
\(=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^3.\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
cảm ơn bạn nhiều, không biết còn cách không? Mong nhận đượ giúp đỡ!
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Đặt : \(P=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Ta có: x+y+z=0
⇔(x+y+z)2=0⇔(x+y+z)2=0
⇔x2+y2+z2+2xy+2yz+2xz=0⇔x2+y2+z2+2xy+2yz+2xz=0(1)
Ta có: K=x2+y2+z2(x−y)2+(y−z)2+(z−x)2K=x2+y2+z2(x−y)2+(y−z)2+(z−x)2
=x2+y2+z2x2−2xy+y2+y2−2yz+z2+z2−2xz+x2=x2+y2+z2x2−2xy+y2+y2−2yz+z2+z2−2xz+x2
=x2+y2+z23x2+3y2+3z2−x2−y2−z2−2xy−2yz−2xz=x2+y2+z23x2+3y2+3z2−x2−y2−z2−2xy−2yz−2xz
=x2+y2+z23(x2+y2+z2)−(x2+y2+z2+2xy+2yz−2xz)=x2+y2+z23(x2+y2+z2)−(x2+y2+z2+2xy+2yz−2xz)
=x2+y2+z23(x2+y2+z2)=13=x2+y2+z23(x2+y2+z2)=13
Vậy: K=13K=13
Bài 1:
a. $=2x(x-3)$
b. $=x^3(x+3)+(x+3)=(x^3+1)(x+3)=(x+1)(x^2-x+1)(x+3)$
c. $=64-(x^2-2xy+y^2)=8^2-(x-y)^2$
$=(8-x+y)(8+x-y)$
Bài 2:
$(x+5)(x+1)+(x-2)(x^2+2x+4)-x(x^2+x-2)$
$=x^2+6x+5+(x^3-2^3)-(x^3+x^2-2x)$
$=x^2+6x+5+x^3-8-x^3-x^2+2x$
$=8x-3$
Ta có đpcm.
a: Xét ΔAHB vuông tại H và ΔCHA vuông tại H có
góc HAB=góc HCA
=>ΔAHB đồng dạng với ΔCHA
b,c: góc FAE+góc FHE=180 độ
=>FAEH nội tiếp
=>góc HFE=góc HAE=góc C
Xét ΔHFE vuông tại H và ΔHCA vuông tại H có
góc HFE=góc HCA
=>ΔHFE đồng dạng với ΔHCA
=>HF/HC=HE/HA
=>HF*HA=HC*HE
Câu 1: A
Câu 2: B
Câu 3: D
Câu 4: A
Câu 5: C
Câu 6: B
Câu 7: A
Câu 9: B
11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)
12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
13)
\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)
14)
\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)
Đặt \(A=2^{17}-2^{16}-2^{15}-...-2^2-2-1\) ta có :
\(A=2^{17}-\left(2^{16}+2^{15}+...+2+1\right)\)
Đặt \(B=2^{16}+2^{15}+...+2+1\) ta có :
\(2B=2^{17}+2^{16}+...+2^2+2\)
\(2B-B=\left(2^{17}+2^{16}+...+2^2+2\right)-\left(2^{16}+2^{15}+...+2+1\right)\)
\(B=2^{17}-1\)
\(\Rightarrow\)\(A=2^{17}-B=2^{17}-\left(2^{17}-1\right)=2^{17}-2^{17}+1=1\)
Vậy \(A=1\)
Chúc bạn iu họk tốt :3
bài này không biết làm á