1+4=5
2+5=12
3+6=21
8+11=?
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\(\frac{2}{3};1;\frac{4}{3};\frac{5}{3};2;\frac{7}{3};\frac{8}{3};9;\frac{10}{3}\)
HT
A=\(x.\left(x+2\right).\left(x^2+2x+2\right)+1\)
\(=x.\left(x+2\right).\left[x.\left(x+2\right)+2\right]+1\)
đặt \(x.\left(x+2\right)=a\) ta có:
\(A=a.\left(a+2\right)+1\)
\(A=a^2+2a+1\)
\(=\left(a+1\right)^2\)
\(=\left[x.\left(x+2\right)+1\right]^2\)
\(=\left(x^2+2x+1\right)^2\)
\(=\left(x+1\right)^4\)
Tổng ba số là :
\(\left(12363+18535+20018\right)\div2=25458.\)
Số thứ ba là :
\(25458-12363=13095\)
gọi 3 số đó lần lượt là \(a;b;c\)
theo đề bài ta có:
\(a+b=12363;b+c=18535;a+c=20018\)
\(\Rightarrow a+b+b+c+a+c=12363+18535+20018=50916\)
\(\Rightarrow2.\left(a+b+c\right)=50916\)
\(\Rightarrow a+b+c=25458\)
\(\Rightarrow12363+c=25458\)
\(\Rightarrow c=25458-12363=13095\)
vậu số thứ 3 là: \(13095\)
phân tích đa thức : \(a^2.\left(b-c\right)+b^2.\left(c-a\right)+c^2.\left(a-b\right)\) thành nhân tử
`Answer:`
`a^2.(b-c)+b^2.(c-a)+c^2.(a-b)`
`=a^2.(b-c)+b^2[(c-b)-(a-b)]+c^2.(a-b)`
`=a^2.(b-c)+b^2.(c-b)+b^2.(a-b)+c^2.(a-b)`
`=(b-c)(a^2-b^2)-(a-b)(b^2-c^2)`
`=(b-c)(a-b)(a+b)-(a-b)(b-c)(b+c)`
`=(a-b)(b-c)(a+b-b-c)`
`=(a-b)(b-c)(a-c)`
\(a^2.\left(b-c\right)+b^2,\left(c-a\right)+c^2.\left(a-b\right)\)
\(=a^2.\left(b-c\right)-b^2.\left(a-c\right)+c^2.\left(a-b\right)\)
\(=a^2.\left(b-c\right)-b^2.\left[\left(a-b\right)+\left(b-c\right)\right]+c^2.\left(a-b\right)\)
\(=a^2.\left(b-c\right)-b^2.\left(a-b\right)-b^2.\left(b-c\right)+c^2.\left(a-b\right)\)
\(=\left(b-c\right).\left(a^2-b^2\right)+\left(a-b\right).\left(c^2-b^2\right)\)
\(=\left(b-c\right).\left(a-b\right).\left(a+b\right)-\left(b-c\right).\left(b+c\right).\left(a-b\right)\)
\(=\left(b-c\right).\left(a-b\right).\left(a+b-b-c\right)\)
\(=\left(b-c\right).\left(a-b\right).\left(a-c\right)\)
vì \(x+y+z=0\) \(\Rightarrow\)\(x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=-z^3\)\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=-3xy.\left(x+y\right)\)
\(\Rightarrow x^3+y^3+z^3=3xyz\) do \(x+y=-z\)
\(\Rightarrow\left(x^3+y^3+z^3\right).\left(x^2+y^2+z^2\right)=3xyz.\left(x^2+y^2+z^2\right)\)
\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(y^2+z^2\right)+y^3.\left(x^2+z^2\right)+z^3.\left(x^2+y^2\right)\)
lại có: \(x^2+y^2=\left(x+y\right)^2-2xy=z^2-2xy\)
tương tự thì: \(y^2+z^2=x^2-2yz\)
\(z^2+x^2=y^2-2xz\)
vì vậy nên \(3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(x^2-2yz\right)+y^3.\left(y^2-2xz\right)+z^3.\left(z^2-2xy\right)\)
\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=2x^5+2y^5+2z^5-2xyz.\left(x^2+y^2+z^2\right)\)
\(\Rightarrow5xyz.\left(x^2+y^2+z^2\right)=2.\left(x^5+y^5+z^5\right)\)
đpcm
VÌ \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^5=-z^5\)
\(\Leftrightarrow x^5+y^{^5}+5\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right)=-z^5\)
\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x^3+y^3+2x^3y^2+2x^2y^3\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x+y\right)+\left(x^2-xy+y^2+2xy\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5-5xyz\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5=5xyz\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(2x^2+2xy+2y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+\left(x+y\right)^2+y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)Vì (x+y=-z)
HT
á , thì ra mày chọn cái sai
đáp án là 1+4=5 vì 4x1+1=5
2+5=12 vì 5x2+2=12
3+6=21 vì 3x6+3=21
vậy 8+11=96 vì 8x11+8=96
mong k đúng
8+11=19
HT