phân tích đa thức thành nhân tử: x3-x2-4
làm xog đầu tiên mình tích:)))
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\(9-x^2+4x-4\)
\(=3^2-\left(x^2-4x+4\right)\)
\(=3^2-\left(x-2\right)^2\)
\(=\left(3-x+2\right).\left(3+x-2\right)\)
\(=\left(5-x\right)\left(1+x\right)\)
\(x^2-xy+x-y\)
\(=x\left(x+1\right)-y.\left(x+1\right)\)
\(=\left(x+1\right)\left(x-y\right)\)
Tham khảo nhé~
Ta có: \(x-y=4;xy=2\)
\(A=x^3-y^3+5\)
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+5\)
\(A=4.\left[\left(x-y\right)^2+3xy\right]+5\)
\(A=4.\left(4^2+3.2\right)+5\)
\(A=4.22+5\)
\(A=88+5\)
\(A=93\)
Tham khảo nhé~
A=2012x2014=2012x(2012+2)=2012^2+4024
B=2013^2=(2012+1)^2=2012^2+2x2012+1=2012^2+2025
=>A<B
chúc bạn học tốt~~~
Bài 1 :
\(a)\)\(A=2012.2014=\left(2013-1\right)\left(2013+1\right)=2013^2-1< 2013^2=B\)
Vậy \(A< B\)
\(b)\)\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(2A=3^{32}-1\)
\(A=\frac{3^{32}-1}{2}< 3^{32}-1=B\)
\(c)\)\(A=2017^2-17^2=\left(2017-17\right)\left(2017+17\right)=2000.2034>2000.2000=2000^2=B\)
Vậy \(A>B\)
\(A=x^3-3xy+y^3\)
\(A=\left(x^3+y^3\right)-3xy\)
\(A=\left(x+y\right)\left(x^2-xy+y^2\right)-3xy\)
\(A=5\left(21-xy\right)-3xy\)
\(A=105-5xy-3xy\)
\(A=105-8xy\)
Từ \(x+y=5\)\(\Leftrightarrow\)\(\left(x+y\right)^2=5^2\)
\(\Leftrightarrow\)\(x^2+2xy+y^2=25\)
\(\Leftrightarrow\)\(2xy=25-\left(x^2+y^2\right)\)
\(\Leftrightarrow\)\(2xy=25-21\)
\(\Leftrightarrow\)\(2xy=4\)
\(\Leftrightarrow\)\(8xy=16\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(A=105-8xy\) ta được :
\(A=105-16\)
\(A=89\)
Vậy \(A=89\)
Chúc bạn học tốt ~
x^3-2x-4
=x^3-4x+2x-4
=x(x^2-4)+2(x-2)
=x(x-2)(x+2)+2(x-2)
=(x^2+2x+2)(x-2)
Nhớ cho mình nha!!!
Dễ thôi
\(x^3-x^2-4\)
\(=\left(x^3-8\right)-\left(x^2-4\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)