Viết các đa thức sau thành tích:
\(16-x^2\)
\(4x^2-9\)
\(a^4-25\)
\(\left(a+b\right)^{2-1}\)
\(\left(a+b\right)^2-\left(m-n\right)^2\)
\(x^3-27\)
\(64x^3+\dfrac{1}{27}\)
Viết các đa thức sau thành tích:
\(16-x^2\)
\(4x^2-9\)
\(a^4-25\)
\(\left(a+b\right)^2-1\)
\(\left(a+b\right)^2-\left(m-n\right)^2\)
\(x^3-27\)
\(64x^3+\frac{1}{27}\)
\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)
\(\left(a+b\right)^2-1=\left(a+b\right)^2-1^2=\left(a+b-1\right)\left(a+b-1\right)\)
\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b-m+n\right)\left(a+b+m-n\right)\)
\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+3^2\right)\)
\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2+\frac{4}{3}x+\frac{1}{9}\right)\)
Tham khảo~
\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)
\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2+5\right)\left(a^2-5\right)\)
\(\left(a+b\right)^2-1=\left(a+b+1\right)\left(a+b-1\right)\)
\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b+m-n\right)\left(a+b-m+n\right)\)
\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)
\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2-\frac{4}{3}x+\frac{1}{9}\right)\)
Tìm điều kiện của x để phân thức sau xác định;
a)\(\dfrac{\dfrac{1}{x-4}}{2x+2}\)
b)\(\dfrac{x^3+2x}{4x^2-25}\)
c)\(\dfrac{2x^2+2x}{8x^3+27}\)
d)\(\dfrac{2x+1}{\left(2x+2\right)\left(4y^2-9\right)}\)
`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
Phân tích đa thức thành nhân tử:
23). \(\left(2a-b\right)^2-4\left(a-b\right)^2\)
24). \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)
25). \(4\left(2a-b\right)^2-16\left(a-b\right)^2\)
26). \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)
27). \(\left(x^2+1\right)^2-4x^2\)
28). \(16x^2-\left(x^2+4\right)\)
29). \(\left(4x^2+\dfrac{1}{4}\right)-4x^2\)
30). \(\left(4x^2+1\right)^2-16x^2\)
23: \(=\left(2a-b\right)^2-\left(2a-2b\right)^2\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
24: \(=\left(3a+3b\right)^2-\left(2a-4b\right)^2\)
\(=\left(3a+3b-2a+4b\right)\left(3a+3b+2a-4b\right)\)
\(=\left(a+7b\right)\left(5a-b\right)\)
25: \(=\left(4a-2b\right)^2-\left(4a-4b\right)^2\)
\(=\left(4a-2b-4a+4b\right)\left(4a-2b+4a-4b\right)\)
\(=2b\left(8a-6b\right)\)
=4b(4a-3b)
Viết các biểu thức sau dưới dạng tích :
a. \(9.x^2+30x+25\)
b. \(\dfrac{4}{9}.x^{^{ }4}-16x^2\)
c. \(a^2y^2+b^2x^2-2axby\)
d. \(100-\left(3x-y\right)^2\)
e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4\)
f.\(64x^2-\left(8a+b\right)^2\)
g.\(27x^3-a^3b^3\)
a. \(9x^2+30x+25=\left(3x+5\right)^2\)
b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)
c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)
d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)
e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)
f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)
bài 1: phân tích đa thức thành nhân tử
a,2x+10y
b,x\(^2+4x+4\)
c,\(x^2-y^2+10y-25\)
bài 2 tìm x, biết
a,\(x^2-3x+x-3=0\)
b,\(2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\)
c,\(x^2-\left(x-3\right)\left(2x-5\right)=9\)
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
Cho 2 đa thức: \(N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\)
\(M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\)
a) Tính \(N\left(x\right)-M\left(x\right)\)
b) Tính \(M\left(x\right)+N\left(x\right)\)
a)
\(\begin{matrix}N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\^-M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\\overline{N\left(x\right)-M\left(x\right)=-3x^4+18x^3-2x^2-4x-1}\end{matrix}\)
b)
\(\begin{matrix}M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\^+N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\\overline{M\left(x\right)+N\left(x\right)=-5x^4+14x+\dfrac{5}{3}}\end{matrix}\)
\(A=\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right):\left(-4\dfrac{1}{6}+3\dfrac{1}{7}\right)+7\dfrac{1}{2}\)
\(B=4\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}:\dfrac{125}{64}\right):\left(-\dfrac{27}{8}\right)\)
giải hộ mk nhanh nhanh nhoa ☺
Tìm nghiệm của các đa thức sau:
\(A=-\dfrac{2}{3}x-\dfrac{5}{9}\) \(B=x^2-\dfrac{4}{25}\) \(C=\dfrac{1}{2}x^3+\dfrac{4}{27}\) \(D=x^3-\dfrac{1}{8}x\)
\(E=-\dfrac{16}{81}+x^4\) \(F=-x.\left(-2x+3\right).\left(1-x^3\right)\) \(I=2.\left(2-x\right)+\dfrac{1}{2}\left(x-2\right)^2\) \(G=x^{100}-8.x^{97}\)
a: Đặt A=0
=>-2/3x=5/9
hay x=-5/6
b: Đặt B(x)=0
=>(x-2/5)(x+2/5)=0
=>x=2/5 hoặc x=-2/5
c: Đặt C(X)=0
\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)
\(\Leftrightarrow x^3=-\dfrac{8}{27}\)
hay x=-2/3
Phân tích đa thức \(18x^3-\dfrac{8}{25}x\) thành nhân tử
a. \(\dfrac{2}{25}x\left(9x^2-4\right)=\dfrac{2}{25}x\left(3x-2\right)\left(3x+2\right)\)
b. \(2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
Cách phân tích nào đúng, a hay b. Giải thích vì sao?