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Mink nghĩ đề này là phân tích đa thức thành nhân tử chứ k phải tìm x^^
a) \(x^2-x-56=x^2-8x+7x-56=x\left(x-8\right)+7\left(x-8\right)=\left(x+7\right)\left(x-8\right)\)
b) \(4x^4+1=\left(4x^4+4x^2+1\right)-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(\left(2x^2+1-2x\right)\left(2x^2+1+2x\right)\)
c) \(5x^2-x-4=5x^2-5x+4x-4=5x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(5x+4\right)\)
d) \(4x^4+81=\left(4x^4+36x^2+81\right)-36x^2=\left(2x^2+9\right)^2-\left(6x\right)^2\)
\(=\left(2x^2+9+6x\right)\left(2x^2+9-6x\right)\)
e) \(64x^4+y^4=\left(64x^4+16x^2y^2+y^4\right)-\left(4xy\right)^2=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
a)\(x^2-x-56\)
\(=x^2+7x-8x-56\)
\(=x\left(x+7\right)-8\left(x+7\right)\)
\(=\left(x-8\right)\left(x+7\right)\)
b)\(4x^4+1\)
\(=\left(2x+1\right)^2-4x^2\)
\(=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
c)\(5x^2-x-4\)
\(=5x^2+4x-5x-4\)
\(=x\left(5x+4\right)-\left(5x+4\right)\)
\(=\left(x-1\right)\left(5x+4\right)\)
d)\(4x^4+81\)
\(=\left(2x^2\right)^2+9^2+36x^2-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
e)\(64x^4+y^4\)
\(=\left(8x^2\right)^2+y^4+16x^2y^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
a) = \(4x^4+4x^2+1\)
= \(\left(2x^2+1\right)^2\)
b) = \(4x^4+36x^2+81-36x^2\)
= \(\left(2x^2+9\right)^2\)
c) = \(64x^4+16x^2y^2+y^4-16x^2y^2\)
= \(\left(8x^2+y^2\right)^2\)
d) = \(x^8+4x^4+4-4x^4\)
= \(\left(x^4+2\right)^2\)
e) = \(\left(x^4+2x^2+1\right)-x^2\)
= \(\left(x^2+1\right)^2-x^2\)
= \(\left(x^2+1-x\right).\left(x^2+1+x\right)\)
f) = \(\left(x^7-x\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)
= \(x.\left(x^3-1\right).\left(x^3+1\right)+x^2.\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(\left(x^2+x+1\right).\left(x-1\right).\left(x^4+x\right)+x^2.\left(x-1\right).\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right).\left(x^5-x^4+x^3-1+1\right)\)
c/=64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2+4xy)(8x^2+y^2-4xy)
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
ĐKXĐ bạn tự tìm nha : )
k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)
\(=\frac{3x\left(1-2x\right)\left(1+2x\right)}{2x\left(x+4\right)\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)
j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)
\(=\frac{3\left(x-y\right)\left(x+y\right)}{x\left(y-x\right)}=\frac{3\left(x-y\right)\left(x+y\right)}{-x\left(x-y\right)}=\frac{-3\left(x+y\right)}{x}\)
i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)
\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)
h, = k,
f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
1. D = 3( x2 - 2x.1/3 + 1/9) -1/3 +1
GTNN D = 5/6
dài quá, nản quá
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)