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\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2\right)^2+2.5xy^2.3z+\left(3z\right)^2=\left(5xy^2+3z\right)^2\)
\(\frac{16}{9}x^2+4xyz^2+\frac{9}{4}y^2z^4=\left(\frac{4}{3}x\right)^2+2.\frac{4}{3}x.\frac{3}{2}yz^2+\left(\frac{3}{2}yz^2\right)^2=\left(\frac{4}{3}x+\frac{3}{2}yz^2\right)^2\)
\(\frac{9}{25}x^2+\frac{12}{35}xy+\frac{4}{49}y^2=\left(\frac{3}{5}x\right)^2+2.\frac{3}{5}x.\frac{2}{7}y+\left(\frac{2}{7}y\right)^2=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2\)( tự thay vào tính nhé )
\(\frac{25}{16}u^4y^2+\frac{1}{5}u^2+y^3+\frac{4}{625}y^4=\left(\frac{5}{4}u^2y\right)^2+2.\frac{5}{4}u^2y.\frac{2}{25}.y^2+\left(\frac{2}{25}y^2\right)^2=\left(\frac{5}{4}u^2y+\frac{2}{25}y^2\right)^2\)( tự thay vào tính nhé )
Tham khảo nhé~
a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)
\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)
a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
a: \(9x^2+30x+25=\left(3x+5\right)^2\)
b: \(\dfrac{4}{9}x^4-16x^2=x^2\left(\dfrac{4}{9}x^2-16\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)
c: \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4\)
\(=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)\)
\(=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
1)\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2+3z\right)^2\)
\(\dfrac{16}{9}x^2+4xyz^2+\dfrac{9}{4}y^2z^4=\left(\dfrac{4}{3}x+\dfrac{3}{2}yz^2\right)^2\)
2)
a)\(\dfrac{9}{25}x^2+\dfrac{12}{35}xy+\dfrac{4}{49}y^2=\left(\dfrac{3}{5}x+\dfrac{2}{7}y\right)^2=\left(\dfrac{3}{5}.5+\dfrac{2}{7}.\left(-7\right)\right)^2=\left(3-2\right)^2=1\)b)\(\dfrac{25}{16}u^4v^2+\dfrac{1}{5}u^2v^3+\dfrac{4}{625}v^4\)
\(=\left(\dfrac{5}{4}u^2v+\dfrac{2}{25}v^2\right)^2=\left(\dfrac{5}{4}.\dfrac{4}{25}.\left(-5\right)+\dfrac{2}{25}.\left(-5\right)^2\right)^2\)
\(=\left(-1+2\right)^2=1\)