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Bài 2:
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)
\(=25x^2+10x+1-\left(2xy-3\right)^2\)
\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)
\(=25x^2+10x+1-4x^2y^2+12xy-9\)
\(=25x^2-4x^2y^2+10x+12xy-8\)
Bài 2:
\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)
\(=x^3-1=x^3-9x^2+2x+6\)
\(=x^3-9x^2+2x+6=x^3-1\)
\(=x^3-9x^2+2x+6+1=x^3-1+1\)
\(=x^3-9x^2+2x+7=x^3\)
\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)
\(=-9x^2+2x+7=0\)
\(\Rightarrow x=-\frac{7}{9};x=1\)
a, - \(\dfrac{1}{3}\).\(xy\).(3\(x^3\).y2 - 6\(x^2\) + y2)
= - \(x^4\).y3 + 2\(x^3\).y - \(\dfrac{1}{3}\).\(xy^3\)
b, (2\(x\) -3).(4\(x\)2 + 6\(x\) + 9)
= (2\(x\))3 - 33
= 8\(x^3\) - 27
Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=x+2-x+5\)
\(\Leftrightarrow18x-2=7\)
\(\Leftrightarrow18x=9\)
hay \(x=\dfrac{1}{2}\)
Giải tiêu biểu câu a nhé.
a/ \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Leftrightarrow19x+5=0\)
\(\Leftrightarrow x=-\frac{5}{19}\)
\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
\(\Leftrightarrow\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}=\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)
\(\Leftrightarrow6x+30-16x+24=18x-3+4x-2\)
\(\Leftrightarrow6x-16x-18x-4x=-2-3-24-30\)
\(\Leftrightarrow-32x=-59\)
\(\Leftrightarrow x=\frac{59}{32}\)