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Ta có: \(x^2=20x-100\)
\(\Leftrightarrow x^2-20x+100=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}+\frac{4032-3ac}{3ac-4032+2016a}\)
\(=\frac{2bc-abc}{3c-2bc+abc}-\frac{2b}{3-2b+ab}+\frac{2abc-3ac}{3ac-2abc+a^2bc}\)
\(=\frac{c\left(2b-ab\right)}{c\left(3-2b+ab\right)}-\frac{2b}{3-2b+ab}+\frac{ac\left(2b-3\right)}{ac\left(3-2b+ab\right)}\)
\(=\frac{2b-ab}{3-2b+ab}-\frac{2b}{3-2b+ab}+\frac{2b-3}{3-2b+ab}\)
\(=\frac{2b-ab-2b+2b-3}{3-2b+ab}=\frac{2b-ab-3}{-\left(2b-ab-3\right)}=-1\)
a: Ta có: \(\left(x-3\right)^2-x\left(x+5\right)=9\)
\(\Leftrightarrow x^2-6x+9-x^2-5x=9\)
\(\Leftrightarrow x=0\)
b: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
\(=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right).\left(x-y+3\right)\)
a. 6x2 - (2x + 5)(3x - 2) = 7
<=> 6x2 - 6x2 + 4x - 15x + 10 = 7
<=> -11x = -3
<=> \(x=\dfrac{3}{11}\)
b. (5 - x)(25 + 5x + x2) + x(x2 - 7) = 25
<=> 125 - x3 + x3 - 7x = 25
<=> -7x = 25 - 125
<=> -7x = -100
<=> \(x=\dfrac{100}{7}\)
c. (7 - 2x)2 + (3 + 2x)(3 - 2x) = 30
<=> 49 - 28x + 4x2 + 9 - 4x2 = 30
<=> 4x2 - 4x2 - 28x = 30 - 49 - 9
<=> -28x = -28
<=> x = 1
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
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