5x3^11+4x3^12 /3^9x5^2-3^9x2^2
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a, 5 x 4 + 9 x 5 - 5 x 3 - 10
= 5 x ( 4 + 9 - 3) - 10
= 5 x ( 13 - 3) - 10
= 5 x 10 - 10
= 40
b, 4 x 7 + 4 x 6 - 4 x 3 - 4
= 4 x 7 + 4 x 6 - 4 x 3 - 4 x1
= 4 x ( 7 + 6 - 3 -1)
= 4 x 9
= 36
P = \(\frac{5.3^{11}+4.3^{12}}{3^9.5^2-3^9.2^2}\)
= \(\frac{\left(5+4.3\right).3^{11}}{3^9.\left(5^2-2^2\right)}\)
=\(\frac{17.3^{11}}{3^9.21}\)
= \(\frac{17.3^2}{7.3}\)
= \(\frac{17.3}{7}=\frac{51}{7}\)
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
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