Giup mk voi
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\(\frac{7}{12}-0,75:\left(4x-3\right)^2=\frac{-3}{2}\)
\(\frac{3}{4}:\left(4x-3\right)^2=\frac{25}{12}\)
\(\left(4x-3\right)^2=\frac{9}{25}=\left(\frac{\pm3}{5}\right)^2\)
+) 4x - 3 = 3/5
4x = 18/5
x = 9/10
+) 4x - 3 = -3/5
4x = 12/5
x = 3/5
Vậy,.........
\(\frac{7}{12}-0,75:\left(4x-3\right)^2=-\frac{3}{2}\)
\(\Leftrightarrow\frac{7}{12}.12\left(4x-3\right)^2-\frac{0,75}{\left(4x-3\right)^2}.12\left(4x-3\right)^2=-\frac{3}{2}.12\left(4x-3\right)^2\)
\(\Leftrightarrow7\left(4x-3\right)^2-9=-18\left(4x-3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{10}\\x=\frac{3}{5}\end{cases}}\)
C2: = a2+2a+a+1=(a+1)2+a
\(A=3+3^2+3^3+3^4+...+3^{100}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{101}\)
\(3A-A=3^{101}-1\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
Vì \(\frac{3^{101}-1}{2}< 3^{101}-1\)
\(\Rightarrow A< B\)
\(\Rightarrow dpcm\)
\(A=1+4+4^2+4^3+...+4^{99}\)
\(4A=4+4^2+4^3+4^4+...+4^{100}\)
\(3A=4^{100}-1\)
\(A=\frac{4^{100}-1}{3}\)
\(B=4^{100}\)
\(\Rightarrow\frac{B}{3}=\frac{4^{100}}{3}\)
Vì \(\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)
\(\Rightarrow A< \frac{B}{3}\)
\(\Rightarrow dpcm\)