(-3).( x+1)= 36
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\(B=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=9.\left(\frac{1}{3}-\frac{1}{675}\right)\)
\(=9.\frac{224}{675}\)
\(=\frac{224}{75}\)
a) \(\left(x^2-3\right)\left(x^2-36\right)=0\)
TH1: \(x^2-3=0\Rightarrow x^2=3\)
Ta thấy không có số nguyên nào mà bình phương nên bằng 3 nên không có giá trị x thỏa mãn.
TH2: \(x^2-36=0\Rightarrow x^2=36=6.6=\left(-6\right).\left(-6\right)\)
Vậy x = 6 hoặc x = -6.
b) \(\left(x^2-3\right)\left(x^2-36\right)< 0\)
Do \(x^2-3>x^2-36\) nên chỉ có thể xảy ra trường hợp \(\hept{\begin{cases}x^2-3>0\\x^2-36< 0\end{cases}}\)
\(\Rightarrow3\le x^2\le36\Rightarrow2\le x\le6\) hoặc \(-6\le x\le-2\)
a.219 - 7(x+1) = 100
7(x+1) = 219 - 100
7(x+1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
b. (3x - 6 ) . 3 = 36
3x - 6 = 36 : 3
3x - 6 = 12
3x = 12 + 6
3x = 18
x = 18 : 3
x = 6
c.716 - ( x-143) = 659
x-143 = 716 - 659
x-143 = 57
x = 57 + 143
x = 200
b. 30 - [4(x-2)+15] = 3
4(x-2) + 15 = 30 - 3
4(x-2)+15 = 27
4(x-2) = 27 - 15
4(x-2) = 12
x-2 = 12 : 4
x-2 = 3
x = 2 + 3 = 5
e.[(8x - 12) : 4] .33 = 36
[(8x - 12) : 4] . 27 = 729
(8x - 12) : 4 = 729 : 27 = 27
8x - 12 = 27 . 4 = 108
8x = 108 + 12 = 120
x = 120 : 8 = 15
a) \(\Leftrightarrow7\left(x+1\right)=119\\ \Leftrightarrow x+1=17\\ \Leftrightarrow x=16\)
b) \(\Leftrightarrow9\left(x-2\right)=36\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\)
c) \(\Leftrightarrow x-143=57\\ \Leftrightarrow x=200\)
d) \(\Leftrightarrow4\left(x-2\right)+15=27\\ \Leftrightarrow4\left(x-2\right)=12\\ \Leftrightarrow x-2=3\\ \Leftrightarrow x=5\)
e) \(\Leftrightarrow\left(2x-3\right).4:4=3^3\\ \Leftrightarrow2x-3=27\\ \Leftrightarrow2x=24\\ \Leftrightarrow x=12\)
Rút gọn.
\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)
ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)
\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)
=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)
Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
Ta có:
\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)
\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)
\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)
\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)
tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
a, \(\frac{2}{x}=\frac{y}{4}\Leftrightarrow xy=8\)
Suy ra : \(x;y\inƯ\left(8\right)=\left\{1;2;4;8\right\}\)tự lập bảng
b, Xét : \(\frac{1}{x}=\frac{5}{15}\Leftrightarrow\frac{5}{5x}=\frac{5}{15}\Leftrightarrow x=3\)
\(\frac{y}{21}=\frac{5}{15}\Leftrightarrow15y=105\Leftrightarrow y=3\)
\(\frac{10}{z}=\frac{5}{15}\Leftrightarrow5z=150\Leftrightarrow z=30\)
c, tương tự b
\(\left(-3\right).\left(x+1\right)=36\)
\(-3x-3=36\)
\(-3x=36+3\)
\(-3x=39\)
\(-x=39:3\)
\(x=-13\)