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\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)
Bài 2:
a, 1/3 + 1/2 : x = -4
=> 1/2 : x = -4 - 1/3
=> 1/2 : x = -13/3
=> x = 1/2 ; -13/3
=> x = -3/26
Vậy x = -3 / 26
Bài 2:
b, x2 - 4x = 0
=> x.(x - 4) =0
=> x=0 hoặc x - 4 = 0
x - 4= 0 => x=4
Vậy x=0 và x=4
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)
\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)
\(8A=1-3=-2\)
A=\(\frac{-2}{8}=\frac{-1}{4}\)
\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)
Vậy B=1
Bài giải
Ta có : \(B=-\frac{1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }B=-\frac{1}{3^0}-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(B=-1-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
Đặt \(C=\frac{1}{3^1}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }3C=1+\frac{1}{3^1}+...+\frac{1}{3^{99}}\)
\(\Rightarrow\text{ }3C-C=2C=1-\frac{1}{3^{100}}\)
\(C=\frac{1-\frac{1}{3^{100}}}{2}=\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
Thay vào biểu thức B ta được :
\(B=-1-\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=-\frac{3}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=\frac{\left(-3\right)^{101}}{2\cdot3^{100}}-\frac{1}{2\cdot3^{100}}=\frac{\left(-3\right)^{101}-1}{2\cdot3^{100}}\)
Bài giải
Ta có : \(B=-\frac{1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }B=-\frac{1}{3^0}-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(B=-1-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
Đặt \(C=\frac{1}{3^1}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }3C=1+\frac{1}{3^1}+...+\frac{1}{3^{99}}\)
\(\Rightarrow\text{ }3C-C=2C=1-\frac{1}{3^{100}}\)
\(C=\frac{1-\frac{1}{3^{100}}}{2}=\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
Thay vào biểu thức B ta được :
\(B=-1-\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=-\frac{3}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=\frac{\left(-3\right)^{101}}{2\cdot3^{100}}-\frac{1}{2\cdot3^{100}}=\frac{\left(-3\right)^{101}-1}{2\cdot3^{100}}\)