Cho biết x ∈ N và (327.x + 2 4).7 3 = 7.7 5
Ta có x = ?
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tc tỉ lệ thức,ta có (ra số hơi to thông cảm )
\(\Leftrightarrow\frac{22990993x+80556557}{1870865100}=-\frac{4}{1}\Rightarrow\left(22990993x+80556557\right)1=-1870865100.4\)
\(\frac{\left(22990993x+80556557\right)1}{22990933x}=-\frac{1870865100.4}{22990993x}\)( chia 2 vế cho 2290933x)
\(\Rightarrow\frac{22990933x+80556557}{22990933x}=-\frac{1870865100.4}{22990933x}\)
=>x=-329
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}=-4\)
<=>\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)=0\)
<=>\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)
<=>\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)
Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\Rightarrow x+329=0\Leftrightarrow x=-329\)
Sửa đề: Tìm x \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
Bài giải
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\Leftrightarrow\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+329+20}{5}=0\)
\(\Leftrightarrow\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+329}{5}+\dfrac{20}{5}=0\)
\(\Leftrightarrow\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+329}{5}+4=0\)
\(\Leftrightarrow\dfrac{x+2}{327}+1+\dfrac{x+3}{326}+1+\dfrac{x+4}{325}+1+\dfrac{x+5}{324}+1+\dfrac{x+329}{5}=0\)
\(\Leftrightarrow\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
\(\Rightarrow x+329=0\). Do \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\ne0\)
\(\Rightarrow x=-329\)
Ta có: \(\left(327x+2^4\right)\cdot7^3=7\cdot7^5\)
\(\Leftrightarrow\left(327x+16\right)\cdot7^3=7^6\)
\(\Leftrightarrow327x+16=7^3\)
\(\Leftrightarrow327x+16=343\)
\(\Leftrightarrow327x=327\)
\(\Rightarrow x=1\)
(327.x + 24).73 = 7.75
=> (327.x + 24) = \(\frac{7\cdot7^5}{7^3}=7^3\)
=> 327.x + 24 = 73
=> 327.x = 73 - 24
=> 327.x = 327
=> x = 1