tìm số tự nhiên x
\(\frac{x-2}{255}=\frac{144}{153}\)
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Ta có:
\(\frac{x-2}{255}=\frac{114}{153}\)
\(\Rightarrow\left(x-2\right).153=114.255\)
\(\Rightarrow\left(x-2\right).153=29070\)
\(\Rightarrow x-2=29070:153\)
\(\Rightarrow x-2=190\)
\(\Rightarrow x=190+2\)
\(\Rightarrow x=192\)
Vậy x = 192
\(\frac{x-2}{255}=\frac{114}{153}\)
\(\Rightarrow\left(x-2\right).153=255.114\)
\(\Rightarrow\left(x-2\right).153=29070\)
\(\Rightarrow x-2=29070:153\)
\(\Rightarrow x-2=190\)
\(\Rightarrow x=190+2=192\)
Vậy \(x=192\)
k mk nhé
a) \(\frac{x-2}{255}=\frac{114}{153}\)
\(\Leftrightarrow\left(x-2\right)\cdot153=114\cdot255\)
\(\Leftrightarrow\left(x-2\right)\cdot153=29070\)
\(\Leftrightarrow x-2=190\)
\(\Leftrightarrow x=192\)
b) \(\frac{50}{19}\cdot\frac{38}{25}\le x\le\frac{69}{17}+\frac{33}{17}\)
\(\Leftrightarrow4\le x\le6\)
Vậy x =4 hoặc x=5 hoặc x=6
\(x-\frac{2}{255}=\frac{144}{153}\)
\(x=\frac{144}{153}+\frac{2}{255}=\frac{252}{255}\)
\(x-\frac{2}{225}=\frac{144}{153}\)
\(x=\frac{144}{153}+\frac{2}{255}\)
\(x=\frac{242}{255}\)
\(x-\frac{2}{255}=\frac{144}{153}\)
\(x=\frac{144}{153}+\frac{2}{255}\)
\(x=\frac{242}{255}\)
nếu đúng thì tích mk nha
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(P=\left(\frac{3x-\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right)\div\frac{1}{x-1}\)
\(P=\frac{3x-3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(x-1\right)\)
\(P=\frac{3x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)
\(P=\frac{\left(3\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
a: \(35=5\cdot7;105=3\cdot5\cdot7\)
=>\(ƯCLN\left(35;105\right)=5\cdot7=35\)
\(35⋮x;105⋮x\)
=>\(x\inƯC\left(105;35\right)\)
=>\(x\inƯ\left(35\right)\)
=>\(x\in\left\{1;5;7;35\right\}\)
mà x>5
nên \(x\in\left\{7;35\right\}\)
b: \(144=2^4\cdot3^2;192=2^6\cdot3;240=2^4\cdot3\cdot5\)
=>\(ƯCLN\left(144;192;240\right)=2^4\cdot3=48\)
\(144⋮x;192⋮x;240⋮x\)
=>\(x\inƯC\left(192;144;240\right)\)
=>\(x\inƯ\left(48\right)\)
=>\(x\in\left\{1;2;3;4;6;8;12;16;24;48\right\}\)
mà 10<=x<=99
nên \(x\in\left\{12;16;24;48\right\}\)
\(dk:x\ne\left\{1,\sqrt{2},4\right\};x\ge0\)dat \(\sqrt{x}=t\)
\(A=\left(\frac{3t^2}{t^2-t-2}+\frac{1}{t-1}+\frac{1}{t-2}\right)\left(t^2-1\right)==\left(\frac{3t^2}{\left(t-2\right)\left(t-1\right)}+\frac{1}{t-1}+\frac{1}{t-2}\right)\left(t^2-1\right)\)
\(=\left(\frac{3t^2}{\left(t-2\right)\left(t-1\right)}+\frac{t-2}{t-1}+\frac{t-1}{t-2}\right)\left(t-1\right)\left(t+1\right)=3t^2+2t-3\)
\(A=3x+2\sqrt{x}-3\)
b
\(\frac{1}{A}=\frac{1}{3x+2\sqrt{x}-3}\Rightarrow\orbr{\begin{cases}3x+2\sqrt{x}-3=-1\\3x+2\sqrt{x}-3=1\end{cases}}\)tư làm tiếp
ta có: 2/x=1x/3x =>2.3x=1x.x Hay:6x=x^2 =>6x-x^2=0 =>x(6-x)=0
Với x=0 =>Loại vì x là mẫu số =>x khác 0
Với 6-x=0 =>x=6 T/M
vậy x cần tìm là 6
=> (x-2)x 153= 144x 255
<=> (x-2). 153= 36720
<=> x-2 = 240
<=> x= 242
\(Vi`\frac{x-2}{255}=\frac{144}{153}\)
\(\Rightarrow\left(x-2\right)153=144.255\)
\(\Rightarrow\left(x-2\right)153=36720\)
\(\Rightarrow\left(x-2\right)=240\)
\(\Rightarrow x=242\)