a,Tìm x B, Tính giá trị của biểu thức
36- x = 8,6 ÷4 (4,35-1,48 )× 1,3
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a,`7,2-x:6,5=5,7`
`x:6,5=1,5`
`x=9,75`
`232,8-x=2,56xx4,5`
`232,8-x=11,52`
`x=221,28`
`x-93,1=72,81:18`
`x-93,1=4,045`
`x=97,145`
b, `8,6-8,33:3,4=8,6-2,45=6,15`
`9,52:6,8+3,86=1,4+3,86=5,26`
`(25,46-56,42:4)+16,5xx23=(25,46-14,105)+379,5=11,355+379,5=390,855`
1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)
2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >4\end{matrix}\right.\)
\(C=B:A\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}-x}{x-4}\right):\dfrac{\sqrt{x}}{2\sqrt{x}-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+3\sqrt{x}-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}-x}{\sqrt{x}+2}\cdot\dfrac{2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)
3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)
=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{3}< 0\)
=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)
=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)
=>\(2\sqrt{x}-8< 0\)
=>\(\sqrt{x}< 4\)
=>\(0< =x< 16\)
Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)
\(A=\frac{12\cdot25+4\cdot175\cdot3}{36\cdot\left(101-1\right)+24\cdot\left(99+1\right)}\)
\(A=\frac{12\cdot25+12\cdot175}{36\cdot100+24\cdot100}\)
\(A=\frac{12\cdot100}{60\cdot100}\)
\(A=\frac{1}{5}\)
Bấm L IKE ủng hộ nhá :)))
a)
A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)
\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
MTC: 5(x-1)(x+1)
\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)
\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)
\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)
\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)
\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)
\(\Leftrightarrow10x+10\)
36 - \(x\) = 8,6: 4
36 - \(x\) = 2,15
\(x\) = 36 - 2,15
\(x\) = 33,85
( 4,35 - 1,48) \(\times\) 1,3
= 2,87 \(\times\) 1,3
= 3,731