\(25\%\cdot x-0,4\cdot x=x+1\)
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DM
2
29 tháng 7 2017
\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{9x^2-6x+1}+\sqrt{9x^2-30x+25}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}\)
\(=\left|3x-1\right|+\left|3x-5\right|\)
\(=\left|3x-1\right|+\left|5-3x\right|\)
\(\ge\left|3x-1+5-3x\right|=4\)
Xảy ra khi \(\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
29 tháng 7 2017
\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}\)
\(=\left|3x-1\right|+\left|3x-5\right|=\left|3x-1\right|+\left|5-3x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A\ge\left|3x-1+5-3x\right|=\left|4\right|=4\)
Dấu " = " khi \(\left\{{}\begin{matrix}3x-1\ge0\\5-3x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(MIN_A=4\) khi \(\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
PN
13 tháng 7 2017
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
2 tháng 9 2017
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
24 tháng 5 2018
1.\(\left(x-5\right).\left(x+5\right)-\left(x+3\right)^2=2x-3\)
\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)
\(\Leftrightarrow-8x-31=0\)
\(\Leftrightarrow x=\dfrac{-31}{8}\)
24 tháng 5 2018
\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-\left(x^3-5^3\right)=\left(x^3+2^3\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-x^3+5^3=x^3+2^3-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)
\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)
\(\Leftrightarrow96x=-117\)
\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)
KK
1 tháng 9 2019
\(25\%.y+50\%.y-\frac{3}{4}.y+4.y=10\)
\(y.\left(\frac{1}{4}+\frac{1}{2}-\frac{3}{4}+4\right)=10\)
\(y.4=10\)
\(y=\frac{5}{2}\)
\(x.\frac{1}{4}-\frac{3}{4}=6:\frac{3}{4}\)
\(x.\frac{1}{4}-\frac{3}{4}=6.\frac{4}{3}\)
\(x.\frac{1}{4}=8+\frac{3}{4}\)
\(x.\frac{1}{4}=\frac{35}{4}\)
\(x=\frac{35}{4}:\frac{1}{4}\)
\(x=35\)
AN
1 tháng 9 2019
25% x y + 50% x y - 3/4 x y + 4 x y = 10
1/4 x y + 1/2 x y - 3/4 x y + 4 x y = 10
y x ( 1/4 + 1/2 - 3/4 + 4 ) = 10
y x 4 = 10
y = 10 : 4
y = 2.5
2 tháng 5 2017
a) \(25\%x+x=-1,25\\ < =>\dfrac{125}{100}x=-1,25\\ =>x=\dfrac{-1,25}{125}=-0,01\)
b) \(75\%x+x:\dfrac{4}{5}=-2010\\ < =>\dfrac{75}{100}x+\dfrac{5}{4}x=-2010\\ < =>\dfrac{75}{100}x+\dfrac{125}{100}x=-2010\\ < =>2x=-2010\\ =>x=\dfrac{-2010}{2}=-1005\)
c) \(25\%x+0,5x-\dfrac{x}{3}=-26\\ < =>\dfrac{1}{4}x+\dfrac{1}{2}x-\dfrac{1}{3}x=-26\\ < =>\dfrac{5}{12}x=-26\\ =>x=\dfrac{-26}{\dfrac{5}{12}}=-62\dfrac{2}{5}\)
`25% . x - 0,4 . x = x + 1`
`1 / 4 . x - 2 / 5 . x = x + 1`
`1 / 4 . x - 2 / 5 . x - x = 1`
`( 1 / 4 - 2 / 5 - 1 ) . x = 1`
`[-23] / 20 . x = 1`
`x = 1 : [-23] / 20`
`x = [-20] / 23`
\(25\%\cdot x-0,4\cdot x=x+1\)
\(\dfrac{1}{4}\cdot x-\dfrac{2}{5}\cdot x=x+1\)
\(\left(\dfrac{1}{4}-\dfrac{2}{5}\right)x=x+1\)
\(-\dfrac{3}{20}x=x+1\)
\(-\dfrac{3}{20}x+\left(-\dfrac{20}{20}\right)x=1\)
[ \(-\dfrac{3}{20}+\left(-\dfrac{20}{20}\right)\text{ ]}x=1\)
\(-\dfrac{23}{20}x=1\)
\(x=1:\left(-\dfrac{23}{20}\right)\)
\(x=-\dfrac{20}{23}\)