\(\text{Tìm x biết:}\)
\(\text{1+3+5+7+...+x=25}\)
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24 tháng 6 2021
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
24 tháng 6 2021
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
H9
HT.Phong (9A5)
CTVHS
10 tháng 6 2023
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
XO
27 tháng 12 2021
a) ĐKXĐ : \(3\le x\le7\)
Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)
\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)
Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)
LT
0
E
1
1 tháng 5 2021
Ta có: \(20\%x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-4}{5}x=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{15}{20}-\dfrac{4}{20}=\dfrac{11}{20}\)
\(\Leftrightarrow x=\dfrac{11}{20}:\dfrac{-4}{5}=\dfrac{11}{20}\cdot\dfrac{5}{-4}=\dfrac{-55}{80}=\dfrac{-11}{16}\)
Vậy: \(x=-\dfrac{11}{16}\)
18 tháng 12 2021
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
Ta có: 25 = 1 + 3 + 5 + 7 + 9
--> x = 9
Trả lời:
x = 9
~HT~