|x-7|+|x-10|=3
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a) \(\dfrac{7}{5}x\div\left(9-6\dfrac{13}{21}\right)=2\dfrac{13}{25}\)
\(\dfrac{7}{5}x\div\left(9-\dfrac{139}{21}\right)=\dfrac{63}{25}\)
\(\dfrac{7}{5}x\div\dfrac{50}{21}=\dfrac{63}{25}\)
\(\dfrac{7}{5}x=6\)
\(x=\dfrac{30}{7}\)
b) \(\left(1,16-x\right)\times5,25\div\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right)\times2\dfrac{2}{17}=75\%\)
\(\left(1,16-x\right)\times5,25\div\left(\dfrac{95}{9}-\dfrac{29}{4}\right)\times\dfrac{36}{17}=\dfrac{75}{100}\)
\(\left(1,16-x\right)\times5,25\div\dfrac{119}{36}=\dfrac{17}{48}\)
\(\left(1,16-x\right)\times5,25=\dfrac{2023}{1728}\)
\(1,16-x=\dfrac{289}{1296}\)
\(x=0,9370061728\)
52 ×.(y : 78) = 3380
y : 78 = 3380 : 52
y : 78 = 65
y = 65 ×. 78
y = 5070
\(\frac{5x+7}{4}+\frac{3x+5}{8}>\frac{9x+4}{5}\)
\(\frac{10\cdot\left(5x+7\right)}{40}+\frac{5\cdot\left(3x+5\right)}{40}>\frac{8\cdot\left(9x+4\right)}{40}\)
10.(5x + 7) + 5.(3x + 5) > 8.(9x + 4)
10.(5x + 7) + 5.(3x + 5) - 8.(9x + 4) > 0
50x + 70 + 15x + 25 - 72x - 32 > 0
- 7x + 63 > 0
- 7.(x - 9) > 0
\(\Rightarrow x-9<0\Rightarrow x<9\)
<=>x-2+5 chia hết cho x-2
<=>5 chia hết cho x-2
<=>x-2 thuộc {1;5}
<=>x thuộc {3;7}
Bài 1 :
a) \(...=5^5:5^4=5\)
b) \(...=7^8:7^9=\dfrac{1}{7}\)
c) \(...=2^{15}:\left(2^6.2^5\right)=2^{15}:2^{11}=2^4=16\)
d) \(...=3^{28}:3^{26}=3^2=9\)
Bài 2 :
a) \(...=3^2.3^3:3^4=3^5:3^4=3\)
b) \(...=10^9-10^9=0\)
c) \(...=5^{10}.5^{30}:5^{12}=5^{40}:5^{12}=5^{28}\)
a, 7x + 10x = 5x
17x = 5x
17x - 5x = 0
12x = 0
x =0
2;
a, 4x + 7x = 22
11x = 22
x = 2
b, 12x - 8x = 25
4x = 25
x = \(\dfrac{25}{4}\)
c, \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\)
(\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)
\(\dfrac{1}{6}\)x = \(\dfrac{4}{5}\)
x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)
x = \(\dfrac{24}{5}\)
\(A=\left(2^5\cdot2^7\right):2^{10}+5\cdot3^4-6^3:6\\ =2^{12}:2^{10}+5\cdot3^4-6^2\\ =2^2+5\cdot3^4-36\\ =4+405-36\\ =373\\ B=\left(5^2-3\right):11\\ =\left(25-3\right):11\\ =22:11\\ =2\\ C=\left(5^{2022}-5^{2021}\right):5^{2021}\\ =5^{2021}\cdot\left(5-1\right):5^{2021}\\ =4\\ D=\left(7^{20}-7^{19}\right):7^{19}\\ =7^{19}\cdot\left(7-1\right):7^{19}\\ =6\)
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\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
\(\left|x-7\right|+\left|x-10\right|=3\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-7=3\\x-10=3\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\x=13\end{array}\right.\)
Vậy \(x\in\left\{10;13\right\}\)
Ta có
\(\left|x-7\right|+\left|x-10\right|=\left|x-7\right|+\left|10-x\right|\)
Vì \(\begin{cases}\left|x-7\right|\ge x-7\\\left|10-x\right|\ge10-x\end{cases}\)\(\Rightarrow\left|x-7\right|+\left|10-x\right|\ge\left(x-7\right)+\left(10-x\right)\)
\(\Rightarrow\left|x-7\right|+\left|10-x\right|\ge3\)
Dấu " = " xảy ra khi \(\begin{cases}x-7\ge0\\10-x\ge0\end{cases}\)\(\Rightarrow\left[\begin{array}{nghiempt}x\ge7\\x\le10\end{array}\right.\)
Vậy \(x\in\left\{7;8;9;10\right\}\)