nếu 1/(3 * 10 * 17) = 1/98 * (a/3 + b/10 + c/17). tìm giá trị a - b + c
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Ta có:
Tập hợp A:
\(A=\left\{1;5;9;13;17;21;25\right\}\)
Tập hợp B:
\(B=\left\{0;1;3;5;10;13\right\}\)
Mà: \(A\cap B\)
\(\Rightarrow A\cap B=\left\{1;5;13\right\}\)
⇒ Chọn B
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
a; A = |-101| + |21| + |-99| - |25|
A = 101 + 21 + 99 - 25
A = (101 + 99) - (25 - 21)
A = 200 - 4
A = 196
b; B = ||17 - 42| - 64|
B = ||-25| - 64|
B = |25 - 64|
B = |-39|
B = 39
c, C = |27 - 72| + |33 - 34| + |103 - 35|
C = |128 - 49| + |27 - 81| + |1000 - 243|
C = |79| + |-54| + | 757|
C = 79 + 54 + 757
C = 133 + 757
C = 890
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
17A=1719+1+16/1719+1
17A=1+16/1719+1
phần in nghiêng mình không hiểu lắm, bn giải thích cho mình được ko?
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
a) \(\left(-17\right)\cdot32+17\cdot\left(-68\right)-17\)
\(=17\cdot\left(-32-68-1\right)\)
\(=17\cdot\left(-101\right)\)
\(=17\cdot\left(-100-1\right)\)
\(=-1700-17\)
\(=-1717\)
b) \(25\cdot\left(34-89\right)+25\cdot89\)
\(=25\cdot\left(34-89+89\right)\)
\(=25\cdot34\)
\(=850\)
c) \(-145\cdot\left(13-57\right)+57\cdot\left(10-145\right)\)
\(=-145\cdot13+145\cdot57+57\cdot10-145\cdot57\)
\(=-145\cdot\left(13+57\right)+57\cdot\left(145+10\right)\)
\(=-145\cdot70+57\cdot155\)
\(=-10150+8835\)
\(=-1315\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\((- 17) . 32 + 17 . (- 68) – 17 \)
`= (-17). (32 + 68 + 1) `
`= (-17). 101`
`= -1717`
`b)`
\(25 . (34 – 89) + 25 . 89\)
`= 25. (34 - 89 + 89)`
`= 25. 34`
`= 850`
`c)`
\(– 145. (13 – 57) + 57. (10 – 145)\)
`= (-145). 13 - (-145). 57 + 57.10 + 57. (-145)`
`= (-145). (13 - 57 + 57) + 57.10`
`= (-145). 13 + 570`
`= -1885 + 570 = -1315`
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
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