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a) Ta có: -|2,5 - x| \(\le\)0 \(\forall\)x
=> -|2,5 - x| - 5,8 \(\le\)-5,8 \(\forall\)x
Hay B \(\le\)-5,8 \(\forall\)x
Dấu "=" xảy ra <=> 2,5 - x = 0 <=> x = 2,5
Vậy MaxB = -5,8 <=> x = 2,5
b) Ta có: 4|x - 2| \(\ge\)0 \(\forall\)x
=> 10 - 4|x - 2| \(\le\)10 \(\forall\)x
hay C \(\le\)10 \(\forall\)x
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy MaxC = 10 <=> x = 2
a) Ta có : \(\left|2,5-x\right|\ge0\forall x\)
\(\Rightarrow-\left|2,5-x\right|\le0\)
\(\Rightarrow-\left|2,5-x\right|-5,8\le-5,8\)
\(\Rightarrow\text{GTLN của B = }-5,8\)
\(\text{Dấu bằng xảy ra khi }\left|2,5-x\right|=0\)
\(\Rightarrow2,5-x=0\)
\(\Rightarrow x=2,5\)
Vậy GTLN của B là -5,8 khi x = 2,5
b) Ta có : \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow4\left|x-2\right|\ge0\)
\(\Rightarrow10-4\left|x-2\right|\le10\)
Dấu " = " xảy ra khi |x - 2| = 0
=> x - 2 = 0
=> x = 2
Vậy GTLN của C là 10 khi x = 2
a. x-36:18=12-15 b.(3x-2^4).7^3=2.7^4 3x-16=14 x-36 =-3 3x-2^4 =2.7^4:7^3 3x =16+14 x =-3+36 3x-16 =2.7^4-3 3x =30 x =33 3x-16 =2.7 3x =30:3 =10
a)
x-36:18=12-15
<=> x-2=-3
<=> x= -1
Vậy x= -1
b)
(3x- 2^4).7^3=2.7^4
<=> 3x-2^4= 2.7^4:7^3
<=> 3x-16= 2.7
<=> 3x-16= 14
<=> 3x= 30
<=> x=10
Vậy x=10
a) \(\frac{113}{13}-\frac{24}{7}-\frac{53}{13}=\frac{60}{13}-\frac{24}{7}\)
=\(\frac{420-312}{91}=\frac{108}{91}\)
b) \(\frac{2}{7}.\left(\frac{51}{4}-\frac{31}{4}\right)=\frac{2}{7}.5=\frac{10}{7}\)
c) \(\frac{3}{5}+1-\frac{1}{20}\)= \(\frac{12}{20}+\frac{20}{20}-\frac{1}{20}=\frac{31}{20}\)
\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{9}{4}:\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+\left(\frac{9}{4}.\frac{16}{3}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.4^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.2^6}{2^7.5^2.2^9}\)
=\(\frac{2^6.\left(27+2\right)}{2^6.5^2.2^{10}}\)
=\(\frac{29}{25600}\)
Đoạn \(\frac{2^6\cdot\left(27+2\right)}{2^6\cdot5^2\cdot2^{10}}\)là sai rồi bn ơi!!!
Bn phải lm như mục trên là:
\(\frac{2^6\left(2+3^3\right)}{2^7\left(5^2+2^2\right)}\)\(=\frac{2^6\cdot29}{2^7\cdot29}=\frac{1}{2}\)
Nhưng dù sao cx c.ơn bn vì đã giúp mk,mk sẽ cho bn 1 ths nka!!!
a, \(2n+5⋮n-1\)
\(2\left(n-1\right)+7⋮n-1\)
\(7⋮n-1\)hay \(n-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
| n - 1 | 1 | -1 | 7 | -7 |
| n | 2 | 0 | 8 | -6 |
b, Công thức tổng quát : \(A\left(x\right).B\left(x\right)=0\Rightarrow\orbr{\begin{cases}A\left(x\right)=0\\B\left(x\right)=0\end{cases}}\)
\(\left(2n+3\right)\left(n-4\right)=0\Leftrightarrow\orbr{\begin{cases}n=-\frac{3}{2}\\n=4\end{cases}}\)
c, \(\left|x-3\right|< 3\Leftrightarrow-3< x-3< 3\)
\(\Leftrightarrow-3+3< x< 3+3\Leftrightarrow0< x< 6\)
Vậy \(x\in\left\{1;2;3;4;5;\right\}\)
\(\frac{3}{4}x-\frac{1}{2}=2\left(x-4\right)+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-\frac{1}{2}=2\text{x}-8+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-2\text{x}-\frac{1}{4}x=-8+\frac{1}{2}\)
\(\Leftrightarrow\frac{3-8-1}{4}x=\frac{-15}{2}\)
\(\Leftrightarrow-\frac{3}{2}x=-\frac{15}{2}\Leftrightarrow x=\frac{-15}{-3}=5\)
Vậy x = 5
\(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)=\frac{16}{9}\div\frac{2}{9}\)
\(\Rightarrow\left(x-1\right)=\frac{16}{9}\cdot\frac{9}{2}\)
\(\Rightarrow x-1=8\Rightarrow x=9\)
Vậy x = 9
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(\Rightarrow\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\div2\)
\(\Rightarrow\frac{x}{x+1}=\frac{2004}{2005}\)
\(\Rightarrow2005\text{x}=2004\left(x+1\right)\)
\(\Rightarrow2005\text{x}=2004\text{x}+2004\)
\(\Rightarrow2005\text{x}-2004\text{x}=2004\)
\(\Rightarrow x=2004\)
Vậy x = 2004
x-3.02=-4.74
x=-4.74+3.02
x=-1.72
\(x-3,02=5,26-4\cdot2,5\)
\(\Rightarrow x-3,02=-4,74\)
\(\Rightarrow x=-4,74+3,02\)
\(\Rightarrow x=-1,72\)