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\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
b: =>3|x-5|=8+4=12
=>|x-5|=4
=>x-5=4 hoặc x-5=-4
=>x=9 hoặc x=1
d: =>2x+6=3-3x-2
=>2x+6=1-3x
=>5x=-5
hay x=-1
e: \(\Leftrightarrow x-3\inƯC\left(70;98\right)\)
\(\Leftrightarrow x-3\in\left\{1;2;7;14\right\}\)
mà x>8
nên \(x\in\left\{10;17\right\}\)
Bài 2:
a: =>|x-3|=x-3
=>x-3>=0
hay x>=3
b: =>-|x+3|<2-7=-5
=>|x+3|>5
=>x+3>5 hoặc x+3<-5
=>x>2 hoặc x<-8
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(3x-2-x-1\right)\left(3x-2+x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x-3\right)\left(4x-1\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{3}{2};\dfrac{1}{4}\right\}\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}x< =7\\\left(x-3-7+x\right)\left(x-3+7-x\right)=0\end{matrix}\right.\)
=>x=5
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
a,
3x + 3 - [7x+4] = 7 + [4x-1]
=> 3x + 3 - x - 4 = 7 + 4x - 1
=> 2x - 1 = 6 + 4x
=> 2x - 4x = 6 + 1
=> -2x = 7
=> x = -7/2
b,
3x+1 + 3x+3 =810
=> 3x+1[1 + 32] = 810
=> 3x+1 = 810 / 10
=> 3x+1 = 81
=> x = 4
c, \(1\frac{1}{2}:\left[\frac{1}{2}-\frac{1}{3}\right]-x=5\)
\(\Rightarrow\frac{3}{2}:\frac{1}{6}-x=5\Leftrightarrow9-x=5\)
\(\Leftrightarrow x=4\)
d,
\(2,4:\left[25\%+\frac{x}{40}\right]-\frac{12}{15}=3\frac{1}{5}\)
\(\Rightarrow\frac{12}{5}:\left[\frac{1}{4}+\frac{x}{40}\right]-\frac{12}{15}=\frac{16}{5}\)
\(\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=\frac{16}{5}+\frac{12}{15}\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=4\)
\(\Rightarrow\frac{10+x}{40}=\frac{12}{5}:4\Leftrightarrow\frac{10+x}{40}=\frac{3}{5}\)
\(\Rightarrow\frac{10+x}{40}=\frac{24}{40}\Leftrightarrow10+x=24\Rightarrow x=14\)
a) 3x + 3 - ( x + 4 ) = 7 + ( 4x - 1 )
3x + 3 - x - 4 = 7 + 4x - 1
2x - 1 = 6 + 4x
-2x = 7
\(\Rightarrow\)x = \(\frac{-7}{2}\)
b) 3x+1 + 3x+3 = 810
3x . 3 + 3x . 33 = 810
3x . ( 3 + 33 ) = 810
3x . 30 = 810
3x = 810 : 30
3x = 27
3x = 33
\(\Rightarrow\)x = 3
c) \(1\frac{1}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\frac{1}{6}-x=5\)
\(9-x=5\)
\(\Rightarrow x=9-5\)
\(\Rightarrow x=4\)
d) 2,4 : ( 25% + \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(3\frac{1}{5}\)
\(\frac{12}{5}\) : ( \(\frac{1}{4}\)+ \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(\frac{16}{5}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=\frac{16}{5}+\frac{12}{15}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{12}{5}:4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{3}{5}\)
\(\frac{x}{40}=\frac{3}{5}-\frac{1}{4}\)
\(\frac{x}{40}=\frac{7}{20}\)
\(\Rightarrow\frac{x}{40}=\frac{14}{40}\)
\(\Rightarrow x=14\)
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
a) 9 - 25 = (7 - x) - (25 + 7)
7 - x - 25 - 7 = 16
-x - 25 = 16
-x = 41
x= -41
a) x=-9
b)+c) bạn hỏi I hay x thế ?