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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 8
Câu a:
\(\frac27\) + \(\frac{-3}{8}\) + \(\frac{11}{7}\) + \(\frac17\) + \(\frac{5}{-8}\)
= (\(\frac27+\frac{11}{7}+\frac17\)) -(\(\frac38\) + \(\frac58\))
= (\(\frac{13}{7}+\frac17\)) - \(\frac88\)
= \(\frac{14}{7}-1\)
= 2 - 1
= 1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 8
Câu b:
\(\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+\frac{17}{-35}+\frac{-8}{13}\)
= (\(\frac{3}{17}+\frac{14}{17}\)) - (\(\frac{5}{13}\) + \(\frac{8}{13}\)) - (\(\frac{18}{35}+\frac{17}{35}\))
= \(\frac{17}{17}-\frac{18}{18}-\) \(\frac{35}{35}\)
= 1 - 1 - 1
= 0 - 1
= -1
3 tháng 3 2019
1. \(-\dfrac{52}{17}+\left(\dfrac{12}{19}+\dfrac{52}{17}\right)=\left[\left(-\dfrac{52}{17}\right)+\left(-\dfrac{52}{17}\right)\right]+\dfrac{12}{19}=\dfrac{12}{19}\)
2. \(\dfrac{21}{35}+\left(-1+\dfrac{14}{35}\right)=\dfrac{3}{5}+\left(-1+\dfrac{2}{5}\right)=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(-1\right)=1-1=0\)
5 tháng 3 2019
1. \(\frac{-52}{17}+\left(\frac{12}{19}+\frac{52}{17}\right)=\frac{-52}{17}+\frac{12}{19}+\frac{52}{17}\)\(=\left(\frac{-52}{17}+\frac{52}{17}\right)+\frac{12}{19}=\frac{0}{17}+\frac{12}{19}=\frac{12}{19}\)
2. \(\frac{21}{35}+\left(-1+\frac{14}{35}\right)=\frac{21}{35}-1+\frac{14}{35}=\left(\frac{21}{35}+\frac{14}{35}\right)-1=1-1=0\)
JH
25 tháng 1 2017
\(\text{a) A = | -x + 8| - 21}\)
Vì | -x + 8| \(\le\) 0 ( với mọi x )
=> A = | -x + 8| - 21\(\ge\) -21
=> Amax = -21 khi | -x + 8| = 0 => -x + 8 = 0 => -x = -8 => x = 8
Vậy với Amin = -21 thì x = 8
b) \(B=\left|-x-17\right|+\left|y-36\right|+12\)
Vì \(\left\{\begin{matrix}\left|-x-17\right|\ge0\\\left|y-36\right|\ge0\end{matrix}\right.\)=> \(\left|-x-17\right|+\left|y-36\right|\ge0\)
=> \(B=\left|-x-17\right|+\left|y-36\right|+12\le12\)
=> Bmin = 12 khi \(\left|-x-17\right|+\left|y-36\right|=0\)
=> \(\left\{\begin{matrix}\left|-x-17\right|=0\\\left|y-36\right|=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}-x-17=0\\y-36=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}-x=17\\y=36\end{matrix}\right.\)=>\(\left\{\begin{matrix}x=-17\\y=36\end{matrix}\right.\)
Vậy Bmin = 12 khi \(\left\{\begin{matrix}x=-17\\y=36\end{matrix}\right.\)
c) \(C=-\left|2x-8\right|-35\)
Vì \(-\left|2x-8\right|\ge0\)
=> \(C=-\left|2x-8\right|-35\ge-35\)
=> Cmin = -35 khi \(-\left|2x-8\right|=0\)=> \(-2x-8=0\)=>\(-2x=8\)=> \(x=4\)
Vậy Cmin = -35 khi x = 4
d) \(D=3\left(3x-12\right)^2-37\)
Vì \(\left(3x-12\right)^2\ge0\)
=> \(3\left(3x-12\right)^2\ge0\)
=> \(D=3\left(3x-12\right)^2-37\ge-37\)
=> Dmin = -37 khi \(3\left(3x-12\right)^2=0\) => \(\left(3x-12\right)^2=0\)=> \(3x-12=0\)=> \(3x=12\)=>\(x=4\)
Vậy Dmin = -37 khi x = 4
25 tháng 1 2017
a, A=|-x+8|-21
Vì |-x+8|>hoặc =0 với mọi x
suy ra |-x+8|-21>hoặc = -21
Dấu = xảy ra khi và chỉ khi |-x+8|=0
Khi và chỉ khi -x+8=0
Khi và chỉ khi-x=-8
khi và chỉ khi x =8
Vậy GTNN của A là -21 tại x=8
22 tháng 2 2017
\(A=\frac{21.99+21}{35.201-35}=\frac{21\left(99+21\right)}{35\left(201-1\right)}=\frac{21.\left(99+1+20\right)}{35.200}=\frac{21.120}{35.200}=\frac{\left(3.7\right).\left(2.2.3.\right)10}{\left(5.7\right).\left(2.2.5\right).10}=\frac{3.3}{5.5}=\left(\frac{3}{5}\right)^2\)
\(B=\frac{\left(-2\right)^{36}.3^{17}}{3^{18}.2^{35}}=\frac{2^{36}.3^{17}}{2^{35}.3^{18}}=2^{\left(36-35\right)}.3^{17-18}=2.3^{-1}=\frac{2}{3}\)
\(C=\frac{7^{100}-7^{99}}{7^{98}-7^{97}}=\frac{7^{99}\left(7^1-7^0\right)}{7^{97}\left(7^1-7^0\right)}=7^{99-97}.\frac{\left(7-1\right)}{7-1}=7^2.1=49\)
15 tháng 2 2020
\(A=\left|-x+8\right|-21\)
\(A=\left|-x+8\right|-21\ge-21\)
\(MinA=-21\Leftrightarrow-x+8=0\)\(\Leftrightarrow x=8\)
\(B=\left|-x-17\right|+\left|y-36\right|+12\)
\(B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)
\(MinB=12\Leftrightarrow\hept{\begin{cases}-x-17=0\\y-36=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-17\\y=36\end{cases}}\)
\(C=-\left|2x+8\right|-35\)
\(C=-\left|2x+8\right|-35\le-35\)
\(MaxC=-35\Leftrightarrow2x+8=0\Leftrightarrow x=-4\)
12 tháng 8 2017
\(\frac{13}{21}\)+ \(\frac{8}{21}\) : \(\frac{16}{21}\)
= \(\frac{13}{21}\)+ \(\left(\frac{8}{21}:\frac{16}{21}\right)\)= \(\frac{13}{21}\)+ \(\frac{1}{2}\)
= \(\frac{47}{42}\)
\(\frac{36}{35}\): \(\frac{8}{3}\)- \(\frac{36}{35}\): \(\frac{6}{11}\)
= \(\frac{27}{70}\)- \(\frac{66}{35}\)= \(-\frac{3}{2}\)
9 tháng 4 2017
a) \(\dfrac{37}{40}-0,64\\ =\dfrac{37}{40}-\dfrac{16}{25}\\ =\dfrac{185}{200}-\dfrac{128}{200}\\ =\dfrac{57}{200}\)
9 tháng 4 2017
b) \(130\dfrac{25}{28}-120\dfrac{12}{35}\\ =\dfrac{3665}{28}-\dfrac{4212}{35}\\ =\dfrac{18325}{140}-\dfrac{16848}{140}\\ =\dfrac{211}{20}\)
26 tháng 12 2018
a, x+7=12
x=12-7
x=5
b, x-15=-21
x=-21+15
x=-6
c, 17-(x-9)=12
x-9=17-12
x-9=5
x=5+9
x=14
d, \(\left|x\right|+\left|-35\right|=41\)
\(\left|x\right|+35=41\)
\(\left|x\right|=41-35\)
\(\left|x\right|=6\)
\(\Rightarrow x=6\) hoặc \(x=-6\)
Vậy \(x\in\left\{6;-6\right\}\)
e, \(\left|x+3\right|=6\)
=> x+3=6 hoặc x+3=-6
+) x+3=6
x=6-3
x=3
+) x+3=-6
x=(-6)-3
x=-3
f, \(\left|-11\right|-\left|2-x\right|=9\)
\(11-\left|2-x\right|=9\)
\(\left|2-x\right|=11-9\)
\(\left|2-x\right|=2\)
\(\Rightarrow2-x=2\) hoặc \(2-x=-2\)
+) 2-x=2
x=2-2=0
+) 2-x=-2
x=2-(-2)=4
27 tháng 3 2017
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
\(\left\lbrack\left(3.x.x-35\right)+21\right\rbrack.17=578\)
\(\left\lbrack\left(3x.x-35\right)+21\right\rbrack=578:17\)
\(\left(3x.x-35\right)+21=34\)
\(\left(3x.x-35\right)=34-21\)
\(3x.x-35=13\)
\(3.x.x=13+35\)
\(3.x.x=48\)
\(3x^2=48\)
\(\rArr3x^2=6^2\)
\(\rArr3.x=6\)
\(\rArr x=6:3\)
\(\rArr x=2\)
\(Vậy\) \(x=2\)
d) [(3 . x . x - 35) + 21] . 17 = 578
(3 . x\(^2\) - 35) + 21 = 578 : 17
(3 . x\(^2\) - 35) + 21 = 34
3 . x\(^2\) - 35 = 34 - 21
3 . x\(^2\) - 35 = 13
3 . x\(^2\) = 13 + 35
3 . x\(^2\) = 48
x\(^2\) = 48 : 3
x\(^2\) = 16 = \(4^2=\left(-4\right)^2\)
=> x ∈ {4; -4}
Vậy x ∈ {4; -4}