Câu 7
3. 24 - 2 . 24
633 - { [ 105 + 3 ) : 3 ] -3 }
bài 9
4 x-3 -3 . 42 = 42
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A) 12/16
2*3; 6/21
42:6; 4/7
B) 6/9
6/8
8/28=10/35
12/20
4/3
6/9=2/3=16/28
1: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x-y}{7-13}=\dfrac{42}{-6}=-7\)
=>x=-48; y=-91
2: x/y=3/4
=>4x=3y
=>4x-3y=0
mà 2x+y=10
nên x=3 và y=4
3: =>7x-3y=0 và x-y=-24
=>x=18 và y=42
4: =>7x-5y=0 và x+y=24
=>x=10 và y=14
1)1/6 - - 5/6 = 1/6 + 5/6 = 1
2)6/13 - -14/39 = 6/13 + 14/39= 32/39
3)4/5 - 4/-18 = 4/5 + 4/18= 46/45
4)7/21 - 9/-36 = 7/21 + 9/36 = 7/12
5)-12/18 - -21/35 = -12/18 + 21/35 = -1/15
6)-3/21 - 6/42 = -2/7
7)-18/24 - 15/21 = -41/28
8)1/6 - 2/5 =-7/30
Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
s: \(=125\cdot100-25\cdot192\)
\(=12500-4800\)
=7700
1, 54 : x - 1 = 5
54 : x = 5+1 = 6
x = 54 : 6 = 9
2, 42 : x + 0 = 8
x = 42 : 8 = 21/4
3, 24 : x - 8 = 0
24 : x = 0 + 8 = 8
x = 24 : 8 = 3
Tk mk nha
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)
\(\left(a\right)\frac{3}{7}-\left(\frac{-2}{7}\right)=\frac{3}{7}+\frac{2}{7}=\frac{5}{7}\)
\(\left(b\right)\frac{-13}{11}+\frac{6}{22}-\frac{36}{33}=\frac{-39}{33}+\frac{9}{33}-\frac{36}{33}=\frac{-66}{33}=-2\)
\(a,\frac{3}{7}-\frac{-2}{7}=\frac{3-\left(-2\right)}{7}=\frac{3+2}{7}=\frac{5}{7}\)
\(b,-\frac{13}{11}+\frac{6}{22}-\frac{36}{33}=-\frac{13}{11}+\frac{3}{11}-\frac{12}{11}=\frac{-13+3-12}{11}=-\frac{22}{11}=-2\)
\(c,-\frac{6}{17}-\frac{5}{-12}+\frac{3}{81}=-\frac{6}{17}+\frac{5}{12}+\frac{1}{27}=\frac{-6.27.12+5.17.27+17.12}{17.12.27}=\frac{185}{1836}\)
\(\frac{2}{70}-\frac{9}{-24}+\frac{-5}{42}=\frac{1}{35}+\frac{3}{8}-\frac{5}{42}=\frac{1.24+105.3-5.20}{840}=\frac{239}{840}\)
Bài 7:
3 . 24 - 2 . 24
= (3 - 2) . 24
= 1 . 24
= 24.
633 - { [(105 + 3) : 3] - 3}
= 633 - { [108 : 3] - 3}
= 633 - {36 - 3}
= 633 - 33
= 600.
Bài 9:
4x - 3 - 3 . 42 = 42
4x - 3 - 3 . 16 = 16
4x - 3 - 48 = 16
4x - 3 = 16 + 48
4x - 3 = 64
\(\Rightarrow\) 4x - 3 = 43
\(\Rightarrow\) x - 3 = 3
x = 3 + 3
x = 6.
\(3.2^4-2.2^4\)
\(=2^4.\left(3-2\right)\)
\(=2^4.1\)
\(=2^4=16\)
.
\(633-\left\{\left[\left(105+3\right):3\right]-3\right\}\)
\(=633-\left\{\left[108:3\right]-3\right\}\)
\(=633-\left\{36-3\right\}\)
\(=633-33\)
\(=600\)
.
\(4^{x-3}-3.4^2=4^2\)
\(4^{x-3}-4^2=3.4^2\)
\(4^{x-3}-4^2=48\)
\(4^{x-3}=48+4^2\)
\(4^{x-3}=64\)
Ta có: \(4^3=64\)
\(\Rightarrow4^{x-3}=4^3\)
\(\Rightarrow4^x:4^3=4^3\)
\(\Rightarrow4^x=4^3:4^3\)
\(4^x=4^0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
. Ko hiểu gì hỏi mình nha <3