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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) \(A=\left(a-2b+c\right)-\left(a-2b-c\right)\)
\(A=a-2b+c-a+2b+c=2c\)
b) \(B=\left(-x-y+3\right)-\left(-x+2-y\right)\)
\(B=-x-y+3+x-2+y=1\)
c) \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)\)
\(C=6a+2b-2-6a-3b+6=4-b\)
a. \(A=\left(a-2b+c\right)-\left(a-2b-c\right)=a-2b+c-a+2b+c=0\)
b. \(B=\left(-x-y+3\right)-\left(-x+2-y\right)=-x-y+3+x-2+y=1\)
c. \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)=6a+2b-2-6b-3b+6=4-3b\)
1.
a.\(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
b. \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\)
c. \(\left(\frac{-3}{5}\right)^5=\frac{-243}{3125}\)
d. \(\left(\frac{-1}{5}\right)^2=\frac{1}{25}\)
e. \(\left(\frac{-1}{6}\right)^3=\frac{-1}{216}\)
Trả lời:
Bài 1:
a, \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{16}\)
b, \(\left(\frac{1}{2}\right)^3=\frac{1^3}{2^3}=\frac{1}{8}\)
c, \(\left(\frac{-3}{5}\right)^2=\frac{\left(-3\right)^2}{5^2}=\frac{9}{25}\)
d, \(\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)
e, \(\left(\frac{-1}{6}\right)^3=\frac{\left(-1\right)^3}{6^3}=\frac{-1}{216}\)
Bài 2:
a, \(\left(\frac{3}{2}\right)^2.\left(\frac{4}{3}\right)^2=\frac{9}{4}.\frac{16}{9}=4\)
b, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)
c, \(\left(-\frac{1}{2}\right)^2.\left(\frac{2}{5}\right)^2=\frac{1}{4}.\frac{4}{25}=\frac{1}{25}\)
d, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)
e, \(\left(-5\right)^3.\frac{1}{5}=-125.\frac{1}{5}=-25\)
f, \(\left(\frac{2}{9}\right)^5.\left(-\frac{27}{4}\right)^5=\frac{2^5}{9^5}.\frac{\left(-27\right)^5}{4^5}=\frac{2^5.\left(-27\right)^5}{9^5.4^5}=\frac{2^5.\left[\left(-3\right)^3\right]^5}{\left(3^2\right)^5.\left(2^2\right)^5}=-\frac{2^5.3^{15}}{3^{10}.2^{10}}=\frac{3^5}{2^5}\)
10 - { [ ( x : 3 + 17 ) : 10 + 3 : 24 ] : 10 } = 5
[ ( x : 3 + 17 ) : 10 + 3 : 24 ] : 10 = 10 - 5 = 5
( x : 3 + 17 ) : 10 + 3 : 24 = 5 x 10
( x : 3 + 17 ) : 10 + 48 = 50
( x : 3 + 17 ) : 10 = 50 - 48
( x : 3 + 17 ) : 10 = 2
x : 3 + 17 = 2 x 10
x : 3 + 17 = 20
x : 3 = 20 - 17 = 3
x = 3 x 3 = 9
a) [(2x+14) : 4 - 3] : 2 = 1
(2x+14) : 4 - 3 = 1/2
(2x+14) : 4 = 1/2 + 3
(2x+14) : 4 = 7/2
2x+14 = 7/2 . 1/4
2x = 7/8 - 1/4
2x = 5/8
x= 5/8.1/2
x= 5/16
a)\(A=\left|x-2\right|+\left|x-3\right|=\left|x-2\right|+\left|3-x\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1\)
Dấu "=" xảy ra khi \(2\le x\le3\)
Vậy \(Min_A=1\) khi \(2\le x\le3\)
b)Ta thấy: \(\left|x-1\right|\ge0\)
\(\Rightarrow\left|x-1\right|-2\ge-2\)
\(\Rightarrow B\ge-2\)
Dấu "=" xảy ra khi \(x=1\)
Vậy \(Min_B=-2\) khi \(x=1\)
c)\(C=\left|x-3\right|+\left|x-4\right|=\left|x-3\right|+\left|4-x\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-3\right|+\left|4-x\right|\ge\left|x-3+4-x\right|=1\)
Dấu "=" xảy ra khi \(3\le x\le4\)
Vậy \(Min_C=1\) khi \(3\le x\le4\)
d)\(D=\left|x-1\right|+\left|x+5\right|+2=\left|x-1\right|+\left|-\left(x+5\right)\right|+2\)
\(=\left|x-1\right|+\left|-x-5\right|+2\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|-x-5\right|+2\ge\left|x-1+\left(-x\right)-5\right|+2=6+2=8\)
Dấu "=" xảy ra khi \(-5\le x\le1\)
Vậy \(Min_D=8\) khi \(-5\le x\le1\)
Cảm ơn bạn đã giải giúp mình bài toán này nhé!
Bạn giải cũng na ná cô giáo mình .
a) 3 - (-6/7)0 + (1/2)2 : 2
= 3 + 1 + 1/4 : 2
= 3 + 1 + 1/8
= 33/8
b) (-2)3 + 22 + (-1)20 + (-2)0
= (-8) + 4 - 1 - 1
= -6
c) [(3)2]2 - [(-5)2]2 - [(-2)3]2
= 81 - 625 - 64
= -608
d) 24 + 8.[(-2)2 : 1/2]0 - 2-2.4 + (-2)2
= 16 + 8.1 - 1/4.4 + 4
= 16 + 8 - 4 + 4
= 27
e) 23 + 3.(1/2)0 - 2-2.4 + [(-2)2 : 1/2].8
= 8 + 3 - 1/4.4 + 8.8
= 8 + 3 - 1 + 64
= 74
a) Vì a \(⋮\) a => \(2⋮a\)
\(\Rightarrow a\inƯ\left(2\right)\Rightarrow a\in\left\{\pm1;\pm2\right\}\)
b) Ta có: a + 5 = (a+1) +4
Do a+ 1 \(⋮a+1\Rightarrow4⋮a+1\)
\(\Rightarrow a+1\inƯ\left(4\right)\)
\(\Rightarrow a+1\left\{\pm1;\pm2;\pm4\right\}\)
Với x + 1 = 1 thì x = 0
Với x + 1 = -1 thì x = -2
...
c) Ta có: \(a^2+3=a\left(a+1\right)-a-1+4\)
\(=a\left(a+1\right)-\left(a+1\right)+4=\left(a-1\right)\left(a+1\right)+4\)
Do \(\left(a-1\right)\left(a+1\right)⋮\left(a+1\right)\Rightarrow4⋮\left(a+1\right)\)
\(\Rightarrow a+1\inƯ\left(4\right)\)
...
d) Làm như trên và loại bớt trường hợp bằng cách lí luận 2a + 1 luôn lẻ.
e) Tương tự.
câu d thì làm như câu nào vậy