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\(A=a+\left(42-70+18\right)-\left(42+18-a\right)\)
\(A=a+42-70+18-42-18-a\)
\(A=\left(a-a\right)+\left(42-42\right)+\left(18-18\right)-70\)
\(A=0+0+0-70=-70\)
a) A = a + (42-70+18) - (42+18+a)
= a + 42 - 70 + 18 - 42 - 18 - a
= (a-a) + (42-42) + (18-18) - 70
= 0 + 0 + 0 - 70 = -70.
Vậy A = -70.
b) B = a + 30 + 12 - (-20) + (-12) - (2+a)
= a + 30 + 12 + 20 - 12 - 2 - a
= (a-a) + (12-12) + (30+20-2)
= 0 + 0 + (50-2)
= 50 - 2 = 48.
Vậy B = 48.
c) C = (x-y+z) - (x-y-z) - (2x+y)
= x - y + z - x + y + z - 2x - y
= (x-x-2x) + (-y+y-y) + (z+z)
= -2x + (-y) + 2z
= -2x - y + 2z.
Vậy C = -2x - y + 2z.
a, (a+b) -( b+c) +(a+c)
= a +b-b-c +a+c
= 2a
tính
a, bỏ ngoặc ta đc giá trị là 70
b giá trị là 50-2
=48
chúc bn học giỏi
a) (a +b) - (b+c) +(a+c)
= a+b-b-c+a+c
= (a+a)+(b-b)+(c-c)
= 2a+0+0
=2a
b) a+(42+70+18)-(42+18+a)
= a+42+70+18-42-18-a
=(a-a)+(42-42)+(18-18)+70
=0+0+0+70
=70
C) a+30+12-(-20)+(-12)-(2+a)
=a+30+12+20-12-2-a
=(a-a)+(12-12)+(30+20-2)
=0+0+48
=48
Bài 1 : em chưa học
Bài 2 : \(A=a+\left(42-70+18\right)-\left(42+18+a\right)\)
\(=a-10-60-a=-70\)
A=a+42-70+18-42-18-a
A=(a-a) + (42-42) +( 18-18) - 70
A= 0 - 0 + 0 - 70
A = -70
A= a+42 +18-70 - 42 -18 -a
A= (a-a) + (42-42)+(18-18) + 72
A= 0+0+0+72
A= 72
\(a)A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)
\(A=\dfrac{2.3}{16}\)
\(=\dfrac{3}{8}\)
\(b)B=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}\)
\(B=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{56}\right)\)
\(B=6-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)
\(B=6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(B=6-\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(B=\dfrac{48}{8}-\dfrac{3}{8}\)
\(B=\dfrac{45}{8}\)
a: \(42=2\cdot3\cdot7;70=2\cdot5\cdot7\)
=>\(BCNN\left(42;70\right)=2\cdot3\cdot5\cdot7=210\)
=>\(BC\left(42;70\right)=B\left(210\right)=\left\{0;210;420;...\right\}\)
b: \(70=2\cdot5\cdot7;180=3^2\cdot5\cdot2^2\)
=>\(BCNN\left(70;180\right)=2^2\cdot3^2\cdot5\cdot7=1260\)
=>\(BC\left(70;180\right)=\left\{1260;2520;...\right\}\)
c: \(5=5;7=7;8=2^3\)
=>\(BCNN\left(5;7;8\right)=5\cdot7\cdot8=280\)
=>\(BC\left(5;7;8\right)=\left\{280;560;...\right\}\)
d: \(12=2^2\cdot3;18=3^2\cdot2\)
=>\(BCNN\left(12;18\right)=2^2\cdot3^2=36\)
=>\(BC\left(12;18\right)=\left\{36;72;...\right\}\)
e: \(15=3\cdot5;18=3^2\cdot2\)
=>\(BCNN\left(15;18\right)=3^2\cdot2\cdot5=90\)
=>\(BC\left(15;18\right)=\left\{90;180;...\right\}\)
f: \(84=2^2\cdot3\cdot7;108=3^3\cdot2^2\)
=>\(BCNN\left(84;108\right)=2^2\cdot3^3\cdot7=756\)
=>\(BC\left(84;108\right)=\left\{756;1512;...\right\}\)
j: \(33=3\cdot11;44=2^2\cdot11;55=5\cdot11\)
=>\(BCNN\left(33;44;55\right)=3\cdot2^2\cdot5\cdot11=660\)
=>\(BC\left(33;44;55\right)=\left\{660;1320;...\right\}\)
g: \(1=1;12=2^2\cdot3;27=3^3\)
=>\(BCNN\left(1;12;27\right)=1\cdot2^2\cdot3^3=108\)
=>\(BC\left(1;12;27\right)=\left\{108;216;...\right\}\)
n: \(5=5;9=3^2;11=11\)
=>\(BCNN\left(5;9;11\right)=5\cdot3^2\cdot11=495\)
=>\(BC\left(5;9;11\right)=\left\{495;990;...\right\}\)
a, A = a + ( 42 - 70 + 18 ) - ( 42 + 18 + a )
A = a + 42 - 70 + 18 - 42 - 18 - a
A = ( a - a ) + ( 42 - 42 ) + ( 18 - 18 ) - 70
A = 0 + 0 + 0 - 70
A = -70
b, B = a + 30 + 12 - ( -20 ) + ( -12 ) - ( 2 + a )
B = a + 30 + 12 + 20 - 12 - 2 - a
B = ( a - a ) + ( 30 + 20 ) + ( 12 - 12 ) - 2
B = 50 - 2
B = 48
Chúc bạn học tốt!
A = a + 42 -70 +18 - 42 - 18 - a
A = a + (-70) - a
A = a - 70 - a
B = a + 30 + 12 + 20 - 12 - 2 - a
B = a + 48 - a