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a, 11/13 - ( 5/42 - x ) = - (5/28 - 11/13)
11/13 - (5/42 - x) = - 5/28 + 11/13
- (5/42 - x) + 5/28 = -11/13 + 11/13
- 5/42 + x + 5/28 = 0
- 5/42 + x = 0 - 5/28
- 5/42 + x = - 5/28
x = -5/28 +5/42
x = - 5/84
b, / x + 4/15 \ - / - 3,75 \ = - / - 2,15 \
./ x + 4/15 \ - 3,75 = - 2,15
/ x + 4/15 \ = -2,15 + 3,75
/ x + 4/15 \ = 1,6
x + 4 / 15 = 1,6 hoặc x+ 4/15 = - 1,6
x = 1,6 - 4/15 x = - 1,6 -4/15
x = 4/3 x = -28/15
Vậy x = 4/3 hoặc x = - 28/15
c, ( 0,25 - 30% x ) . 1/3 = 1/4 - 31/6
( 1/4 - 3/10 x ) . 1/3 = - 59/12
( 1/4 - 3/10 x ) = - 59/12 : 1/3
1/4 - 3/10 x = - 59/4
3/10 x = 1/4 + 59/4
3/10 x = 15
x = 15 : 3/10
x = 50
d, ( x - 1/2 ) : 1/3 + 5/7 = 68/7
( x - 1/2 ) : 1/3 = 68/7 - 5/7
( x - 1/2 ) : 1/3 = 63/7
( x - 1/2 ) = 63/7 . 1/3
x -1/2 = 3
x = 3 + 1/2
x = 7/2
a, ( 1/2 - 13/14 ) : 5/7 - ( -2/21 + 1/7 ) : 5/7
= ( 7/14 - 13/14 ) : 5/7 - ( -2/21 + 3/21 ) : 5/7
= -3/7 : 5/7 - 1/21 : 5/7
= -3/7 . 7/5 - 1/21 . 7/5
= ( -3/7 - 1/21 ) . 7/5
= ( -9/7+ -1/21 ) . 7/5
= -10/21 . 7/5
= -2/3
b, (68/5 + 19/4 ) - 43/5
= 68/5 + 19/4 - 43/5
= ( 68/5 - 43/5 ) + 19/4
= 5 +19/4
= 39/4
c, ( 93/11 + 29/8 ) - 38/11
= 93/11 + 29/8 - 38/11
= ( 93/11 - 38/11 ) + 29/8
= 5 + 29/8
=69/8
d, 4/9 : ( -1/7 ) + 59/9 : ( -1/7 )
= 4/9 . -7 + 59/9 . -7
= ( 4/9 + 59/9 ) . -7
= 7 . -7
= -49
Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
Khi phá ngoặc của của đa thức f(x) ta sẽ được đa thức \(f\left(x\right)=a_1x^n+a_2x^{n-1}+a_3x^{n-2}+...+a_{n-1}x+a_n\)(với n là bậc của đa thức)
Ta có:\(f\left(1\right)=a_1+a_2+a_3+...+a_{n-1}+a_n\)
Mà \(f\left(1\right)=\left(3-12+8\right)^{111}\cdot\left(4+3+2+1-12+1\right)^{2222}\)\(=-1\)
Suy ra:\(a_1+a_2+a_3+...+a_{n-1}+a_n=-1\)
Vậy tổng các hệ số của đa thức sau khi phá ngoặc là -1
a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247x.(1233+20203333+303030333333+4242424233333333)=32
74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247x.(1233+2033+3033+4233)=32
74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247x.(3.433+4.533+5.633+6.733)=32
74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247x.33.(31−41+41−51+51−61+61−71)=32
74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247x.33.(31−71)=32
74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31+61+101+151+...+x.(x−1)2=20092007
26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62+122+202+302+...+(x−1).x2=20092007
22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32+3.42+4.52+5.62+...+(x−1).x2=20092007
2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−31+31−41+41−51+51−61+...+x−11−x1)=20092007
2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−x1)=20092007
1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2=20092007
2x=22009\frac{2}{x}=\frac{2}{2009}x2=20092
=> x = 2009