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\(\left(a-b+c\right)-\left(a+c\right)\)
\(=a-b+c-a-c\)
\(=\left(a-a\right)+\left(c-c\right)-b\)
\(=0+0-b\)
\(=0-b\)
\(=-b\)
1) (a - b + c) - (a + c)
= a - b + c - a - c
= (a - a) - b + (c - c)
= 0 - b + 0 = -b
2) (a + b) - (b - a) + c
= a + b - b + a + c
= (a + a) + (b - b) + c
= 2a + 0 + c = 2a + c
3) -(a + b - c) + (a - b - c)
= -a - b + c + a - b - c
= (-a + a) - (b + b) + (c - c)
= 0 - 2b + 0 = -2b
4) a(b + c) - a(b + d)
= ab + ac - ab - ad
= (ab - ab) + a(c - d)
= 0 + a(c - d) = a(c - d)
5) tự lm
a) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=2\)
Vậy x=2
b) \(x+4=2^0+1^{2019}\)
\(\Leftrightarrow x+4=1+1\)
\(\Leftrightarrow x+4=2\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
a)\(9-3x< 0\)
\(9< 3x\)
\(x>3\)
b)\(\left(2x-4\right)\left(x+5\right)< 0\)
\(\Leftrightarrow2x-4>0\Rightarrow2x>4\Rightarrow x>2\)
\(x+5< 0\Rightarrow x< -5\)
\(2< x< -5\)
\(x\in\left\{\varnothing\right\}\)
\(\Leftrightarrow2x-4< 0\Rightarrow2x< 4\Rightarrow x< 2\)
\(x+5>0\Rightarrow x>-5\)
\(-5< x< 2\)
\(x\in\left\{-4;-3;-2;-1;0;1\right\}\)
c)
\(\dfrac{x+4}{x-2}\in Z^-\)
\(\Leftrightarrow x+4< 0\Rightarrow x< -4\)
\(x-2>0\Rightarrow x>2\)
\(2< x< -4\)
\(x\in\left\{\varnothing\right\}\)
\(\Leftrightarrow x+4>0\Rightarrow x>-4\)
\(x-2< 0\Rightarrow x< 2\)
\(-4< x< 2\)
\(x\in\left\{-3;-2;-1;0;1\right\}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
\(x+2=41\)
\(x=41-2\)
\(x=39\)
Tìm x
a) \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{x\times\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+1\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{40}{41}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{\left(x+2\right)}=\frac{40}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Rightarrow x+2=41\)
\(\Rightarrow x=41-2\)
\(\Rightarrow x=39\)
Vậy x = 39
2^x = 4 . 128 = 512
2^x = 2^9
=> x = 9
b) 2^x . 3^x = 4.9
=> 2^x . 3^x = 36
=> x = 2
x+x+x+82=-2-x
=>x+x+x+x=-2-82
4x=-84
x-84/4
x=-21
(a-b+c-d)-9a+b+c+d)=a-b+c-d-a-b-c-d
=(a-a)-(b+b)+(c-c)-(d-d)
=0-2b+0-2d
=-2b-2d
(-a+b-c)+(a-b)-(a-b+c)
=-a+b-c+a-b-a+b-c
=(-a+a-a)+(b-b+b)-(c+c)
=-a+b-2c
-(a-b-c)+(b-c+d)-(a+b+d)
=-a+b+c+c-c+d-a-b-d
=(-a-a)+(b-b)+(c+c-c)+(d-d)
=-2a+0+c+0
=-2a+c
x+x+x+82=-2-x
3x+82=-2-x
3x+x=-2-82
4x=-84
x=-84:4=-21
5.(-40).x=-100
-200.x=-100
x=-100:(-200)
x=0,5
-1.(-3).(-6).x=36
-18.x=36
x=36:(-18)
x=-2
|1-4x|=7
Xảy ra hai trường hợp:1-4x=7=>4x=1-7=-6=>x=-6:4=-1,5
1-4x=-7=>4x=1+7=8=>x=8:4=2
Vì x là số nguyên nên ta chỉ có một đáp án đó là:2
x.(x-2)=0+> x=0 hoặc x=2
x.(x-2)>0=>x=-1;-2;-3;-4;...