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2(x+1)^2-5=45
2(x+1)^2=45+5
2(x+1)^2=50
(x+1)^2=50/2
(x+1)^2=25
(x+1)^2=52
=>x+1=5
x=5-1
x=4
A, 70 - 5 . ( x - 3 ) = 45
5.( x - 3 ) = 70 - 45
5. ( x - 3 ) = 25
x - 3 = 25 : 5
x - 3 = 5
x = 5 + 3
x = 8
B, 52x - 3 - 2.52 = 52.3
52x - 3 - 2.25 = 25.3
52x - 3 - 50 = 75
52x - 3 = 75 + 50
52x - 3 = 125
52x - 3 = 53
2x - 3 = 3
2x = 3 + 3
2x = 6
x = 6 : 2
x = 3
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
\(6^2-\left(x+3\right)=45\)
\(36-\left(x+3\right)=45\)
\(x+3=35-45\)
\(x+3=-10\)
\(x=-13\)
Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
5.(7+48:x)=45
(7+48:x)=45:5
(7+48:x)=9
48:x=9-7
48:x=2
x=48:2
x=24
99 - 3(x + 17) = 45
=> 3(x + 17) = 99 - 45
=> 3x + 51 = 54
=> 3x = 54 - 53
=> 3x = 3
=> x = 3 : 3
=> x = 1
vậy_
a,99-3x-51=45
3x=45+51-99
3x=-3
x=-1
b,x5:8=4
x5=32
x=2
c,3.x3=375
x3=125
x=5
d,12.x-33=35
12x=276
x=23
e,4x5=128
x5=32
x=2
g,2x:4=16
2x=64
x=6
a) -45:5(-3-2x)=3
5(-3-2x)=-45:3
5(-3-2x)=-15
-3-2x=-15:5
-3-2x=-3
2x=(-3)-(-3)
2x=-6
x=-6:2
x=-3