x*13+x*95+x+x+x=5300
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a, x:0,5+x:0,125 đáp án đâu baaa
b,(3,5x10)x(53x11-5300x0,1-53)
(3,5x10)x(53x11-53x10-53x1)
(3,5x10)x[53x(11-10-1)]
(3,5x10)x(53x0)
35x0
=0
tick hộ anh vs
\(5300x=340\left(x+\frac{1}{15}\right)\)
\(\Leftrightarrow5300x=340x+\frac{68}{3}\)
\(\Leftrightarrow5300x-340x=\frac{68}{3}\)
\(\Leftrightarrow4960x=\frac{68}{3}\)
\(\Leftrightarrow x=4960:\frac{68}{3}\)
\(\Leftrightarrow x=....\)(tự tính)
5300x=340x +\(\frac{68}{3}\)
4960x=\(\frac{68}{3}\)
x=\(\frac{34}{7440}\)
=>2/13x-95/26x=-1/3-1/4
=>-91/26x=-7/12
=>-7/2x=-7/12
hay x=1/6
\(\frac{x-11}{95}+\frac{x-13}{93}=\frac{x-15}{91}+\frac{x-17}{89}\) => \(\frac{x-11}{95}-1+\frac{x-13}{93}-1=\frac{x-15}{91}-1+\frac{x-17}{89}-1\)
=>\(\frac{x-106}{95}+\frac{x-106}{93}=\frac{x-106}{91}+\frac{x-106}{89}\)
=>\(\left(\frac{1}{95}+\frac{1}{93}\right)\left(x-106\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\left(x-106\right)=0\)
<=>\(\left[\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\right]\left(x-106\right)=0\).Vì\(\frac{1}{95}< \frac{1}{91};\frac{1}{93}< \frac{1}{89}\) nên\(\frac{1}{95}+\frac{1}{93}< \frac{1}{91}+\frac{1}{89}\)
=>\(\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)< 0\) hay khác 0.Vậy x - 106 = 0, tìm được x = 106
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
2) x(x+1)(x+2)(x+3)= 24
⇔ x.(x+3) . (x+2).(x+1) = 24
⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24
Đặt \(x^2\)+ 3x = b
⇒ b . (b+2)= 24
Hay: \(b^2\) +2b = 24
⇔\(b^2\) + 2b + 1 = 25
⇔\(\left(b+1\right)^2\)= 25
+ Xét b+1 = 5 ⇒ b=4 ⇒ \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0
⇒(x-1)(x+4)=0⇒x=1 và x=-4
+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0
⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\) Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)
⇒x= 1 và x= 4
a) 95 + 11 x 206
= 95 + 2266 = 2361
b) 95 x 11 + 206
= 1045 + 206 = 1251
c) 95 x 11 x 206
= 1045 x 206 = 215270
\(x\cdot13+x\cdot95+x+x+x=5300\)
\(x\cdot\left(13+95+1+1+1\right)=5300\)
\(x\cdot111=5300\)
\(x=5300:111\)
\(x=\frac{5300}{111}\)
x*13 + x*95 + x*3 = 5300
x*( 13 + 95 + 3 ) = 5300
x* 111 = 5300
x = 5300/111