Ta có x-y=8; y - z=10=>x - z=8+10=18 =>x=(12+18):2=30:2=15

=>y=15-8=7

=>x+y+z=(x+z)+y=12+7=19

Mình đã kiểm tra và đây là kết quả đúng.

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=9900\)

\(\Leftrightarrow99x+\left(\frac{99-1}{1}+1\right)=9900\)

\(\Leftrightarrow99x=9900-99\)

\(\Leftrightarrow x=99\)

k mk nha

(x + 1) +( x + 2) + ... + (x + 99 )= 9900

=>99x +(99-1/1  + 1 )=9900

=>99x=9900-99

=>x=90

ấn chậm quá

@@

làm lại nha . làm theo này chăc chắn hơn

(x+1)+(x+2)+...+(x+98)+(x+99)=9900

=>99x + ( 1+2+3+...+99)=9900

=>99x + 4950 = 9900

=> 99x = 4950

=> x= 50

vậy ...

Chắc chắn #

\(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{5}=0\\x^2+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{1}{5}\\x^2=-\dfrac{1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{2}{15}\)

\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Rightarrow\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\)

\(\Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\)

\(\Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

\(\Rightarrow x=-100\)(do \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}>0\))

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)=15\)

⇔ \(\left(x^3-3.x^2.3+3.x.3^2-3^3\right)-\left(x^3-3^3\right)+9x+9=15\)

⇔ \(x^3-9x^2+27x-27-x^3+27+9x+9=15\)

⇔ \(36x-9x^2+9=15\)

⇔ \(9x\left(4-x\right)=6\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 210 ) + ( x + 211 ) = 23632

x + 1 + x + 2 + x +3 + .... + x + 210 + x + 211 = 23632

( x + x + x + .... + x ) + ( 1 + 2 + 3 + ... + 210 + 211 ) = 23632

211x + 22366 = 23632

             211x  = 23632 - 22366

             211x  = 1266

=> x = 1266 : 211 = 6

Vậy x = 6

ta có (x+1) + (x+2) + (x+3) + ... + (x+210) + (x+211) = 23632

     = X + ( 1 + 2 +3 + 4 + .... + 210 + 211 ) = 23632

     = X + ( 1 + 211 ) + ( 2 + 210 ) + .... =23632

     = X + 212 x 105 =23632

     = X + 22260 = 23632

     = X          = 23632 - 22260

     = X          = 1372

khó nhìn quá

Đề là như này à bạn ?

\(\frac{-1}{7}\) . \(2^3\)- \(2^x\) : \(\frac{7}{3}\)= \(2^x\) - 1 

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