a, 3.2^x+1-2=94

B,  (3x-1)³=125

C,  2^x+2^x+1+...........+2^x+99=2¹⁰⁰-1

. là dấu nhân 

MIK CẦN GẤP

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\hept{\begin{cases}x-\left(y+z\right)=\frac{-1}{12}\\y-\left(x+z\right)=\frac{-1}{2}\\z-\left(x+y\right)=\frac{-7}{12}\end{cases}}\)

\(\Leftrightarrow-\left(x+y+z\right)=\frac{-7}{6}\)

\(\Leftrightarrow\hept{\begin{cases}-\left(x+y\right)=z-\frac{7}{6}\\-\left(x+z\right)=y-\frac{7}{6}\\-\left(y+z\right)=x-\frac{7}{6}\end{cases}}\)

Thay vô tinh tiếp, đc chứ??

Ta suy ra :x - (y + z) + y - (x + z) + z - (x+y) = \(\frac{-1}{12}\)+\(\frac{-1}{2}\)+\(\frac{-7}{12}\)= \(\frac{-7}{6}\)

=> x - y -z + y - x - z + z - x - y = -(x + y + z)= \(\frac{-7}{6}\)

Ta có x - (y + z) = x + (-(y +z)) = -(y + z) - x + 2x = -(x + y + z) + 2x = \(\frac{-7}{6}\)+ 2x = \(\frac{-1}{12}\)

=> 2x = \(\frac{13}{12}\)

x =\(\frac{13}{12}\) : 2 = \(\frac{13}{24}\)

Tương tự : y - (z + x) = - (x + y + z) + 2y = \(\frac{-7}{6}\)+ 2y = \(\frac{-1}{2}\)

=> 2y = \(\frac{4}{6}\)=\(\frac{2}{3}\)

=> y = \(\frac{2}{3}\): 2 = \(\frac{2}{6}\)=\(\frac{1}{3}\)

Vậy : - (x + y + z) = -(\(\frac{13}{24}\)+ \(\frac{1}{3}\)+ z) = -(\(\frac{7}{8}\)+ z) = -\(\frac{7}{6}\)

=> z = \(\frac{7}{6}\)- \(\frac{7}{8}\)= \(\frac{7}{24}\)

lười học thế

suốt ngày chép mạng

d, gt=> (x-3)³-(x-3)=0

<=>(x-3)(x²-6x+9-1)=0 <=>x=3 hoặc x=4 hoặc x=2

bn o dau minh moi biet de lam

\(\left(x-2\right)^5-\left(x-2\right)^3=0\)

\(\Rightarrow\left(x-2\right)^3\left(\left(x-2\right)^2-1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\\left(x-2\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

Vậy \(x\in\left\{1;2;3\right\}\)

⇒ ( x - 2)³ . (x - 2)² - (x - 2)³  . 1 = 0                                                          ⇒ ( x - 2)³ . [( x - 2)² - 1] = 0 

Bạn Nguyễn Đức Trí ơi, cảm ơn bạn đã trả lời giúp mik nhưng mik thấy cô mik dạy là x chỉ thuộc { 2 ; 3 } thôi nhé bạn!

11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)

12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

13)

\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)

14)

\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)

\(x.(x^2-2x+1)=x(x-1)^2\)

\(\left(x+1\right)^2+\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x^2+1\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x^2=-1\end{cases}\Leftrightarrow}x=-1}\)

Vậy x=-1