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

Ta có: \(\hept{\begin{cases}\left(2x+1\right)^{2008}\ge0\forall x\\|3y-1|^{2007}\ge0\forall y\end{cases}}\)\(\Rightarrow\left(2x+1\right)^{2008}+|3y-1|^{2007}\ge0\forall x,y\)

Do đó \(\left(2x+1\right)^{2008}+|3y-1|^{2007}=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=0\\3y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{1}{3}\end{cases}}}\)

Vậy \(\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{1}{3}\end{cases}}\)

ko hiểu thì hỏi nhá 

1/

Ta có: \(x-y=xy\Rightarrow x=xy+y=y\left(x+1\right)\Rightarrow x:y=x+1\left(y\ne0\right)\)

Mà x - y =  x:y

\(\Rightarrow x-y=x+1\Rightarrow-y=1\Rightarrow y=-1\)

Thay y = -1 vào x - y = xy ta được:

\(x-\left(-1\right)=x.\left(-1\right)\Rightarrow x+1=-x\Rightarrow2x=-1\Rightarrow x=\frac{-1}{2}\)

Vậy...

2/ tương tự bài 1    x = 1/2, y = -1

\(a,\left(y^{54}\right)^2=y\)\(\Rightarrow y^{108}=y\)\(\Rightarrow y=\pm1\)

\(b,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)

\(\Rightarrow x\in\left\{0;1;2\right\}\)

\(c,x\left(6-x\right)^{2019}=\left(6-x\right)^{2019}\)

\(\Rightarrow\left(6-x\right)^{2019}\left(x-1\right)=0\)

\(\Rightarrow x\in\left\{1;6\right\}\)

\(\left(y^{54}\right)^2=y\)

\(\Rightarrow y^{108}=y\)

\(\Rightarrow y^{108}-y=0\)

\(\Rightarrow y\cdot\left(y^{107}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^{107}-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^{107}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

a) \(\left(y^{54}\right)^2=y\)

\(\Leftrightarrow y^{108}=y\)

\(\Leftrightarrow y^{108}-y=0\)

\(\Leftrightarrow y.\left(y^{107}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\\y^{107}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)

Vậy \(y\in\left\{0;1\right\}\)

Ta có: (x-2) (xy-1) = 5

Suy ra: x-2; xy-1 thuộc Ư₍₅₎={-1; 1; -5; 5}

Lập bảng:

Vậy(x;y) = (1; -4) ; (-3 ; 0) ; (3 ; 2)

Bạn ơi chứng minh nhỏ hơn hoặc bằng 0 nhé

\(=-y^{2018}-\left(x^2-x+1\right)\)

\(=-y^{2018}-\left(x+1\right)^2\)

Vì \(\hept{\begin{cases}-y^{2018}\le0;\forall x,y\\-\left(x+1\right)^2\le0;\forall x,y\end{cases}}\)

\(\Rightarrow-y^{2018}-\left(x+1\right)^2\le0;\forall x,y\left(đpcm\right)\)

\(xy=x-y+3\)

\(\Leftrightarrow xy-x+y=3\)

\(\Leftrightarrow x\left(y-1\right)+\left(y-1\right)=2\)

\(\Leftrightarrow\left(x+1\right)\left(y-1\right)=2\)

\(\Leftrightarrow x+1;y-1\inƯ\left(2\right)\)

Ta có: \(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)

Lập bảng:

Vậy các cặp số nguyên (x; y) thỏa mãn là (-2; -1); (0;3); (-3; 0) và (1; 2)

Ta có: \(x^4+y^4+\left(x+y\right)^4\)

\(=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

\(=2\left(x^4+y^4+2x^2y^2\right)+4xy\left(x+y\right)+2x^2y^2\)

\(=2\left[\left(x^2+y^2\right)+2xy\left(x+y\right)+x^2y^2\right]\)

\(=2\left(x^2+xy+y^2\right)^2\left(đpcm\right)\)

Em xem lại dòng thứ 3 và 4, chưa đúng rồi em !