7,3 x Y + Y x 2,7=5

Y x ( 7,3 + 2,7 ) =5

Y x 10 = 5

        Y = \(\frac{5}{10}=\frac{1}{2}\)

Vay............................

7,3 x Y + Y x 2,7=5

Y x (7,3 + 2,7) = 5

Y x 10 = 5

Y         = 5 : 10

Y         = 0,5.

Hok tốt ^_^ !

7,3 x Y + Y x 2,7=5
Y x ( 7,3 + 2,7 ) =5
Y x 10 = 5=> y=1/2

\(xy-5x+y=17\)

\(\Leftrightarrow x\left(y-5\right)+\left(y-5\right)=12\)

\(\Leftrightarrow\left(y-5\right)\left(x+1\right)=12\)

Đến đây bạn lập bảng làm nốt nha !

\(xy-5x+y=17\)

\(=>x.\left(y-5\right)+\left(y-5\right)=12\)

\(=>\left(x+1\right)\left(y-5\right)=12\)

\(=>x+1;y-5\in\)cặp \(Ư\left(15\right)\)

Ta có bảng sau :

Vậy ...

\(2010x^2+2011y^2-4020x+4022y+4021=0\)

\(\Leftrightarrow\left(2010x^2-4020x+2010\right)+\left(2011y^2+4022y+2011\right)=0\)

\(\Leftrightarrow2010\left(x^2-2x+1\right)+2011\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow2010\left(x-1\right)^2+2011\left(y+1\right)^2=0\)

Vì \(2010\left(x-1\right)^2\ge0\forall x;2011\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow2010\left(x-1\right)^2+2011\left(y+1\right)^2\ge0\)

Để \(2010\left(x-1\right)^2+2011\left(y+1\right)^2=0\Leftrightarrow\hept{\begin{cases}2010\left(x-1\right)^2=0\\2011\left(y+1\right)^2=0\end{cases}}\)

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

Vậy ..........

mk biết nè k mk đi mk nói cho

sorry mk ko biet

bik có biết nhưng bạn ý off lâu lắm rồi ak

3^x+2-5.3^x=36

3^x(3^2-5)=36

3^x.4=36

3^x=9

x=2

\(3^{x+2}-5\cdot3^x=36\)

\(3^x\cdot3^2-5\cdot3^x=36\)

\(3^x\left(3^2-5\right)=36\)

\(3^x\left(9-5\right)=36\)

\(3^x\cdot4=36\)

\(3^x=36:4\)

\(3^x=9\)có \(3^2=9\)

\(\Rightarrow x=2\)

\(\Rightarrow x\in\left\{2\right\}\)

x³ - 2x² + 6x = 12

x³ - 2x² + 6x - 12 = 0

x²(x - 2) + 6(x - 2)=0

(x - 2)(x² + 6) = 0

\(\Leftrightarrow \begin{bmatrix} x - 2 = 0 & & \\ x^{2} + 6 = 0& & \end{bmatrix}\) bỏ dấu ngoặc bên phải nha pn

\(\Leftrightarrow \begin{bmatrix} x = 2 & & \\ x^{2} = - 6 & & \end{bmatrix}\) không tìm được giá trị của x (pn ghi cái này kế pn chỗ x² = - 6 nhé

Vậy x = 2

\(x^3-2x^2+6x=12\)

\(\Rightarrow\) \(x^3-2x^2+6x-12=0\)

\(\Rightarrow x^2\left(x-2\right)+6\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

Vậy $x=2$

\(x^3-2x^2+6x=12\)

\(\Rightarrow x^3-2x^2+6x-12=0\)

\(\Rightarrow x^2\left(x-2\right)+6\left(x-2\right)-0\)

\(\Rightarrow\left(x^2+6\right)\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+6=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)

Vậy \(x=2\)

Cái đề bài k hỉu j luôn sao lại bằng 1 và 1/3 !!

( 1/3 + 1/15 + 1/35 + 1/63 ) * x = 1 + 1/3

( 1/1x3 + 1/3x5 + 1/5x7 + 1/7x9 ) * x = 1 +1/3

1/2 x ( 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 ) * x = 1 + 1/3

 1/2 x ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 ) * x = 1 + 1/3

1/2 x ( 1 - 1/9 ) * x = 4/3

1/2 x 8/9 * x = 4/3

4/9 * x = 4/3

x = 4/3 : 4/9

x = 9/3

x =3 

Vậy x = 3