\(3^{2x-2}\).5\(^{2y}\) = 15\(^{x}\)

3\(^{2x-2}\).5\(^{2y}\) = 3\(^{x}\).5\(^{2y}\)

\(\begin{cases}2x-2=x\\ 2y=x\end{cases}\)

\(\begin{cases}2x-x=2\\ 2y=x\end{cases}\)

\(\begin{cases}x=2\\ 2y=2\end{cases}\)

\(\begin{cases}x=2\\ y=2:2\end{cases}\)

\(\begin{cases}x=2\\ y=1\end{cases}\)

Vậy (x; y) = (2; 1)

Câu 1: Đề thiếu

Câu 2: D

Câu 3: C

Câu 4: B

Câu 5: C

a. P= (-8.x^2.y-2.5+6x^2)-(-7x^2y+2x^2)

b. P= vế phải cộng vế trái là ra

a. P=-x^2.y-10+4.x^2

b.P= 7xy+x^2-11y

Ta có : 

\(\left|2x+3\right|\ge0\)

\(\left|2y-7\right|\ge0\)

\(\Rightarrow\)\(A=\left|2x+3\right|+\left|2y-7\right|+15\ge15\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}2x+3=0\\2y-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-3\\2y=7\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{-3}{2}\\y=\frac{7}{2}\end{cases}}}\)

Vậy GTNN của \(A\) là \(15\) khi \(x=\frac{-3}{2}\) và \(y=\frac{7}{2}\)

Chúc bạn học tốt ~ 

Rút gọn biểu thức : 3x (x-2) - 5x(1-x) -8(x^2 - 3)
 

a) Ta có: \(\left|2x+5\right|+\left|2x-3\right|=8\)

\(\Rightarrow\left|2x+5\right|+\left|3-2x\right|=8\)

Nhận thấy \(\left[{}\begin{matrix}\left|2x+5\right|\ge2x+5\forall x\\\left|3-2x\right|\ge3-2x\forall x\end{matrix}\right.\)

\(\Rightarrow\left|2x+5\right|+\left|3-2x\right|\ge2x+5+3-2x\forall x\)

\(\Rightarrow\left|2x+5\right|+\left|3-2x\right|\ge8\)

Dấu \("="\) xảy ra khi \(\left[{}\begin{matrix}2x+5\ge0\\3-2x\ge0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x\ge-5\\2x\le3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{-5}{2}\\x\le\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{-5}{2}\le x\le\dfrac{3}{2}\)

Vậy \(\dfrac{-5}{2}\le x\le\dfrac{3}{2}.\)

\(\dfrac{x}{3}=\dfrac{y-5}{7}=\dfrac{z+2}{3}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}\)

\(x+2y=5z\Leftrightarrow x+2y-5z=0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}=\dfrac{x+2y-10-5z-10}{3+14-15}\)

\(=\dfrac{-20}{2}=-10\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-30\\y=-65\\z=-32\end{matrix}\right.\)

Vậy...

\(C = 2.(x-y)+13x^3y^2.(x-y)+15.xy.\)

\((y-x) +1\)

\(C = 2.( x- y )+13x^3y^2.(x-y)-15.xy.\)

\(( x - y )+1\)

\(C = (x - y)(2 + 13x^3y^2 - 15 ) +1\)

\(C =(x- y)(13x^3y^2 - 13 )+ 1\)

C=2.(x−y)+13x3y2.(x−y)+15.xy.

\(\left(\right. y - x \left.\right) + 1\)

\(C = 2. \left(\right. x - y \left.\right) + 13 x^{3} y^{2} . \left(\right. x - y \left.\right) - 15. x y .\)

\(\left(\right. x - y \left.\right) + 1\)

\(C = \left(\right. x - y \left.\right) \left(\right. 2 + 13 x^{3} y^{2} - 15 \left.\right) + 1\)

\(C = \left(\right. x - y \left.\right) \left(\right. 13 x^{3} y^{2} - 13 \left.\right) + 1\) đây nhé