\(P=\dfrac{4}{5}\left(x+y\right)+\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)\ge\dfrac{4}{5}.10+2\sqrt{\dfrac{180x}{5x}}+2\sqrt{\dfrac{5y}{5y}}=22\)

\(P_{min}=22\) khi \(x=y=5\)

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)

a) Ta có: \(M=-x^2-4x+20\)

\(=-\left(x^2+4x-20\right)\)

\(=-\left(x^2+4x+4-24\right)\)

\(=-\left(x+2\right)^2+24\le24\forall x\)

Dấu '=' xảy ra khi x=-2

\(4^x=32^{10}\)

\(\Rightarrow4^x=4^{80}\)

\(\Rightarrow x=80\)

\(4^x=32^{10}\)

\(\Rightarrow2^{2x}=2^{50}\)

\(\Rightarrow2x=50\Rightarrow x=25\)

\(2\left(x-1\right)-5\left(x+2\right)=-10\\ \Rightarrow2x-2-5x-10=-10\\ \Rightarrow-3x=2\\ \Rightarrow x=\dfrac{-2}{3}\)

Vậy \(x=\dfrac{-2}{3}\)

Mấy câu này khá giống nhau nhé anh (câu 1 giống câu 4 và 5, cấu 2 giống câu 3) =)))

Câu 1: 2x - 7 + (x - 14) = 0

<=> 3x -21 = 0

<=> 3x = 21 => x = 7

Câu 2:

x² - 6x = 0 <=> x.(x - 6) = 0 => \(\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

Chúc anh học tốt !!!

Câu 1, 2 có người làm rồi nên mik làm tiếp cho mấy câu tiếp. Cứ áp dụng A.B = 0 => A = 0 hoặc B = 0

3; ( x - 3 )( 16 - 4x ) = 0

=> x - 3 = 0 hoặc 16 - 4x = 0

=> x = 3 hoặc x = 4

Vậy x = 3 hoặc x = 4.

4; ( x - 3 ) - ( 16 - 4x ) = 0

=> x - 3 - 16 + 4x = 0

=> ( x + 4x ) - ( 3 + 16 ) = 0

=> 5x - 19 = 0

=> x = 19/5

Vậy x = 19/5

5; ( x + 3 ) + ( 16 - 4x ) = 0

=> x + 3 + 16 - 4x = 0

=> ( x - 4x ) + ( 16 + 3 ) = 0

=> 3x + 19 = 0

=> x = 19/3

Vậy x = 19/3

-14 -x =-25

x=-14-(-25)

x=11

(-14)-x=(-10)-(-15)

(-14)-x=5

x=-14-5

x=-19

x=11

a) \(\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)

\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)

\(=\dfrac{16+x-18}{x^2-2x}\)

\(=\dfrac{x-2}{x\left(x-2\right)}\)

\(=\dfrac{1}{x}\)

a) \(\sqrt[]{x^2-2x+4}=2x-2\)

\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)

Giải pt \(3x^2-6x=0\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=2\)

c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

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