1, CÓ( X+1,5)⁸ VÀ (2,7 -Y)¹²> HOẶC = 0

         MÀ (X+1,5)⁸ + (2,7-Y)¹²  =0

    SUY RA   \(\hept{\begin{cases}X+1,5=0\\2,7-Y=0\end{cases}}\)

      SUY RA\(\hept{\begin{cases}X=-1,5\\Y=2,7\end{cases}}\)

\(\left(x+1.5\right)^8+\left(2.7-y\right)^{10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1.5=0\\2.7-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1.5\\y=2.7\end{matrix}\right.\)

Vậy : phương trình có cặp nghiệm \(\left(x,y\right)=\left(-1.5,2.7\right)\)

(x+1,5)⁸+(2,7-y)¹⁰=0

⇒\(\left[{}\begin{matrix}\left(x+1,5\right)^8\\\left(2,7-y\right)^{10}\end{matrix}\right.=0\)

⇒\(\left[{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\) ⇒\(\left[{}\begin{matrix}x=0-1,5\\y=2,7-0\end{matrix}\right.\) ⇒\(\left[{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

Ta có: \(\left(x+1.5\right)^8\ge0\forall x\)

\(\left(2,7-y\right)^{10}\ge0\forall y\)

Do đó: \(\left(x+1.5\right)^8+\left(2.7-y\right)^{10}\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-1,5;2,7\right)\)

=>x+1,5=0 và 2,7-y=0

=>x=-1,5(loại) và y=2,7(loại)

Vậy: Không có cặp số nguyên x,y nào thỏa mãn

=>x+1,5=0 và 2,7-y=0

=>x=-1,5(loại) và y=2,7(loại)

Vậy: Không có cặp số nguyên x,y nào thỏa mãn

a, đề sai

b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=-1\\x-3=0\\x-3=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)

Vậy .....

c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{12}=0\)

Vì: \(\left\{{}\begin{matrix}\left(x+1,5\right)^8\ge0\forall x\\\left(2,7-y\right)^{12}\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\) để bt = 0

=>\(\left\{{}\begin{matrix}\left(x+1,5\right)^8=0\\\left(2,7-y\right)^{12}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

Vậy.............

\(\left(x-1,2\right)^2=4\)

⇔\(x^2-2.x.1,2+1,2^2=4\)

⇔\(x^2-2,4x+1,44=4\)

⇔\(x^2-2,4x=4-1,44\)

⇔\(x\left(x-2,4\right)=2,56\)

⇔\(x=2,56\) hoặc \(x-2,4=2,56\)

⇔\(x=2,56\) hoặc \(x=4,96\)

a) \(\left(x-1,2\right)^2=4=2^2\)

\(\Leftrightarrow x-1,2=4\)

\(\Leftrightarrow x=5,2\)

b) \(\left(x+1\right)^3=-125=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=-6\)

c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

b: \(\Leftrightarrow3x-36-5x-175+x-15=0\)

\(\Leftrightarrow-x-226=0\)

hay x=-226

Ta có :

\(\left\{{}\begin{matrix}\left(x+1,5\right)^8\ge0\\\left(2,7-y\right)^{12}\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+1,5\right)^8+\left(2,7-y\right)^{12}\ge0\)

Mà \(\left(x+1,5\right)^8+\left(2,7-y\right)^{12}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1,5\right)^8=0\\\left(2,7-y\right)^{12}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

Vậy...