tìm các giá trị biểu thức có giá trị tại 0 . A= 2xy B= ( 2x – 3 ) . ( 4 x- 5) C =( 2x-1)mũ 2 + ( 3y-2) mũ 4 D = |x-1|+ |3y+3|

0 bình luận về “tìm các giá trị biểu thức có giá trị tại 0 . A= 2xy B= ( 2x – 3 ) . ( 4 x- 5) C =( 2x-1)mũ 2 + ( 3y-2) mũ 4 D = |x-1|+ |3y+3|”

1. Đáp án + giải thích bước giải :

   `A = 2xy`

   `⇔ 2xy = 0`

   `⇔` \(\left[ \begin{array}{l}2x=0\\y=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=0\\x=0\end{array} \right.\) 

   `B = (2x – 3) (4x – 5)`

   `⇔ (2x – 3) (4x – 5) = 0`

   `⇔` \(\left[ \begin{array}{l}2x-3=0\\4x-5=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}2x=3\\4x=5\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=\frac{5}{4}\end{array} \right.\) 

   `C = (2x – 1)^2 + (3y – 2)^4`

   `⇔ (2x – 1)^2 + (3y – 2)^4 = 0`

   Ta có :

   \(\left\{ \begin{array}{l}(2x-1)^2≥0∀x\\(3y-2)^4≥0∀y\end{array} \right.\)

   `⇔ (2x – 1)^2 + (3y – 2)^4 ≥0∀x,y`

   Dấu “`=`” xảy ra khi :

   `⇔` \(\left\{ \begin{array}{l}2x-1=0\\3y-2=0\end{array} \right.\)

   `⇔` \(\left\{ \begin{array}{l}x=\dfrac{1}{2}\\y=\dfrac{2}{3}\end{array} \right.\)

   `E = (x + 3)^2 + (y – 1)^2`

   `⇔ (x + 3)^2 + (y – 1)^2=0`

   Ta có :

   \(\left\{ \begin{array}{l}(x+3)^2≥0∀x\\(y-1)^2≥0∀y\end{array} \right.\)

   `⇔ (x+3)^2+(y-1)^2≥0∀x,y`

   Dấu “`=”` xảy ra khi :

   `⇔` \(\left\{ \begin{array}{l}x+3=0\\y-1=0\end{array} \right.\)

   `⇔` \(\left\{ \begin{array}{l}x=-3\\y=1\end{array} \right.\)

   Trả lời

2. $\text{Đáp án + Giải thích các bước giải:}$

   `a//2xy=0`

   `=>xy=0`

   `=>` \(\left[ \begin{array}{l}x=0\\y=0\end{array} \right.\) 

   `\text{Vậy}` `x=y=0`

   `b//(2x-3)(4x-5)=0`

   `=>` \(\left[ \begin{array}{l}2x-3=0\\4x-5=0\end{array} \right.\) 

   `=>` \(\left[ \begin{array}{l}2x=3\\4x=5\end{array} \right.\) 

   `=>` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{5}{4}\end{array} \right.\) 

   `\text{Vậy}` `x∈{(3)/(2);(5)/(4)}`

   `c//(2x-1)^{2}+(3y-2)^{4}=0`

   `\text{Vì}` $\left\{\begin{matrix}(2x-1)^2≥0& \\(3y-2)^{4}≥0& \end{matrix}\right.$

   `=>(2x-1)^{2}+(3y-2)^{4}≥0`

   `\text{Mà theo đề bài :}` `(2x-1)^{2}+(3y-2)^{4}=0`

   `=>` $\left\{\begin{matrix}(2x-1)^2=0& \\(3y-2)^{4}=0& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}2x-1=0& \\3y-2=0& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}2x=1& \\3y=2& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}x=\dfrac{1}{2}& \\y=\dfrac{2}{3}& \end{matrix}\right.$

   `\text{Vậy}` `(x;y)=((1)/(2);(2)/(3))`

   `d//|x-1|+|3y+3|=0`

   `\text{Vì}` $\left\{\begin{matrix}|x-1|≥0& \\|3y+3|≥0& \end{matrix}\right.$

   `=>|x-1|+|3y+3|≥0`

   `\text{Mà theo đề bài :}` `|x-1|+|3y+3|=0`

   `=>` $\left\{\begin{matrix}|x-1|=0& \\|3y+3|=0& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}x-1=0& \\3y+3=0& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}x=1& \\3y=-3& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}x=1& \\y=-1& \end{matrix}\right.$

   `\text{Vậy}` `(x;y)=(1;-1)`

   `e//(x+3)^{2}+(y-1)^{2}=0`

   `\text{Vì}` $\left\{\begin{matrix}(x+3)^2≥0& \\(y-1)^2≥0& \end{matrix}\right.$

   `=>(x+3)^{2}+(y-1)^{2}≥0`

   `\text{Mà theo đề bài :}` `(x+3)^{2}+(y-1)^{2}=0`

   `=>` $\left\{\begin{matrix}(x+3)^2=0& \\(y-1)^2=0& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}x+3=0& \\y-1=0& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}x=-3& \\y=1& \end{matrix}\right.$

   `\text{Vậy}` `(x;y)=(-3;1)`

   Trả lời