Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải:
a) \(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2+x=0\)
\(\Leftrightarrow2x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2}{12^2}-\dfrac{49^2}{12^2}=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2-49^2}{144}=0\)
\(\Leftrightarrow2x+\dfrac{961-2401}{144}=0\)
\(\Leftrightarrow2x+\dfrac{-1440}{144}=0\)
\(\Leftrightarrow2x+\left(-10\right)=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Mà \(x+\left(-\dfrac{31}{12}\right)^2=y^2\)
\(\Leftrightarrow5+\dfrac{961}{144}=y^2\)
\(\Leftrightarrow y^2=\dfrac{1681}{144}\)
\(\Leftrightarrow y=\pm\dfrac{41}{12}\)
Vậy ...
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Vì \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0;\forall x\)
và \(\left(y^2-\dfrac{1}{4}\right)^{10}\ge0;\forall y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
Ta có:
\(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Rightarrow2x+\frac{961}{144}=\frac{2401}{144}\)
\(\Rightarrow2x=\frac{2401}{144}-\frac{961}{144}\)
\(\Rightarrow2x=\frac{1440}{144}\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11)
TH1: x-7=0 => x=7 => 0^8=0^18 (TM)
TH2: x-7=1 => x=8 (TM)
TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại
KL: x = 7 hoặc x=8
( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
tới đây dễ òi
x+(-31/12)^2=(49/12)^2-x
x+x=(49/12)^2-(-31/12)^2
tính x
từ x tìm ra y
b)x(x-y):[y(x-y)]=3/10:(-3/50)=...
=>x/y=... =>x=...;y=...
\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
Xét \(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Rightarrow2x=\left(\frac{49}{12}\right)^2-\left(\frac{-31}{12}\right)^2=\frac{2401}{144}+\frac{961}{144}\)
\(\Rightarrow2x=\frac{3362}{144}\)
\(\Rightarrow x=\frac{3362}{144}.\frac{1}{2}=\frac{1681}{144}\)
Ta lai xét :
\(x+\left(\frac{-31}{12}\right)^2=y^2\)
\(\Rightarrow\frac{1681}{144}+\frac{-961}{144}=y^2\)
\(\Rightarrow\frac{720}{144}=y^2\)
\(\Rightarrow y^2=5\)
\(\Rightarrow y=2,236067977\)
X+(-31/12)^2 = (49/12)^2 -x=y
(-31/12)^2 - (49/12)^2 = -x-x = y
961/144 - 2410/144 = -2x
-10=-2x
10=2x
10:2=x
5=x
X+961/144=y^2
5+961/144=y^2
1681/144=y^2
=>y=41/144
Dấu phân số mình ký hiệu là / đó nha
Ta có:\(49-y^2\le49\Rightarrow12\left(x-2001\right)^2\le49\)
\(\Rightarrow\left(x-2001\right)^2\le4\)
\(\Rightarrow\left(x-2001\right)^2\in\left\{0;1;4\right\}\)
\(\Rightarrow x-2001\in\left\{0;1;2\right\}\)
\(\Rightarrow x\in\left\{2001;2002;2003\right\}\)