Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left|x-1\right|>=0\forall x;\left(x+y-2\right)^{2024}>=0\forall x,y\)
Do đó: \(\left|x-1\right|+\left(x+y-2\right)^{2024}>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\x+y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x+2=-1+2=1\end{matrix}\right.\)
Thay x=1;y=1 vào Q, ta được:
\(Q=1^{2024}+1^{2024}=1+1=2\)
\(\left|x-1\right|+\left(x+y-2\right)^{2024}=0\)
Do \(\left|x-1\right|\ge0;\left(x+y-2\right)^{2024}\ge0,\forall x;y\in R\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(Q=x^{2024}+y^{2024}=1^{2024}+1^{2024}=2\)
Lấy mẫu số, chứ không phải lấy tử số em nhé!
Độ dài cạnh huyền là:
\(\sqrt{3^2+7^2}=\sqrt{9+49}=\sqrt{58}\left(cm\right)\)
Bình phương cạnh huyền là:
32 + 72 = 58(cm2)
Cạnh huyền là: \(\sqrt{58}\) m
1: \(\left(x^2+2xy-3\right)\left(-xy^2\right)\)
\(=-xy^2\cdot x^2-xy^2\cdot2xy+3\cdot xy^2\)
\(=-x^3y^2-2x^2y^3+3xy^2\)
2: \(3x\left(x+2\right)-3x^2-12=0\)
=>\(3x^2+6x-3x^2-12=0\)
=>6x-12=0
=>6x=12
=>x=2
3: \(\left(2x^3-\dfrac{9}{2}x^2+\dfrac{1}{xy}\right)\cdot x^2y^3\)
\(=2x^3\cdot x^2y^3-\dfrac{9}{2}x^2\cdot x^2y^3+\dfrac{x^2y^3}{xy}\)
\(=2x^5y^3-\dfrac{9}{2}x^4y^3+xy^2\)
2; 3\(x\)(\(x+2\)) - 3\(x^2\) - 12 = 0
3\(x^2\) + 6\(x\) - 3\(x^2\) - 12 = 0
(3\(x^2\) - 3\(x^2\)) + 6\(x\) - 12 = 0
0 + 6\(x\) - 12 = 0
6\(x\) = 12
\(x\) = 12 : 6
\(x=2\)
Vậy \(x=2\)
Bài 1:Ta có: \(\widehat{xAm}+\widehat{yAn}=130^0\)
mà \(\widehat{xAm}=\widehat{yAn}\)(hai góc đối đỉnh)
nên \(\widehat{xAm}=\widehat{yAn}=\dfrac{130^0}{2}=65^0\)
Ta có: \(\widehat{xAm}+\widehat{xAn}=180^0\)(hai góc kề bù)
=>\(\widehat{xAn}+65^0=180^0\)
=>\(\widehat{xAn}=115^0\)
=>\(\widehat{yAm}=115^0\)
Bài 2:
Ta có: \(\widehat{xOz}+\widehat{zOt}+\widehat{tOy}=220^0\)
=>\(\widehat{xOz}+\widehat{tOy}=40^0\)
mà \(\widehat{xOz}=\widehat{tOy}\)(hai góc đối đỉnh)
nên \(\widehat{xOz}=\widehat{tOy}=\dfrac{40^0}{2}=20^0\)
Ta có: \(\widehat{xOz}+\widehat{xOt}=180^0\)(hai góc kề bù)
=>\(\widehat{xOt}=180^0-20^0=160^0\)
=>\(\widehat{yOz}=160^0\)
Bài 1:
a: \(\dfrac{2}{3}-\dfrac{7}{6}+\dfrac{5}{2}=\dfrac{4}{6}-\dfrac{7}{6}+\dfrac{15}{6}=\dfrac{12}{6}=2\)
b: \(9-2023^0+\sqrt{\dfrac{1}{25}}=9-1+\dfrac{1}{5}=8+\dfrac{1}{5}=8,2\)
c: \(\dfrac{4^{1010}\cdot9^{1010}}{3^{2019}\cdot16^{504}}=\dfrac{4^{1010}}{4^{1008}}\cdot\dfrac{3^{2020}}{3^{2019}}=\dfrac{3}{4^8}\)
Bài 3:
Tổng số tiền phải trả cho 1 bánh cỡ to, 2 bánh cỡ vừa, 1 bánh cỡ nhỏ là:
\(300000+250000\cdot2+200000=1000000\left(đồng\right)\)
=>bác Lan đủ tiền mua
Bài 2:
a: \(x-0,5=\dfrac{5}{6}\)
=>\(x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{5}{6}+\dfrac{3}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)
b: \(\left|x-1\right|+\dfrac{1}{2}=\dfrac{3}{2}\)
=>\(\left|x-1\right|=\dfrac{3}{2}-\dfrac{1}{2}=\dfrac{2}{2}=1\)
=>\(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
\(|x^2|x+\dfrac{3}{4}||=x^2\)
=>\(x^2\cdot\left|x+\dfrac{3}{4}\right|=x^2\)
=>\(\left|x+\dfrac{3}{4}\right|=1\)
=>\(\left[{}\begin{matrix}x+\dfrac{3}{4}=1\\x+\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
|\(x^2\).|\(x+\dfrac{3}{4}\)| |= \(x^2\)
\(x^2\).|\(x+\dfrac{3}{4}\)| = \(x^2\)
\(x^2\).|\(x+\dfrac{3}{4}\)| - \(x^2\) = 0
\(x^2\).(|\(x+\dfrac{3}{4}\)| - 1) = 0
\(\left[{}\begin{matrix}x=0\\\left|x+\dfrac{3}{4}\right|=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x+\dfrac{3}{4}=-1\\x+\dfrac{3}{4}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{7}{4}\); 0; \(\dfrac{1}{4}\)}
\(\dfrac{5}{x}-\dfrac{2}{y}=\dfrac{3}{2}\)
=>\(\dfrac{5x-2y}{xy}=\dfrac{3}{2}\)
=>2(5x-2y)=3xy
=>10x-4y-3xy=0
=>10x-3xy-4y=0
=>x(10-3y)-4y=0
=>\(-3x\left(y-\dfrac{10}{3}\right)-4y+\dfrac{40}{3}=0\)
=>\(-3x\left(y-\dfrac{10}{3}\right)-4\left(y-\dfrac{10}{3}\right)=0\)
=>\(\left(-3x-4\right)\left(y-\dfrac{10}{3}\right)=0\)
=>\(\left\{{}\begin{matrix}-3x-4=0\\y-\dfrac{10}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\y=\dfrac{10}{3}\end{matrix}\right.\)
\(\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|>=0\forall x\)
=>\(\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|-\dfrac{2}{11}>=-\dfrac{2}{11}\forall x\)
Dấu '=' xảy ra khi \(\dfrac{4}{3}x-\dfrac{1}{4}=0\)
=>\(\dfrac{4}{3}x=\dfrac{1}{4}\)
=>\(x=\dfrac{1}{4}:\dfrac{4}{3}=\dfrac{3}{16}\)
A = |\(\dfrac{4}{3}\)\(x\) - \(\dfrac{1}{4}\)| - \(\dfrac{2}{11}\)
Vì |\(\dfrac{4}{3}\)\(x\) - \(\dfrac{1}{4}\)| ≥ 0 ∀ \(x\)
|\(\dfrac{4}{3}x\) - \(\dfrac{1}{4}\)| - \(\dfrac{2}{11}\) ≥ - \(\dfrac{2}{11}\) dấu bằng xảy ra khi : \(\dfrac{4}{3}x\) - \(\dfrac{1}{4}\) = 0
⇒ \(\dfrac{4}{3}\)\(x\) = \(\dfrac{1}{4}\) ⇒ \(x\) = \(\dfrac{1}{4}\) : \(\dfrac{4}{3}\) ⇒ \(x\) = \(\dfrac{3}{16}\)
Vậy giá trị nhỏ nhất của biểu thức là - \(\dfrac{2}{11}\) khi \(x=\dfrac{3}{16}\)
\(a,-0,25+\dfrac{2}{3}=-\dfrac{3}{4}+\dfrac{2}{3}=-\dfrac{9}{12}+\dfrac{8}{12}=-\dfrac{1}{12}\\ b,1\dfrac{4}{23}+\dfrac{-5}{21}-\dfrac{4}{23}+0,5-\dfrac{16}{21}\\ =\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(-\dfrac{5}{21}-\dfrac{16}{21}\right)+0,5\\ =\dfrac{23}{23}-\dfrac{21}{21}+0,5\\ =1-1+0,5\\ =0,5\\ c,2-\left[\left(1-\dfrac{1}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1\dfrac{4}{9}-2024^0\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-\dfrac{13}{9}-1\right]\\ =2-\left[\dfrac{4}{9}-\dfrac{13}{9}-\dfrac{9}{9}\right]\\ =2-\left(-2\right)\\ =4\)
\(a,-0,25+\dfrac{2}{3}\\ =-\dfrac{1}{4}+\dfrac{2}{3}\\ =\dfrac{-3}{12}+\dfrac{8}{12}\\ =\dfrac{5}{12}\\ b,1\dfrac{4}{23}+\dfrac{-5}{21}-\dfrac{4}{23}+0,5-\dfrac{16}{21}\\ =1+\left(\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{-5}{21}-\dfrac{16}{21}\right)+\dfrac{1}{2}\\ =1+\dfrac{-21}{21}+\dfrac{1}{2}\\ =1-1+\dfrac{1}{2}\\ =\dfrac{1}{2}\\ c,2-\left[\left(1-\dfrac{1}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1\dfrac{4}{9}-2024^0\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1-\dfrac{4}{9}-1\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^2-2-\dfrac{4}{9}\right]\\ =2-\left(\dfrac{4}{9}-2-\dfrac{4}{9}\right)\\ =2+2\\ =4\)