tìm min \(\left(\frac{\left(t-\frac{1}{2}\right)^2-\frac{5}{4}}{t-2}\right)\)trong khoảng [-1,1]
1.tìm max A=(\(\frac{x}{x+2020}\))\(^2\) với x>0
2. tìm min C= \(\frac{\left(4x+1\right)\left(4+x\right)}{x}\) với x dương
3.cho 3a+5b=12. tìmmin B=ab
4.tìm min \(x^2-x+4+\frac{1}{x^2-x}\)
5. cho x,y là 2 số thỏa mãn \(2x^2+\frac{1}{x^2}+\frac{y}{4}=4\).tìm min max của xy
6. cho a,b>0 và a+b=1. tìm min M=\(\left(1+\frac{1}{a}\right)^2\left(1+\frac{1}{b}\right)^2\)
tìm min \(\frac{\left(t-1\right)^2+t}{-t}\) t thuộc [-1,1].help
Tìm MIn \(P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\)
Trong đó a,b,c dương t/mãn a+b+c=<3/2
Gọi S có n số hạng sao cho S = 1+ 2+ 3 + ...+ n = aaa ( a là chữ số)
=> (n + 1).n : 2 = a.111
=> n(n + 1) = a.222
=> n(n + 1) = a.2.3.37
a là chữ số mà n; n + 1 là hai số tự nhiên liên tiếp nên a = 6
=> n(n + 1) = 36.37
=> n = 36
Vậy cần 36 số hạng
cho mình nha
Áp dụng BĐT AM-GM ta có:
\(\frac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\frac{1}{2}\)
Áp dụng BĐT Holder ta có:
\(VT=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\)
\(\ge\left(3+\frac{1}{\sqrt[3]{abc}}+\frac{1}{\sqrt[3]{abc}}\right)^3\ge(3+2+2)^3=343\)
Xảy ra khi \(a=b=c=\frac{1}{2}\)
Bài 1: Tính a) \(\left(\frac{11}{12}:\frac{44}{16}\right)\cdot\left(\frac{-1}{3}+\frac{1}{2}\right)\) b) \(\frac{\left(-5^2\right)\cdot\left(-5\right)^3\cdot16}{5^4\cdot\left(-2\right)^4}\) c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\)d) \(\left(\frac{-1}{2}+\frac{1}{3}\right)\cdot\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{4}{5}\)
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45
Bài 1:
1. Tính: \(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
2. Tìm và tính tổng các số nguyên x thỏa mãn: \(\frac{21}{5}\left|x\right|< 2019\)
3. Tìm x, biết: \(\frac{2^{24}\left(x-3\right)}{\left(3\frac{5}{7}-1,4\right)\left(6\cdot2^{24}-4^{13}\right)}=\left(\frac{5}{3}\right)^2\)
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)
\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)
\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)
\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)
\(\Rightarrow-480\le x\le480\)
\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)
\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)
\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)
\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)
\(\Leftrightarrow x-3=\frac{90}{7}\)
\(\Rightarrow x=\frac{111}{7}\)
Bài 4: Tính hợp lý
a) \(A=\left(\left|\frac{-3}{4}\right|+\left|\frac{-2}{5}\right|\right):\frac{3}{7}+\left(\frac{-3}{5}+\left|\frac{-1}{4}\right|\right):\frac{3}{7}\)
b) \(B=2\frac{5}{23}-\left(\frac{-7}{9}\right)-\left|\frac{-5}{23}\right|+\frac{12}{9}+\left|-0,75\right|\)
Bài 5 : Tìm x :
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\left|\frac{-2}{3}\right|\)
b) \(\left|x\right|:\left(\frac{1}{9}-\frac{2}{5}\right)=\frac{-1}{2}\)
c) \(\left(x-\frac{1}{5}\right).\left(1\frac{3}{5}+2x\right)=0\)
Tìm min của \(B=\frac{1}{\left(x+1\right)^2}+\frac{4}{\left(y+2\right)^2}+\frac{8}{\left(z+3\right)^2}.\)
cho \(a+b+c\le3b;a,b,c\ge0\) tìm Min \(A=\frac{1}{\left(a+1\right)^2}+\frac{4}{\left(b+2\right)^2}+\frac{8}{\left(c+3\right)^2}\)
cho a^2+b^2+c^2 <= 3b nhé
Tìm x,y biết
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
Giúp t vs ạ
phá ngoặc tính BT , nên kết quả sẽ ko ra con số nhận định !!! tui thử thui nha bà !
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)
\(3x+y-\frac{47}{12}=\frac{1}{4}\)
\(3x+y=\frac{25}{6}\)
\(3x=\frac{25}{6}-y\)
\(x=\frac{25-25y}{18}\)
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)
\(3x+y-\frac{47}{12}=\frac{1}{4}\)
\(3x+y=\frac{25}{6}\)
\(y=\frac{25}{6}-3x\)
Vậy \(x=\frac{25-25y}{18}\)
\(y=\frac{25}{6}-3x\)
Ta có:
\(|x+\frac{1}{2}|\ge x+\frac{1}{2}\forall x;|x+\frac{1}{3}|\ge x+\frac{1}{3}\forall x;|y-5|\ge y-5\forall y;|x+\frac{1}{4}|\ge x+\frac{1}{4}\forall x\)
\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)
Mà \(|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|=\frac{1}{4}\)
\(\Rightarrow\frac{1}{4}\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)
\(\Rightarrow\frac{1}{4}\ge3x+y-\frac{47}{12}\)
\(\Rightarrow3x+y\le\frac{25}{6}\)
\(\Rightarrow x\le\frac{\frac{25}{6}-y}{3}\)
Thay vào tính y
Làm phiền bạn Quỳnh
Bạn bảo thay vào tính y? Vậy bạn trình bày cho mình phần cuối.
Bạn đang nhầm vấn đề nhé. (2 dòng cuối)
Vd: a + b = 5
<=> a = 5 - b
Thay a vào: 5 - b + b = 5
<=> 0 = 0 ???
Bạn tính ra r bạn đưa lên bth bên trên??? Mình thấy bạn làm vậy thì k thể tìm ra a, b. Có thể mình k hiểu ý bạn. Nếu v bạn trình bày giúp mình phần cuối vì mình không hiểu ý bạn. Thankbạn.