\(\dfrac{2\dfrac{1}{2}x-1}{\dfrac{2}{3}}=\dfrac{-\dfrac{2}{3}}{1-2\dfrac{1}{2}x}\)
hu hu giúp vs, mai nộp rùi
\(\dfrac{2\dfrac{1}{2}x-1}{\dfrac{2}{3}}=\dfrac{-\dfrac{2}{3}}{1-2\dfrac{1}{2}x}\)
hu hu giúp vs, mai nộp rùi
\(\frac{2\frac{1}{2}x-1}{\frac{2}{3}}=\frac{\frac{-2}{3}}{1-2\frac{1}{2}x}\) ĐKXĐ \(x\ne\frac{2}{5}\)
\(\Leftrightarrow\)\(\frac{\frac{5}{2}x-1}{\frac{2}{3}}=\frac{\frac{2}{3}}{\frac{5}{2}x-1}\)\(\Leftrightarrow\)\(\left(\frac{5}{2}x-1\right)^2=\frac{4}{9}\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+1=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+\frac{5}{9}=0\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-\frac{25}{6}x-\frac{5}{6}x+\frac{5}{9}=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{4}x^2-\frac{25}{6}x\right)-\left(\frac{5}{6}x-\frac{5}{9}\right)=0\)\(\Leftrightarrow\)\(\frac{25}{2}x\left(\frac{1}{2}x-\frac{1}{3}\right)-\frac{5}{3}\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{2}x-\frac{5}{3}\right)\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{2}{15}\end{cases}}\)
Tìm GTNN của :
A = \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\) ( x , y , z > 0 )
Biết : x2 + y2 + z2 ≤ 3
Làm ơn giúp vs hu hu
Áp dụng BĐT :
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ≥ 9
Trong đó : a = xy ; b = yz ; c = xz
⇒ ( xy + yz + xz )\(\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\) ≥ 9 ( * )
Áp dụng BĐT cô - si :
x2 + y2 ≥ 2xy ( x > 0 ; y > 0) ( 1 )
y2 + z2 ≥ 2yz ( y > 0 ; z > 0 ) ( 2)
z2 + x2 ≥ 2xz ( z >0 ; x > 0) ( 3)
Cộng từng vế của ( 1 ; 2 ; 3) ⇒ x2 + y2 + z2 ≥ xy + yz + xz ( **)
Từ ( * ; **)
⇒(x2 + y2 + z2).A ≥ ( xy + yz + xz). A ≥ 9
⇒ 3A ≥ 9
⇒ A ≥ 3
⇒ AMIN = 3 ⇔ x = y = z
\(\dfrac{7}{1\cdot2}+\dfrac{7}{2\cdot3}+\dfrac{7}{3\cdot4}+...+\dfrac{7}{99\cdot100}\)
giúp mình với mai mình phải nộp rùi!!
Ta đặt
\(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)
\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)
\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)
\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)
\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)
\(A=\dfrac{693}{100}\)
= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)
= 7.(1 - 1/100)
= 7 . 99/100
= 693/100
\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(\Rightarrow A=7x\dfrac{99}{100}=6,93\)
\(\dfrac{x+3}{x+1}\)-2=\(\dfrac{1-x}{x}\)
CÍU VS MAI NỘP CMNR =(((((
Lời giải:
Điều kiện: $x\neq 0; -1$
$\frac{x+3}{x+1}-2=\frac{1-x}{x}$
$1+\frac{2}{x+1}-2=\frac{1}{x}-1$
$\frac{2}{x+1}-1=\frac{1}{x}-1$
$\frac{2}{x+1}=\frac{1}{x}$
$\Rightarrow 2x=x+1$
$\Leftrightarrow x=1$ (thỏa mãn)
Cho M =\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\)
\(Help\)\(me\) Hu hu hu !
tính M hay chứng minh M ko là stn hay đầu bài là j vậy bn????
Cho D=\(\dfrac{5}{6X37}\)+\(\dfrac{1}{6X43}\)+\(\dfrac{6}{7X43}\)+\(\dfrac{10}{7X59}\)
và E=\(\dfrac{8}{9X37}\)+\(\dfrac{2}{9X47}\)+\(\dfrac{3}{10X47}\)+\(\dfrac{9}{10X59}\)
Em hãy tính \(\dfrac{D}{E}\)
giúp minh với mai mình phải nộp rùi
=>\(D=7\left(\dfrac{5}{42\cdot37}+\dfrac{1}{42\cdot43}+\dfrac{6}{43\cdot49}+\dfrac{10}{49\cdot59}\right)\)
\(=7\left(\dfrac{1}{37}-\dfrac{1}{42}+\dfrac{1}{42}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{59}\right)\)
=7(1/37-1/59)
=7*22/2183
\(E=5\left(\dfrac{8}{37\cdot45}+\dfrac{2}{45\cdot47}+\dfrac{3}{47\cdot50}+\dfrac{9}{50\cdot59}\right)\)
\(=5\left(\dfrac{1}{37}-\dfrac{1}{45}+\dfrac{1}{45}-\dfrac{1}{47}+...+\dfrac{1}{50}-\dfrac{1}{59}\right)\)
=5(1/37-1/59)
=>D/E=7/5
Rút gọn
a)\(\dfrac{x}{x+1}+\dfrac{1}{x-1}-\dfrac{2x}{1-x^2}\)
b)\(\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}\)
c)\(\dfrac{2x^2-3x-9}{x^2-9}-\dfrac{x}{x+3}-\dfrac{x+3}{3-x}\)
d)\(\dfrac{x+3}{x-2}+\dfrac{x+2}{1-x}-\dfrac{4x-x^2}{x^2-3x+2}\)
giúp mik vs
cảm ơn <3
a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)
\(1+\dfrac{1}{2}.\dfrac{3.2}{2}+\dfrac{1}{3}.\dfrac{4.3}{2}+...+\dfrac{1}{500}.\dfrac{501.500}{2}\)
Các bn giải hộ mk vs ạ. Mai mik pk nộp rùi
CẢM ƠN CÁC BN RẤT NH!!!
\(1+\dfrac{1}{2}.\dfrac{3.2}{2}+\dfrac{1}{3}.\dfrac{4.3}{2}+...+\dfrac{1}{500}.\dfrac{501.500}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{501}{2}\)
\(=\dfrac{2+3+4+...+501}{2}\)
\(=\dfrac{\left(501-2+1\right).\left(501+2\right)}{4}\)
\(=\dfrac{\left(501-2+1\right).\left(501+2\right)}{4}=62875\)
Bài 2:Tìm x
a) |x-2|-\(\dfrac{3}{5}\)=\(\dfrac{1}{2}\)
b) (x-\(\dfrac{7}{3}\)):\(\dfrac{-1}{3}\)=0,4
c) |x-3|=5
d) (2x+3)\(^2\)=25
e) \(\dfrac{3}{4}\)+\(\dfrac{1}{4}\): x=\(\dfrac{2}{5}\)
f) (x-\(\dfrac{1}{2}\))\(^3\)=\(\dfrac{1}{27}\)
Làm nhanh giúp mik vs ạ(đang rất cần, cảm ơn nhiều ạaa!!!!!)
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)